Math in Focus

This handy Math in Focus Grade 5 Workbook Answer Key Chapter 11 Graphs and Probability provides detailed solutions for the textbook questions.

Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability

Put On Your Thinking Cap!

Challenging Practice

Complete.

Question 1.
The table shows the conversion from gallons to pints. Complete the table.

Answer:

Explanation:
1 gallon = 8 pints
if number of gallons (x) = 1,
number of pints (y) = 1 x 8 = 8
So, substitute the x value and multiply with 8 to complete the table.

Question 2.
Write the equation relating the number of pints (y) to the number of gallons (x).
______________
Answer:
y = x × 8
Explanation:
x – Number of Gallons
y – Number of Pints
y = x × 8

Question 3.
Draw the graph of the equation. Label the axes and the equation.

Answer:

Explanation:
Above graph shows the conversion between pints and gallons.
Take number of gallons on x-axis and number of pints on y-axis.
Take the difference between each pint as 10.

Use the graph to answer the questions.

Question 4.
How many pints is 3\(\frac{1}{2}\) gallons?
Answer:
28 pints
Explanation:
y = 3\(\frac{1}{2}\) × 8 = 28 gallons

Question 5.
How many pints is 4\(\frac{1}{2}\) gallons?
Answer:
36 gallons
Explanation:
y = 4\(\frac{1}{2}\) × 8 = 36 pints

Question 6.
How many gallons is 20 pints?
Answer:
2 \(\frac{1}{2}\) gallons
Explanation:
20 =x × 8 = 28 gallons
x = \(\frac{20}{8}\)
= 2 \(\frac{4}{8}\)

Question 7.
How many gallons is 44 pints?
Answer:
5\(\frac{1}{2}\) gallons
Explanation:
y = x × 8

Complete.

Question 8.
The table shows the conversion from quarts to cups. Complete the table.

Answer:

Explanation:
1 Quarts = 4 cups
if number of Quarts (x) = 1,
number of cups (y) = 1 x 4 = 4
So, substitute the x value and multiply with 4 to complete the table.

Question 9.
Write the equation relating the number of cups (y) to the number of quarts (x).
Answer:
y = x × 4
Explanation:
x is Number of Quarts
y is Number of Cups
y = x × 4

Put On Your Thinking Cap!

Problem Solving

Solve.

Question 1.
Jim has a dime, a nickel, and a quarter. How many different amounts of money can he form using one or more of these coins?
Answer:
These are the possible coins combinations

Explanation:
A probability is a number that reflects the chance or likelihood that a particular event will occur.
So, Jim has the above combinations of coins.

Question 2.
There are an equal number of red, blue, and green beads in a bag. One bead is picked, its color is noted and the bead is replaced. Then a second bead is picked.
a. Draw a tree diagram to show the outcomes.
Answer:

Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking different color beads, is as shown in the diagram.
As we know probability is a number that reflects the chance or likelihood that a particular event will occur.

b. What is the probability of picking two red beads?
Answer:
\(\frac{1}{9}\)
Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking two red beads, is as shown in the diagram below.

c. What is the probability of picking one red and one green bead?
Answer:
\(\frac{2}{9}\)
Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking one red and one green bead, is as shown in the diagram below.

d. What is the probability of picking no red beads?
Answer:
\(\frac{4}{9}\)

Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking no red, means first select blue and green.

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.4 Area of Composite Figures to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.4 Answer Key Area of Composite Figures

Math in Focus Grade 6 Chapter 10 Lesson 10.4 Guided Practice Answer Key

Use graph paper. Copy the hexagon and solve.

Question 1.
Divide the hexagon into two identical triangles and a rectangle.
Answer:

Explanation:
As shown in the figure, a regular hexagon is divided into two identical traingles and a rectangle.
The two identical traingle are AFE and BCD.
The rectangle formed is ABDE.

Question 2.
Divide the hexagon in another way. Name the polygons that make up the hexagon.

Answer:

Explanation:
Mark the center of the hexagon.
Draw a line from the center point to a vertex.
Draw lines from the center point to the third and fifth vertices.
Each part is a rhombus, which is a four-sided figure with equal side lengths and two sets of parallel sides.
A hexagon can form 3 identical rhombus.

Complete.

Question 3.
Trapezoid ABDE is made up of square A BCE and triangle ECD. triangle ECD is 60 square inches. The length of \(\overline{C D}\) is 12 inches.

a) Find the height of triangle ECD.

Area of triangle ECD = in.²
Area of triangle ECD = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) ∙ ∙ EC
= ∙ EC
÷ = ∙ EC ÷
= EC
The height of triangle ECO is î.. inches,
Answer:
In triangle ECD, the base length is equal to 12 in and the area is 60 sq.in.
Area of triangle ECD = \(\frac{1}{2}\)bh
60 = \(\frac{1}{2}\)×12×EC
60 = 6×EC
60÷6 = 6×EC÷6
10 = EC
The height of triangle ECD is 10 in.

b) Find the area of square ABCE.
Area of square ABCE = l²
=
The area of square ABCE is  square inches.
Answer:
A square has all equal sides, so the length of it will be 10 in.
Area of square ABCE = length×length
= 10×10
= 100 sq.in
The area of square ABCE is 100 sq.in

c) Find the areà of trapezoid ABDE.
Area of the trapezoid ABDE
= area of square ABCE + area of triangle ECD
= +
=  in.²
The area of trapezoid ABDE is square inches.
Answer:
Area of the trapezoid ABDE
= \(\frac{1}{2}\)×(sum of bases)×height
Length of BD = BC+CD = 10+12 = 22 in
= \(\frac{1}{2}\)×(10+22)×10
= \(\frac{1}{2}\)×32×10
= 16×10
= 160 sq.in

Complete.

Question 4.
The area of parallelogram ABEF is 84 square meters.

a) Find the area of triangle BDE.
Area of parallelogram ABEF = bh
= ∙ h
÷ = h ÷
= h
The height of parallelogram ABEF is also the height of triangle BDE.
Area of triangle BDE = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) • BD • h
= \(\frac{1}{2}\) • ( + ) •
= m²
The area of triangle BDE is square meters.
Answer:
Given that area of parallelogram ABEF is 84 sq.m and base is 5m.
Area of parallelogram ABEF = bh
84 = 5×h
84÷5 = 5×h÷5
16.8 = h
The height of parallelogram ABEF is also the height of triangle BDE which is 16.8 m high.
Area of triangle BDE = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) × BD × h
= \(\frac{1}{2}\) × (BC+CD) × h
= \(\frac{1}{2}\) × (12+8) × h
= \(\frac{1}{2}\) × 20 × 16.8
= \(\frac{1}{2}\) × 336
= 168 sq.m

b) Find the area of trapezoid CDEF.
Parallelogram ABEF and trapezoid CDEF have the same height.
Area of trapezoid CDEF = \(\frac{1}{2}\)h(b₁ + b₂)
= • • ( + )
The area of trapezoid CDEF is square meters.
Answer:
Parallelogram ABEF and trapezoid CDEF have the same height. Therefore, the height will be 16.8 m
Since ABEF is a parallelogram, AB and FE will have the same length.
Height = 16.8 m
Sum of bases = 8+5 = 13
Area of trapezoid CDEF = \(\frac{1}{2}\)h(b₁ + b₂)
= \(\frac{1}{2}\)×16.8×13
= \(\frac{1}{2}\)×218.4
= 109.2 sq.m
The area of trapezoid CDEF is 109.2 sq.m.

Math in Focus Course 1B Practice 10.4 Answer Key

Copy each figure and draw straight lines to divide. Describe two ways to find the area of each figure.

Question 1.
Divide the figure into a rectangle and two right triangles.

Answer:

Divide the given figure into two equal parts.
It will form two identical rectangles.
Divide the first identical rectangle diagonally into two equal parts, which will form an equilateral triangle.

Question 2.
Divide the figure into a rectangle and two right triangles.

Answer:

Explanation:
Draw a perpendicular segment between the base and the opposite vertex. Similarly, draw a line segment to the other end also.
The figure is divided now into three parts.
Two parts will form an right angle triangle and the third will form a rectangle.

Question 3.
Divide the figure into a rectangle and a right triangle.

Answer:

Explanation:
Draw a line segment parallel to the base length.
The first part formed is the right angle triangle and the second part is the rectangle.

Copy each figure and draw straight lines to divide. Describe a way to find the area.

Question 4.

Answer:

Explanation:
If the given figure is divided using straight line, it will form a triangle and a rectangle.
Area of polygon will be the sum of areas of rectangle and area of triangle.
Area of polygon = Area of triangle+Area of rectangle
= (\(\frac{1}{2}\) × b × h) + (length×width)
Here, the base of triangle and the width of rectangle will be of same length.

Question 5.

Answer:

Explanation:
If the given figure is divided using straight line, it will form 5 identical triangles.
Area of polygon = n×\(\frac{1}{2}\)×b×h
Here n equals 5
Area of polygon = 5×\(\frac{1}{2}\)×b×h
Here, the base of trinagle will be equal to side length of the pentagon.

Find the area of each figure.

Question 6.
Parallelogram ABDE is made up of square ACDF, triangle ABC, and triangle FDE. Triangle ABC and triangle FDE are identical. The area of square ACDF is 36 square meters. Find the area of triangle ABC. Then find the area of parallelogram ABDE.

Answer:
Given that ACDF is a square and its area is 36 sq.m
Area of ACDF = side×side
36 = side×side
6×6 = side×side
Therefore, side of a sqaure will be 6 m.
Here, the side of square is equal to the height of the trinagle.
The height of triangle is 6m and the base is 4m.
Area of triangle = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×4×6
= 24÷2
= 12 sq.m
Area of triangle ABC and DFE will be same as both have the same dimensions.
Area of parallelogram ABDE = Area of ABC + Area of DFE + Area of ACDF
= 12+12+36
= 60 sq.m

Question 7.
Math Journal Describe how you would divide the figure with straight lines. Which sides of the figure would you measure to find its area? Explain your answer.

Answer:

Explanation:
Draw a vertical line segment between the two vertex and divide it in two parts.
Two triangles will be formed. Since both the trinagles lie on same base, the base length of these triangles will be same. The perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Measure the drawn line segment to get the base length and the perpendicular line segment to get the height.
Area of triangle can be found using the formula \(\frac{1}{2}\)×b×h.
Area of polygon will be the sum of areas of two trinagles formed.
Area of polygon = Area of first triangle + Area of second trinangle

Find the area of triangle EBC.

Question 8.
In the figure below, trapezoid ABDE is made up of three triangles, and figure ABCE is a parallelogram. Find the area of triangle EBC if the area of trapezoid ABDE is 180 square centimeters.

Answer:
Given that trapezoid ABDE is made up of three triangles, ABE BEC and ECD and its area is 180 sq.cm
Figure ABCE  is a parallelogram.
Area of triangle ECD:
In triangle ECD, the base measures 6cm and the height measures 12cm.
Area of ECD = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×6×12
= 72÷2
= 36 sq.cm
To find the area of triangle EBC, we need the length of BC, which is the base of the parallelogram ABCE.
To calculate the base of parallelogram ABCE , first we need to find the area of ABCE.
Area of trapezoid = Area of parallelogram ABCE + Area of triangle ECD
180 = Area of parallelogram ABCE + 36
180-36 = Area of parallelogram ABCE + 36 – 36
Area of parallelogram ABCE = 144 sq.cm
We know the height of the polygon is 12cm.
Area of parallelogram ABCE = base × height
144 = base × 12
144 ÷ 12 = (base × 12) ÷ 12
12 = base
Thus, the length of BC will be 12 cm.
Area of triangle EBC = \(\frac{1}{2}\)×b×h.
= \(\frac{1}{2}\)×12×12
= 144÷2
= 72 sq.cm

Solve. Use graph paper.

Question 9.
a) Plot the points P (-2, 2), Q (-2, -2), R (-4, -5), S (1, -5), and T (3, -2) on a coordinate plane. Connect the points in order to form figure PQRST.
Answer:
Plot the points P (-2, 2), Q (-2, -2), R (-4, -5), S (1, -5), and T (3, -2) on a coordinate plane.
When connected the points, it forms a polygon as shown below.

b) Find the area of figure PQRST.
Answer:
When the figure PQRST is divided using a line segment. It forms a triangle and a parallelogram.
Area of the polygon PQRST = Area of QTSR + Area of PQT
In the figure, QTSR forms a parallelogram. Let us assume RS as base and the perpendicular segment between the base and the opposite side will be the height.
Base = 5 units
Height = 3 units
Area of QTSR = base×height
= 5×3
= 15 sq.units
In the figure, PQT forms a parallelogram. Let us assume QT as base and the perpendicular segment between the base and the opposite vertex will be the height.
Base = 5 units
Height = 4 units
Area of PQT = \(\frac{1}{2}\)×b×h
= \(\frac{1}{2}\)×5×4
= \(\frac{1}{2}\)×20
= 10 sq.units
Area of the polygon PQRST = 15 + 10
= 25 sq.units

c) Point V lies on along \(\overline{Q T}\). The area of triangle PQV is \(\frac{2}{5}\) the area of triangle PQT. Give the coordinates of point V. Plot point V on the coordinate plane.
Answer:
Given that the point V lies on along \(\overline{Q T}\). The area of triangle PQV is \(\frac{2}{5}\) the area of triangle PQT.
Area of PQV = \(\frac{2}{5}\)×area of triangle PQT
As calculated above, Area of PQT = 10 sq.units
= \(\frac{2}{5}\)×10
= 20÷5
= 4 sq.units
Let us assume PQ as base which is 5 units.
Area of PQV = \(\frac{1}{2}\)×b×h
4 = \(\frac{1}{2}\)×4×h
4 = 2×h
4÷2 = (2×h)÷2
2 = h
The height will be 2 units long.
Therefore the point V can be placed 2 units high from the base along \(\overline{Q T}\), which will be (0,-2)

Find the area of trapezoid MNRS.

Question 10.
In the figure below, trapezoid MNRS is made up of trapezoid MNPT, triangle TPQ, and parallelogram TQRS. The area of triangle TPQ is 84 square feet. The lengths of \(\overline{N P}\), \(\overline{P Q}\), and \(\overline{Q R}\) are in the ratio 2 : 1.5 : 1.
Find the area of trapezoid MNRS.

Answer:
Given that the trapezoid MNRS made up of trapezoid MNPT, triangle TPQ, and parallelogram TQRS.
Area of triangle TPQ = 84 sq.ft and the length of PQ is 12 ft and MS is 18 ft.
Area of TPQ = \(\frac{1}{2}\)×b×h, where b is length of PQ and h is unknown.
84 = \(\frac{1}{2}\)×12×h
84 = 6×h
84÷6 = (6×h)÷h
14 = h
The height of triangle will be 14 ft.
The lengths of \(\overline{N P}\), \(\overline{P Q}\), and \(\overline{Q R}\) are in the ratio 2 : 1.5 : 1
Let p be the proportion.
1.5p=12
1.5p÷1.5 = 12÷1.5
p = 8
Length of NP = 2p = 2×8 = 16 ft
Length of QR = 1p = 1×8 = 8 ft
NR = NP+PQ+QR
= 16+12+8
= 36 ft
In trapezoid MNRS, MS and NR will be the two parallel bases.
Area of trapezoid MNRS = \(\frac{1}{2}\)×(sum of bases)×height
= \(\frac{1}{2}\)×(18+36)×14
= \(\frac{1}{2}\)×54×14
= 756÷2
= 378 sq.ft
Thus, the area of trapezoid MNRS = 378 sq.ft

Find the area of triangle BDE.

Question 11.
The figure below is made up of two trapezoids ABEFand BCDE. The area of triangle FGE is 26 square inches, and the area of trapezoid BCDE is 82.5 square inches. BG is equal to GE. Find the area of triangle BDE.

Answer:
The figure is made up of two trapezoids ABEFand BCDE, two triangles FGE,BCD and BDE.
Area of triangle FGE is 26 sq.in and height is 8 in.
Area of trapezoid BCDE is 82.5 sq.in
BG is equal to GE and length of CD is 2 in.
First we need to find the base length, GE.
Area of triangle FGE = \(\frac{1}{2}\)×base×height
26 = \(\frac{1}{2}\)×base×8
26 = 4×base
26÷4 = (4×base)÷4
6.5 = base
Length of GE = Length of BG = 6.5 in
Length of BE = Length of GE + Length of BG
= 6.5+6.5 =13 in
Area of trapezoid BCDE = \(\frac{1}{2}\)×(sum of bases)×height
82.5  =  \(\frac{1}{2}\)×(BE+CD)×height
82.5  =  \(\frac{1}{2}\)×15×height
82.5 = 7.5×height
82.5÷7.5 = (7.5×height)÷7.5
height = 11 in
The height of tarpezoid and triangle will be of same length.
Now, we can find the area of triangle BDE.
Area of triangle BDE = \(\frac{1}{2}\)×base×height
= \(\frac{1}{2}\)×BE×height
= \(\frac{1}{2}\)×BE×height
= \(\frac{1}{2}\)×13×11
= 143÷2
= 71.5 sq.in

Find the area of the shaded region.

Question 12.
\(\frac{3}{8}\) of the triangle is shaded.

Answer:
Given trinagle is 28cm wide and 35cm high.
Area of trinagle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 28 × 35
= \(\frac{1}{2}\) × 980
= 980÷2
= 490 sq.cm
The shaded region is \(\frac{3}{8}\) of the triangle.
Therefore, area of shaded region will be \(\frac{3}{8}\) × 490
= 1470÷8
= 183.75 sq.cm

Find the height of trapezoid PQRS.

Question 13.
Trapezoid PORS s made up of isosceles triangle PQS and triangle SQR. The area of triangle PQS is 16.5 square inches. The areas of triangle PQS and triangle SQR are in the ratio 2 : 3, Find the height of trapezoid PQRS.

Answer:
Given that the trapezoid PORS s made up of isosceles triangle PQS and triangle SQR.
Area of triangle PQS is 16.5 square inches.
Since PQS is an isosceles triangle, PS will be equal to SQ.
The areas of triangle PQS and triangle SQR are in the ratio 2 : 3.
\(\frac{PQS}{SQR}\) = \(\frac{2}{3}\)
3×PQS = 2×SQR
3×16.5 = 2×SQR
49.5 = 2×SQR
49.5÷2 = (2×SQR)÷2
24.7 = SQR
Area of triangle SQR = 24.7 sq.in
Area of trapezoid = Area of triangle PQS + Area of triangle SQR
= 16.5+24.7
= 41.2 sq.in
Area of trapezoid = \(\frac{1}{2}\)×(sum of bases)×height
41.2 = \(\frac{1}{2}\)×(6+9)×height
41.2 = \(\frac{1}{2}\)× 15 ×height
41.2 = 7.5×height
41.2 ÷ 7.5 = (7.5×height)÷7.5
5.4= height
Thus, the height of trapezoid PQRS = 5.4 in

Find the area of each figure.

Question 14.
In the figure below, trapezoid ABCD is made up of square BCDE and triangles ABF and APE. The area of square BCDE is 576 square feet. The ratio of BF to FE is 2: 1. Find the area of triangle ABF.

Answer:
Given that the trapezoid ABCD is made up of square BCDE and triangles ABF and APE.
Area of square BCDE = 576 sq.ft
Area of square BCDE = length×length
576 = length×length
24×24 = length×length
Length of square = 24 ft.
The ratio of BF to FE is 2: 1.
\(\frac{BF}{FE}\) = \(\frac{2}{1}\)
1×BF = 2×FE
BE = 2×FE + FE
24 = 3×FE
24÷3 = 3×FE÷3
FE = 8ft
BF = 2×8 = 16ft
Area of triangle ABF = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 16 × 12
= 192÷2
= 96 sq.ft

Brain @ Work

Question 1.
Figure ABCD is made up of square PQRS and four identical triangles. The area of triangle APD
is 49 square feet. The lengths of \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{PD}}\) are in the ratio 1: 2. Find the area of figure ABCD.

Answer:
Given that the figure ABCD is made up of square PQRS and four identical triangles. The area of triangle APD
is 49 square feet.
The lengths of \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{PD}}\) are in the ratio 1: 2.
Let the proportion be ‘p’
\(\frac{AP}{PD}\) = 1: 2
PD = 2×AP
Area of triangle APD = \(\frac{1}{2}\) × base × height
49 = \(\frac{1}{2}\) × PD × AP
49 = \(\frac{1}{2}\) × 2×AP×AP
49 = AP×AP
7×7 = AP×AP
AP = 7ft
PD = 2×7 = 14ft
As all are identical triangles, the longest side or base will be twice of the height.
SC = 2×DS, where DS = 7ft
PD = PS+SD
14 = PS+7
14-7 = PS+7-7
PS = 7ft
PQRS is a square, so all sides will be equal.
Area of square PQRS = length×length
= 7×7
= 49 sq.ft
Area of ABCD = Area of four identical triangles + Area of square PQRS
= 4×49 + 49
= 196+49
= 245 sq.ft

Question 2.
The figure is made up of squares BCDE and AEFH. The length of \(\overline{D E}\) is 6 centimeters, and the length of \(\overline{E F}\) is 12 centimeters.
a) Write the length of \(\overline{F G}\) in terms of x.
Answer:
Length of DE = 6 cm
Length of EF = 12 cm
Since AEFH is a square, all sides will be equal.
Therefore, the length of HF will be 12cm
HF = HG + GF
12 = x+GF
12-x = x+GF-x
GF = 12-x
Therefore, the length of \(\overline{F G}\) will be 12-x

b) Find the area of the shaded region in terms of x.
Answer:

c) Give the value of x for which the shaded region has the greatest area. What is the shape of the shaded region for the value of x you have given?

Answer:

Question 3.
Figure ABCD is a square. Point S is in the middle of \(\overline{A D}\), and point T is in the middle of and \(\overline{C D}\). What fraction of the square is shaded?

Answer:
Given that the figure ABCD is a square.
Point S is in the middle of \(\overline{A D}\), and point T is in the middle of and \(\overline{C D}\).
To find the fraction of shaded square we need to find the area of trinagle STB.
Let us assume each side length of the square as 1m.
AD=DC=CB=BA=1m
Since S and T are the midpoints of AD and DC.
Length of DS = Length of SA = \(\frac{1}{2}\) m
Length of DT = Length of TC = \(\frac{1}{2}\) m
ASB forms a right angle triangle; AB²+AS² = SB²
1²+(\(\frac{1}{2}\))² = SB²
1+\(\frac{1}{4}\) = SB²
\(\frac{5}{4}\) = SB²
SB = √\(\frac{5}{4}\)
SB = \(\frac{√5}{2}\)
In triangle DST
ST²=(\(\frac{1}{2}\))² + (\(\frac{1}{2}\))²
= \(\frac{1}{4}\) + \(\frac{1}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
ST = √\(\frac{1}{2}\)
Draw a perpendicular segment from BU on ST

Now BUS will be a right angle triangle,
BS²=BU²+US²
Length of SU will be \(\frac{1}{2}\) of ST
Length of SU = \(\frac{1}{2}\) × \(\frac{1}{√2}\) = \(\frac{1}{2√2}\)
(\(\frac{√5}{2}\))² = BU²+ (\(\frac{1}{2√2}\) )²
\(\frac{5}{4}\) = BU²+\(\frac{1}{8}\)
\(\frac{5}{4}\)–\(\frac{1}{8}\) = BU²
(\(\frac{5}{4}\)×\(\frac{2}{2}\))-\(\frac{1}{8}\) = BU²
\(\frac{10}{8}\)–\(\frac{1}{8}\) = BU²
\(\frac{9}{8}\)= BU²
BU=√\(\frac{9}{8}\)
BU=\(\frac{3}{2√2}\)
Area of STB:
Length of ST=\(\frac{1}{√2}\)
Height = \(\frac{3}{2√2}\)
Area of triangle STB=\(\frac{1}{2}\)×b×h
=\(\frac{1}{2}\)×\(\frac{1}{√2}\)×\(\frac{3}{2√2}\)
= 2×√2×2√2
= 2×2×2=8(√2×√2=2)
= \(\frac{3}{8}\) sq.m

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Review Test Answer Key

Concepts and Skills

Identify a base and a height of each triangle.

Question 1.

Answer:
Assume a side as the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Let us assume the side BC as the base, then the segment AD which is drawn perpendicular between the base and the opposite vertex will be the height.
Thus, base of triangle = BC
Height of the triangle = AD.

Question 2.

Answer:
Assume a side as the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Let us assume the side NP as the base, then the segment MQ which is drawn perpendicular between the base and the opposite vertex will be the height.
Thus, base of triangle = NP
Height of the triangle = MQ

Find the area of each figure.

Question 3.

Answer:
If we assume the line segment measuring 28cm as the base, then the height will be the perpendicular line segment between the base and the opposite vertex which is 8cm.
Area of triangle = \(\frac{1}{2}\)×28×8
= \(\frac{1}{2}\)×224
= 224÷2
= 112 sq.cm
Thus, the area for the given polygon will be 112 sq.cm

Question 4.

Answer:
The given polygon is parallelogram.
Let us assume the base of the parallelogram as 46.3 cm, then the height will be the perpendicular line segment between the base and the opposite vertex which is 15 cm.
Area of parallelogram = base×height
= 46.3×15
= 694.5 sq.cm
Thus, the area for the given polygon will be 694.5 sq.cm

Question 5.

Answer:
The given polygon is similar to trapezium.
A trapezium has a pair of parallel sides, measuring 50cm and 18cm.
The height will be the perpendicular line segment between the base and the opposite vertex which is 31.5 cm.
Area of trapezoid = \(\frac{1}{2}\)×(sum of bases)×height
= \(\frac{1}{2}\)×(50+18)×31.5
= \(\frac{1}{2}\)×68×31.5
= 2142÷2
= 1071 sq.cm
Thus, the area for the given polygon will be 1071 sq.cm

Find the area of the shaded region.

Question 6.
The area of the regular octagon below is 560 square inches.

Answer:
A regular octagon will be made up of her 8 identical triangles.
Thus, the area of octagon will be equal to sum of the 8 identical triangles.
Given that the area of the regular octagon is 560 sq.in
Area of octagon = 8 × Area of triangle
560 = 8 × Area of triangle
560÷8 = (8 × Area of triangle)÷8
Area of triangle = 70sq.in
The shaded region is half of the triangle.
Area of shaded region = \(\frac{1}{2}\)×Area of triangle
= \(\frac{1}{2}\)×70
= 35 sq.in
Thus, the area of shaded region is 35 sq.in

Problem Solving

Find the area of the shaded region.

Question 7.

Answer:
The given figure has two triangles and a square.
Length of square is 15 cm.
Area of square will be length×length
= 15×15
= 225 sq.cm
The area of shaded region can be calculated by subtracting the area of square from the area of parallelogram.
So, calculate the area of parallelogram.
Area of parallelogram = base×height
The base of parallelogram is 15+20=35 cm and
the height will be 15 cm.
= 35×15
= 525 sq.cm
Area of shaded region = Area of parallelogram – Area of square
= 525 – 225
= 300 sq.cm

Solve.

Question 8.
Figure ABCD is a parallelogram. BC is 16 centimeters, CD is 12 centimeters, and AH is 10 centimeters.
a) Find the area of parallelogram ABCD.
Answer:
Given that figure ABCD is a parallelogram.
BC is 16 centimeters and AH is 10 centimeters.
Area of parallelogram ABCD = base × height
= BC × AH
= 16×10
= 160 sq.cm

b) Find the length of \(\overline{\mathrm{AK}}\). Round your answer to the nearest tenth of a centimeter.

Answer:
ABCD is a parallelogram with two triangles ABC and ACD.
Area of parallelogram = Area of triangle ABC + Area of triangle ACD
The area of parallelogram is 160 sq.cm
Area of triangle ABC= \(\frac{1}{2}\)×base×height
= \(\frac{1}{2}\)×BC×AH
= \(\frac{1}{2}\)×16×10
= 160÷2
= 80 sq.cm
Area of parallelogram = 80 + Area of triangle ACD
160 = 80 + Area of triangle ACD
160-80 = 80 + Area of triangle ACD – 80
80 = Area of triangle ACD
Length of AD = 12cm
Area of triangle ACD = \(\frac{1}{2}\)×base×height
80 = \(\frac{1}{2}\)×12×AK
80 = 6×AK
80÷6 = (6×AK)÷6
13.3 = AK
13.3 when rounded of to nearest tenth will be 13
The length of AK = 13cm

Question 9.
Figure ABCD is a trapezoid. The length of \(\overline{\mathrm{BC}}\) is 36 centimeters. The areas of triangles ABC and ACD are in the ratio 1.5:1. Find the length of \(\overline{\mathrm{AD}}\).

Answer:
Given that figure ABCD is a trapezoid.
The length of \(\overline{\mathrm{BC}}\) is 36 centimeters.
Area of triangle ABC = \(\frac{1}{2}\)×base×height
= \(\frac{1}{2}\)×36×10
= 360÷2
= 180 sq.cm
The areas of triangles ABC and ACD are in the ratio 1.5:1
\(\frac{ABC}{ACD}\) = \(\frac{1.5}{1}\)
Area of triangle ABC = 1.5×Area of triangle ACD
180 = 1.5×Area of triangle ACD
180÷1.5 = (1.5×Area of triangle ACD)÷1.5
120 = Area of triangle ACD
Area of triangle ACD = \(\frac{1}{2}\)×base×height
120 = \(\frac{1}{2}\)×AD×10
120 = AD×5
120 ÷ 5  = (AD×5)÷5
24 = AD
Thus, the length of \(\overline{\mathrm{AD}}\) is 24 cm.

Question 10.
Parallelogram PRTVis made up of triangle PQV, triangle QUV, and trapezoid QRTU. The area of parallelogram PRTV is 96 square feet. The lengths of \(\overline{\mathrm{TU}}\) and \(\overline{\mathrm{UV}}\) are equal. Find the area of triangle QUV.

Answer:
Parallelogram PRTV is made up of triangle PQV, triangle QUV, and trapezoid QRTU.
The area of parallelogram PRTV is 96 sq.ft. The lengths of TU is equal to UV.
Area of parallelogram PRTV = 96
Area of parallelogram PRTV = base×height
96 = base×8
96÷8 = (base×8)÷8
12 = base
TV is the base of the parallelogram PRTV which is equal to 12 ft.
TV = VU+UT
12 = VU+VU
12 = 2×VU
12÷2 = (2×VU)÷2
6 = VU
Length of VU is 6 ft.
Area of triangle QUV = \(\frac{1}{2}\)×base×height
= \(\frac{1}{2}\)×UV×8
= \(\frac{1}{2}\)×6×8
= 48÷2
= 24 ft

Question 11.
Charles drew a regular hexagon and divided it into two identical trapezoids. The side length of the hexagon is 16 centimeters, and the length of the diagonal shown is 32 centimeters. Charles measured the height of one of the trapezoids and found that the height was 13.9 centimeters. Find the area of the hexagon.

Answer:
A regular hexagon and divided it into two identical trapezoids.
The side length of the hexagon is 16 cm and the height of one of the trapezoids is 13.9 cm. The length of the diagonal shown is 32 cm.
The side length makes with the height of the trapezium.
Pythagoras theorem is applied to get the base of the triangle.
base² = 16²-(13.9)²
base² = 256-193.21
base² = 62.79
base will be the square root of 62.79
Base length will be 7.9 cm
The length of opposite diagonal can be calculated by subtracting the sum of base length of two triangle from the length of diagonal.
Length of the parallel side = 32-(7.9+7.9)
= 16.2 cm
Area of trapezoid = \(\frac{1}{2}\)×(sum of bases)×height
= \(\frac{1}{2}\)×(32+16.2)×13.9
= \(\frac{1}{2}\)×48.2×13.9
= \(\frac{1}{2}\)×669.9
= 334.9 sq.cm
Area of hexagon = Area of trapezoid + Area of trapezoid
= 334.9+334.9
= 669.8 sq.ft
Thus, the area of hexagon is 669.8 sq.ft

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.1 Solving Algebraic Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.1 Guided Practice Answer Key

Complete each with = or ≠, and each with the correct value.

Question 1.
For what value of x will x + 3 = 7 be true?
If x = 1, x + 3 = + 3
= ( 7)
If x = 2, x + 3 = + 3
= ( 7)
If x = 4, x + 3 = + 3
=
x + 3 = 7 is true when x = .
Answer:
x + 3 = 7 is true when x = 4,

Explanation:
Given to find for what value of x will x + 3 = 7 be true, so
if x =1, x + 3 = 1 + 3 = 4 ≠ 7,
If x = 2, x + 3 = 2 + 3 = 5 ≠ 7,
if x = 3, x + 3 = 3 + 3 = 6 ≠ 7,
if x = 4, x + 3 = 4 + 3 = 7 = 7, therefore x + 3 = 7 is true when x = 4.

Solve each equation using the substitution method.

Question 2.
p + 6 = 13
Answer:
p = 7, p + 6 = 13 is true,

Explanation:
Given equation p + 6 = 13, so solving p = 13 – 6 = 7, So when p = 7 means 7 + 6 = 13 is true therefore p = 7, p + 6 = 13 is true.

Question 3.
r + 4 = 12
Answer:
r = 8, r + 4 = 12 is true,

Equation::
Given equation r + 4 = 12 so solving r = 12 – 4 = 8 therefore when r = 8 means 8 + 4 = 12 is true, therefore r = 8, r + 4 = 12 is true.

Question 4.
k – 10 = 7
Answer:
k = 17 means k – 10 = 7 is true,

Explanation:
Given equation k – 10 = 7 so solving k = 7 + 10 = 17 therefore when k = 17 means 17 – 10 = 7 is true so k = 17., k – 10 = 7 is true.

Question 5.
2m = 6
Answer:
m = 3 means 2m = 6 is true,

Explanation:
Given equation 2m = 6 so solving m =6/2 = 3 therefore when m = 3 means 2 x 3 = 6 is true so m = 3.

Question 6.
4n = 20
Answer:
n = 5 means 4n = 20 is true,

Explanation:
Given equation 4n = 20 so solving n = 20/4 = 5 therefore when n = 5 means 4 X 5 = 20 is true so n = 5.

Question 7.
\(\frac{1}{5}\)z = 3
Answer:
z = 15 means \(\frac{1}{5}\)z = 3,

Explanation:
Given equation \(\frac{1}{5}\)z = 3 so solving x = 3 X 5 = 15 therefore when z = 15 means \(\frac{1}{5}\) X 15 = 3 is true so z = 3.
Complete each with + or —, and each with the correct value.

Question 8.
Solve x + 8 = 19.
x + 8 = 19
x + 8 =
x =
Answer:
x = 11,

Explanation:
Given to solve x + 8 = 19,
x + 8 – 19 = 19 – 19,
x – 11 = 0,
x = 11.

Solve each equation.

Question 9.
f + 5 = 14
Answer:
f = 9,

Explanation:
Given to solve f + 5 = 14,
subtracting both sides by 12 as f + 5 – 14 = 14 – 14,
f – 9 = 0,
therefore f = 9.

Question 10.
26 = g + 11
Answer:
g = 15,

Explanation:
Given to solve 26 = g + 11,
subtracting by 11 on the both sides as 26 – 11 = g + 11 – 11,
15 = g or g = 15.

Question 11.
w – 6 = 10
Answer:
w = 16,

Explanation:
Given to solve w – 6 = 10,
Adding 6 both the sides we get w – 6 + 6 = 10 + 6,
therefore w = 16.

Question 12.
z – 9 = 21
Answer:
z = 30,

Explanation:
Given to solve z – 9 = 21,
Adding 9 both the sides we get z – 9 + 9 = 21 + 9,
therefore z = 30.

Complete each with × or ÷, and with the correct value.

Question 13.
Solve 3x = 27.
3x = 27
3x = 27
x =
Answer:
x = 9,

Explanation:
Given to solve 3x = 27,
Dividing both sides by 3 as 3x ÷ 3 = 27 ÷ 3,
therefore x = 9.

Solve each equation.

Question 14.
6a = 42
Answer:
a = 7,

Explanation:
Given to solve 6a = 42,
Dividing both sides by 6 as 6a ÷ 6 = 42 ÷ 6,
we get a = 7.

Question 15.
65 = 13b
Answer:
b = 5,

Explanation:
Given to solve 65 = 13b,
Dividing both sides by 13 as 65 ÷ 13 = 13b ÷ 13,
we get 5 = b or b = 5.

Question 16.
\(\frac{m}{8}\) = 9
Answer:
m = 72,

Explanation:
Given to solve \(\frac{m}{8}\) = 9,
Multiplying both sides by 8,
\(\frac{m}{8}\) X 8 = 9 X 8,
We get m = 72.

Question 17.
12 = \(\frac{n}{7}\)
Answer:
n = 84,

Explanation:
Given to solve 12 = \(\frac{n}{7}\),
Multiplying both sides by 7,
12 X 7 = \(\frac{n}{7}\) X 7,
we get 84 = n or n = 84.

Complete each with +, —, × or ÷, and with the correct value.

Question 18.
Solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x =
Answer:
x = \(\frac{2}{7}\),

Explanation:
Given to solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\) subtracting \(\frac{3}{7}\) both sides as x + \(\frac{3}{7}\) – \(\frac{3}{7}\) = \(\frac{5}{7}\) – \(\frac{3}{7}\) we have common denominator so x = \(\frac{5 – 3}{7}\) = \(\frac{2}{7}\).

Solve each equation. First tell which operation you will perform on each side of the equation. Write your answer in simplest form.

Question 19.
k + \(\frac{1}{8}\) = \(\frac{7}{8}\)
Answer:
k = \(\frac{6}{8}\) or \(\frac{3}{4}\),

Explanation:
Given to solve k + \(\frac{1}{8}\) = \(\frac{3}{4}\) subtracting \(\frac{1}{8}\) both sides as k + \(\frac{1}{8}\) – \(\frac{1}{8}\) = \(\frac{7}{8}\) – \(\frac{1}{8}\) we have common denominators so x = \(\frac{7 – 1}{8}\) = \(\frac{6}{8}\) further can be divided as both numerator and denominator goes by 2 we get \(\frac{3}{4}\).

Question 20.
4p = \(\frac{3}{4}\)
Answer:
p = \(\frac{3}{16}\),

Explanation:
Given to solve 4p = \(\frac{3}{4}\), Dividing both sides by 4 as
4p ÷ 4 = \(\frac{3}{4}\) ÷ 4,
p = \(\frac{3}{4 X 4}\) = \(\frac{3}{16}\) .

Math in Focus Course 1B Practice 8.1 Answer Key

Solve each equation using the substitution method.

Question 1.
b + 7 = 10
Answer:
b = 3 means b + 7 = 10 is true,

Explanation:
Given equation b + 7 = 10 so solving b = 10 – 7 = 3 when b = 3 means 3 + 7 = 10 is true therefore b = 3.

Question 2.
17 = e + 9
Answer:
e = 8,

Explanation:
Given 17 = e + 9 so solving e = 17 – 9 = 8 when e = 8 means 17 = 8 + 9  is true therefore e = 8.

Question 3.
k – 4 = 11
Answer:
k = 15,

Explanation:
Given k – 4 = 11 so solving k = 11 + 4 = 15 when k = 15 means 15 – 4 = 11 is true therefore k = 15.

Question 4.
42 = 3p
Answer:
p = 14,

Explanation:
Given 42 = 3p so solving 42/3 = p when p = 14 means 42 = 3 X 14 is true therefore p = 14.

Question 5.
8t = 56
Answer:
t = 7,

Explanation:
Given 8t = 56 so solving t = 56/8 = 7 when t = 7 means 8 X 7 = 56 is true therefore t = 7.

Question 6.
\(\frac{1}{4}\)v = 5
Answer:
v = 20,

Explanation:
Given \(\frac{1}{4}\)v = 5 solving v = 5 X 4 = 20 when v = 4 means \(\frac{1}{4}\) X 4 = 5 x 4 = 20 is true therefore v = 4.

Solve each equation using the concept of balancing.

Question 7.
k + 12 = 23
Answer:
k = 11,

Explanation:
Given to solve k + 12 = 23,
subtracting by 12 on the both sides as k + 12 – 12 = 23 – 12,
k = 11.

Question 8.
x- 8 = 17
Answer:
x = 25,

Explanation:
Given to solve x – 8 = 17,
adding 8 by both sides as x – 8 + 8 = 17 + 8,
x = 25.

Question 9.
24 = f – 16
Answer:
f = 40,

Explanation:
Given to solve 24 = f – 16,
adding both sides by 16 as 24 + 16 = f – 16 + 16,
40 = f or f =40.

Question 10.
5j = 75
Answer:
j = 15,

Explanation:
Given to solve 5j = 75,
dividing both sides by 5 as 5j/5 = 75/5,
j = 15.

Question 11.
81 = 9m
Answer:
m = 9,

Explanation:
Given to solve 81 = 9m,
dividing both sides by 9 as 81/9 = 9m/9,
9 = m or m = 9.

Question 12.
\(\frac{r}{6}\) = 11
Answer:
r = 66,

Explanation:
Given to solve \(\frac{r}{6}\) = 11,
multiplying both sides by 6 as \(\frac{r}{6}\) X 6 = 11 X 6,
r = 66.

Solve each equation using the concept of balancing. Write all fraction answers in simplest form.

Question 13.
\(\frac{5}{6}\) = c + \(\frac{1}{6}\)
Answer:
c = \(\frac{4}{6}\) or \(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{5}{6}\) = c + \(\frac{1}{6}\) so subtracting both sides by \(\frac{1}{6}\) we get \(\frac{5}{6}\) – \(\frac{1}{6}\) = c + \(\frac{1}{6}\) – \(\frac{1}{6}\) as we have common denominators we get \(\frac{5 – 1}{6}\) = c,
c = \(\frac{4}{6}\) we further divide as both numerator and denominators goes by 2 we get c = \(\frac{2 X 2}{3 X 2}\), c = \(\frac{2}{3}\).

Question 14.
h + \(\frac{5}{14}\) = \(\frac{11}{14}\)
Answer:
h = \(\frac{6}{14}\) or \(\frac{3}{7}\),

Explanation:
Given to solve h + \(\frac{5}{14}\) = \(\frac{11}{14}\) so subtracting both sides by \(\frac{5}{14}\) we get h + \(\frac{5}{14}\) – \(\frac{5}{14}\) = \(\frac{11}{14}\) – \(\frac{5}{14}\) as we have common denominators we get h = \(\frac{11 – 5}{6}\),
h = \(\frac{6}{14}\) we further divide as both numerator and denominators goes by 2 we get h = \(\frac{3 X 2}{7 X 2}\), h = \(\frac{3}{7}\).

Question 15.
q – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Answer:
q = 1,

Explanation:
Given to solve q – \(\frac{3}{10}\) = \(\frac{7}{10}\) so adding both sides by \(\frac{3}{10}\) we get q – \(\frac{3}{10}\) + \(\frac{3}{10}\) = \(\frac{7}{10}\) + \(\frac{3}{10}\) as we have common denominators we get q = \(\frac{7 + 3}{10}\),
q = \(\frac{10}{10}\) we further divide as both numerator and denominators goes by 10 we get q = \(\frac{1 X 10}{1 X 10}\), q = 1.

Question 16.
7k = \(\frac{4}{7}\)
Answer:
k = \(\frac{4}{49}\),

Explanation:
Given to solve 7k = \(\frac{4}{7}\) dividing both sides by \(\frac{1}{7}\) we get 7k X \(\frac{1}{7}\) = \(\frac{4 X 1}{7 X 7}\), k = \(\frac{4}{49}\).

Question 17.
\(\frac{5}{12}\) = 5d
Answer:
d = \(\frac{1}{12}\),

Explanation:
Given to solve \(\frac{5}{12}\) = 5d so dividing both sides by \(\frac{1}{5}\) we get \(\frac{5}{12}\) X \(\frac{1}{5}\) = 5d X \(\frac{1}{5}\), d = \(\frac{1}{12}\).

Question 18.
\(\frac{1}{2}\)x = \(\frac{1}{4}\)
Answer:
x = \(\frac{1}{2}\),

Explanation:
Given to solve \(\frac{1}{2}\)x = \(\frac{1}{4}\) so multiplying both sides by 2 we get x X \(\frac{2}{2}\) = \(\frac{2}{4}\) we get x = \(\frac{2}{4}\) we further divide as both numerator and denominators goes by 2 we get x = \(\frac{1 X 2}{2 X 2}\), x = \(\frac{1}{2}\).

Question 19.
\(\frac{8}{9}\) = \(\frac{1}{3}\)f
Answer:
f = \(\frac{8}{3}\),

Explanation:
Given to solve \(\frac{8}{9}\) = \(\frac{1}{3}\)f so multiplying both sides by 3 we get \(\frac{8}{9}\) = \(\frac{1}{3}\)f X 3, \(\frac{8 X 3}{3 X 3}\)= f, So f = \(\frac{8}{3}\).

Question 20.
r + 2.1 = 4.7
Answer:
r = 2.6,

Explanation:
Given to solve r + 2.1 = 4.7 subtracting both sides by 2.1 we get r + 2.1 – 2.1 = 4.7 – 2.1, so r = 2.6.

Question 21.
9.9 = x + 5.4
Answer:
x = 4.5,

Explanation:
Given to solve 9.9 = x + 5.4 subtracting both sides by 5.4 we get 9.9 – 5.4 = x – 5.4, so 4.5 = x.

Question 22.
11.2 = f – 1.8
Answer:
f = 13,

Explanation:
Given to solve 11.2 = f – 1.8 adding both sides by 1.8 as 11.2 + 1.8 = f – 1.8 + 1.8 we get 13 = f or f = 13.

Question 23.
j – 3.7 = 20.4
Answer:
j = 24.1,

Explanation:
Given to solve j – 3.7 = 20.4 adding both sides by 3.7 as j – 3.7 + 3.7 = 20.4 + 3.7 we get j = 24.1.

Question 24.
4w = 6.8
Answer:
w = 1.7,

Explanation:
Given to solve 4w = 6.8, dividing both sides by 4 as 4w/4 = 6.8/4 we get w = 1.7.

Question 25.
13.9 = 2.5z
Answer:
z = 5.56,

Explanation:
Given to solve 13.9 = 2.5z, dividing both sides by 2.5 as 13.9/2.5 = 2.5z/2.5 we get 5.56 = z or z = 5.56.

Question 26.
3.2d = 40.8
Answer:
d = 12.75,

Explanation:
Given to solve 3.2d = 40.8 dividing both sides by 3.2 as 3.2d/3.2 = 40.8/3.2 we get d = 12.75.

Question 27.
x + \(\frac{1}{2}\) = 1\(\frac{3}{4}\)
Answer:
x = \(\frac{5}{4}\) or 1\(\frac{1}{4}\),

Explanation:
Given to solve x + \(\frac{1}{2}\) = 1\(\frac{1}{4}\) so subtracting both sides by \(\frac{1}{2}\) we get x = \(\frac{1 X 4 + 3}{4}\) – \(\frac{1}{2}\) we get x = \(\frac{7}{4}\) – \(\frac{1}{2}\) , x = \(\frac{7 –  1 X 2}{4}\) , x = \(\frac{5}{4}\) as numerator is more we write in mixed fraction as x = \(\frac{1 X 4 + 1}{4}\), x = 1\(\frac{1}{4}\).

Question 28.
g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\)
Answer:
g = 2,

Explanation:
Given to solve g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\) so subtracting both sides by \(\frac{5}{3}\) we get g = \(\frac{3 X 3 + 2}{3}\) – \(\frac{5}{3}\) we get g = \(\frac{11}{3}\) – \(\frac{5}{3}\) , g = \(\frac{11 –  5}{3}\) , g = \(\frac{6}{3}\) as numerator and denominator both goes by 3 we get g = \(\frac{2 X 3}{3 X 1}\), g = 2.

Question 29.
2\(\frac{5}{7}\) = p – \(\frac{2}{7}\)
Answer:
p = 3,

Explanation:
Given to solve 2\(\frac{5}{7}\) = p – \(\frac{2}{7}\) so adding both sides by \(\frac{2}{7}\) we get \(\frac{2 X 7 + 5}{7}\) + \(\frac{2}{7}\)= p – \(\frac{2}{7}\) + \(\frac{2}{7}\), we have common denominators so \(\frac{19 + 2}{7}\) = p, \(\frac{21}{7}\) = p as numerator and denominator both goes by 7 we get p = \(\frac{7 X 3}{7 X 1}\), p = 3.

Question 30.
e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\)
Answer:
e = \(\frac{35}{11}\) or 3\(\frac{2}{11}\),

Explanation:
Given to solve e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\) so adding both sides by \(\frac{18}{11}\) we get e = \(\frac{11 X 1 + 6}{11}\) + \(\frac{18}{11}\) we get e = \(\frac{17}{11} \) + \(\frac{18}{11}\) both have common denominators we get e = \(\frac{17 + 18}{11}\) , e = \(\frac{35}{11}\) as numerator is greater than denominator we write in mixed fraction as e = \(\frac{3 X 11 + 2}{11}\), so e = 3\(\frac{2}{11}\).

Question 31.
\(\frac{4}{3}\)y = 36
Answer:
y = 27,

Explanation:
Given to solve \(\frac{4}{3}\)y = 36 multiplying both sides by \(\frac{3}{4}\) we get y = 36 X \(\frac{3}{4}\), y = \(\frac{36 X 3}{4}\) = \(\frac{4 X 9 X 3}{4}\) we get y = 9 X 3 = 27.

Question 32.
\(\frac{9}{10}\) = \(\frac{5}{6}\)v
Answer:
v = \(\frac{54}{50}\) or 1\(\frac{4}{50}\),

Explanation:
Given to solve \(\frac{9}{10}\) = \(\frac{5}{6}\)v multiplying both sides by \(\frac{6}{5}\) we get \(\frac{9}{10}\) X \(\frac{6}{5}\) =v, v = \(\frac{9 X 6}{10 X 5}\) = \(\frac{54}{50}\) as numerator is greater we can further write in mixed fraction as \(\frac{1 X 50 + 4}{50}\) = 1\(\frac{4}{50}\).

Question 33.
\(\frac{2}{3}\)k = 28 ∙ \(\frac{4}{9}\)
Answer:
k = \(\frac{56}{3}\) or 18\(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{2}{3}\)k = 28 . \(\frac{4}{9}\) multiplying both sides by \(\frac{3}{2}\) we get k = \(\frac{28 X 4}{9}\) X \(\frac{3}{2}\), k = \(\frac{28 X 2}{3}\), k = \(\frac{56}{3}\) as numerator is greater we can further write in mixed fraction as \(\frac{18 X 3 + 2}{3}\) = 18\(\frac{2}{3}\).

Solve.

Question 34.
Find five pairs of whole numbers, such that when they are inserted into the equation below, the solution of the equation is 3,
x + =
Answer:
(0,3), (1,4), (2,5), (3,6) and (4,7),

Explanation:
Given to find five pairs of whole numbers such that when they are inserted into the equation x + __ =___  the solution of the equation is 3 so
1. When (0,3) substituting x + 0 = 3 yes true as x = 3,
2. When (1,4) substituting x + 1 = 4 solving we get x = 4 – 1 = 3 which is true,
3. When (2,5) substituting x + 2 = 5 solving we get x = 5 – 2 = 3 which is true,
4. When (3,6) substituting x + 3 = 6 solving we get x = 6 – 3 = 3 which is true,
5. When (4,7) substituting x + 4 = 7 solving we get x = 7 – 4 = 3 which is true therefore 5 pairs of whole numbers are (0,3), (1,4), (2,5), (3,6) and (4,7).

Question 35.
Find five pairs of numbers, such that when they are inserted into the equation below, the solution of the equation is \(\frac{2}{5}\).
x =
Answer:
(\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5),

Explanation:
Given to find five pairs of numbers such that when they are inserted into the equation ___x = ____ , the solution of the equation is \(\frac{2}{5}\) so
1. When (\(\frac{2}{5}\),1) substituting \(\frac{2}{5}\) X 1 = \(\frac{2}{5}\),
2. When (\(\frac{1}{5}\), 2) substituting \(\frac{1}{5}\) X 2 = \(\frac{2}{5}\),
3. When (\(\frac{2}{15}\), 3) substituting \(\frac{2}{15}\) X 3 = \(\frac{2}{5}\),
4. When (\(\frac{2}{20}\),4) substituting \(\frac{2}{20}\) X 4 = \(\frac{2}{5}\),
5. When (\(\frac{2}{5}\),5) substituting \(\frac{2}{25}\) X 5 = \(\frac{2}{5}\) therefore 5 pairs of numbers are (\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5).

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.2 Writing Linear Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.2 Guided Practice Answer Key

Complete.

Question 1.
Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah.
a) Write an expression for the number of baseball cards that Miguel has in terms of h.

Miguel has baseball cards.

b) If Miguel has k baseball cards, express k in terms of h.
k = +
c) State the independent and dependent variables.
Independent variable: , dependent variable:
Answer:
a) Miguel has h + 7 baseball cards,
b) k = h + 7,
c) Independent variable: h, dependent variable: k,

Explanation:
Given Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah,
a) Wrote an expression for the number of baseball cards that Miguel has in terms of h as Miguel has h + 7 baseball cards.
b) If Miguel has k baseball cards expressing k in terms of h as k = h + 7.
c) The independent and dependent variables are independent variable is h and dependent variable are k which depends on value of h.

Write an equation for each of the following. Then state the independent and dependent variables for each equation.

Question 2.
Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park, express t in terms of p.
Answer:
t = p + 12,
Independent  variable : p, Dependent variable: t,

Explanation:
Given Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park,To jog around the park it is p minutes, and Sofia took 12 more minutes if Sofia took t minutes expressing t in terms of p it is t = p + 12. Here p is independent variable and t is dependent variable on the value of p.

Question 3.
A bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, express n in terms of m.
Answer:
n = $30 – m,
Independent variable : m, Dependent variable: n,

Explanation:
Given a bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, expressing n in terms of m as n = $30 – m, Here m is independent variable and n is dependent variable on value of m.

Question 4.
Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, express c in terms of b.
Answer:
c = 7b,
Independent variable : b, Dependent variable: c,

Explanation:
Given Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, expressing c in terms of b as c = 7b as Nathan has 7 boxes of b marbles each here b is independent variable and c is dependent variable on the value of b.

Question 5.
A motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars, express y in terms of x.
Answer:
x = 12y,
Independent variable : b, Dependent variable: c,

Explanation:
Given a motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars expressing y in terms of x as it is x = 12y where y is independent variable and x is dependent variable on the value of y.

Copy and complete the table. Then use the table to answer the questions.

Question 6.
The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters, write an equation relating p and q.

Answer:
Equation: q = p – 2,

Explanation:
Given The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters so an equation relating p and q is q = p – 2,
If p =3 then q = 3 – 2 = 1,
If p = 4 then q = 4 – 2 = 2,
If p = 5 then q = 5 – 2 = 3,
If p = 7 then q = 7 – 2 = 5,
If p = 8 then q = 8 – 2 = 6,
Completed the table as shown above.

b) Use the data from a) to plot the points on a coordinate plane. Connect the points with a line.
Answer:

Explanation:
Used the data from a) to plot the points on a coordinate plane as length p meters on x – axis and width q meters on y axis. Connected the points with a line as shown above.

c) The point (5.5, 3.5) is on the line you drew in b). Does this point make sense in the situation?
Answer:
Yes,

Explanation:
The point(5.5,3.5) is on the line which i drew yes this point makes sense as p = 5.5 we have q = p – 2 = 5.5 – 2 = 3.5 which is true.

Copy and complete each table. Then express the relationship between the two variables as an equation.

Question 7.
Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee.

Answer:
Equation: y = x + 6,

Explanation:
Given Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee. If x is number of books Lee borrowed and y be books Paul borrowed so the equation is y = x + 6, so
if x = 1 then y = 1 + 6 = 7 books,
if x = 2 then y = 2 + 6 = 8 books,
if x = 3 then y = 3 + 6 = 9 books,
if x = 4 then y = 4 + 6 = 10 books and
if x = 5 then y = 5 + 6 = 11 books shown in the above table.

Question 8.
At a crafts store, Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads.

Answer:
Equation: r = 4b,

Explanation:
Given at a crafts store Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads. Let read beads be r and blue beads be b so the equation is r = 4b,
if b = 2 then r = 4 X 2 = 8 beads,
if b = 3 then r = 4 X 3 = 12 beads,
if b = 4 then r = 4 X 4 = 16 beads,
if b = 5 then r = 4 X 5 = 20 beads shown above in the table.

Use the data in the table to plot points on a coordinate plane. Connect the points to form a line. Then write an equation to show the relationship between the variables.

Question 9.

Answer:

Equation:
d = 50t,

Explanation:
Used the data in the table to plot points on a coordinate plane. Connected the points to form a line as shown above  here d is the distance traveled in miles on y – axis and t is the time taken in t hours on x – axis. As distance traveled is increasing by 50 each hour an equation to show the relationship between the variables is d = 50t.

Math in Focus Course 1B Practice 8.2 Answer Key

Solve.

Question 1.
Joshua is w years old. His brother is 3 years older than he is.
a) If his brother is x years old, express x in terms of w.
Answer:
x = w + 3,

Explanation:
Given Joshua is w years old. His brother is 3 years older than he is means brother is w + 3, So if his brother is x years old expressing x in terms of w as x = w + 3.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : w,
Dependent variable : x,

Explanation:
Here Joshua is w years old his brother is 3 years older than he is and  if his brother is x years old then the independent variable is w and dependent variable is x on the value of w.

Question 2.
Rita has b markers. Sandy has 11 fewer markers than she has.

a) If Sandy has h markers, express h in terms of b.
Answer:
h = b – 11,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Sandy has b – 11 markers, If Sandy has h markers expressing h in terms of b as h = b – 11.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : b,
Dependent variable : h,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Independent variable is b and dependent variable is h on the value of b.

Question 3.
A small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much.

a) If the weight of the jumbo box of cereal is m grams, express m in terms of k.
Answer:

m = 5k,

Explanation:
Given a small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much means jumbo box is 5k, If the weight of the jumbo box of cereal is m grams, expressing m in terms of k as m = 5k.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : k,
Dependent variable : m,

Explanation:
As a small box of cereal weighs k grams and a jumbo box of cereal weighs 5 times as much here the independent variable is k and dependent variable is m on the value of k.

Question 4.
The area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm.

a) If s represents the area of Stan’s farm, express s in terms of n.
Answer:
n = 8s,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n is n = 8s.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : s,
Dependent variable : n,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n here Independent variable is s and dependent variable is n which depends on s.

Question 5.
Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points.

a) Write an equation relating x and y.
Answer:
y = x + 8,

Explanation:
Given Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points. So an equation relating x and y is y = x + 8.

b) Copy and complete the table to show the relationship between x and y.

Answer:

Explanation:
Completed the table to show the relationship between x and y as y = x + 8,
if x = 10 then y = 10 + 8 = 18,
if x = 11 then y = 11 + 8 = 19,
if x = 12 then y = 12 + 8 = 20,
if x = 13 then y = 13 + 8 = 21,
if x = 14 then y = 14 + 8 = 22 and
if x = 15 then y = 15 + 8 = 23 as shown above.

Question 6.
There are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.
a) Express y in terms of x.
Answer:
y = x + 50,

Explanation:
Given there are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.So expressing y in terms of x is y = x + 50.

b) Make a table to show the relationship between y and x. Use values of x = 10, 20, 30, 40, and 50 in your table.
Answer:

Explanation:
Asked to make a table to show the relationship between y and x. Used values of x = 10, 20, 30, 40, and 50 in my table and completed as shown above in the equation y = x + 50,
When x = 10 then y = 10 + 50 = 60,
when x = 20 then y = 20 + 50 = 70,
when x = 30 then y = 30 + 50 = 80,
when x = 40 then y = 40 + 50 = 90,
when x = 50 then y = 50 + 50 = 100.

c) Graph the relationship between y and x in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between y and x in a coordinate plane as shown above as sparrows x on x -axis and total number of sparrows y on y – axis.

Question 7.
A rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
a) Express P in terms of b.
Answer:
P = 6b,

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressing P in terms of b as P = 2(2b + b). = 2(3b) = 6b.

b) Copy and complete the table to show the relationship between P and b.

Answer:

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressed P in terms of b as P = 6b completed the table to show the relationship between P and b above as
if b= 1 then P = 6 X 1 = 6,
if b = 2 then P = 6 X 2 = 12,
if b = 3 then P = 6 X 3 = 18,
if b= 4 then P = 6 X 4 = 24,
if b = 5 then P = 6 X 5 = 30 and
if b = 6 then P = 6 X 6 = 36.

Question 8.
Every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month.
a) Express r in terms of n.
Answer:

r = n – 0.6n,

Explanation:
Given every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month. Expressing r in terms of n as r = n – n X 60/100 = n – 0.6n.

b) Make a table to show the relationship between r and n. Use values of n = 100, 200, 400, and 500 in your table.
Answer:

Explanation:
Made a table above to show the relationship between r and n. Using values of n = 100, 200, 400, and 500 in my table and substituting in r = n – 0.6n as
if n = 100 then r = 100 – 0.6 X 100 = 100 – 60 = 40,
if n = 200 then r = 200 – 0.6 X 200 = 200 – 120 = 80,
if n = 300 then r = 300 – 0.6 X 300 = 300 – 180 = 120,
if n = 400 then r = 400 – 0.6 X 400 = 400 – 240 = 160 and
if n = 500 then r = 500 – 0.6 X 500 = 500 – 300 = 200.

c) Graph the relationship between n and r in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between n and r in a coordinate plane as  r = n – 0.6n where earnings n dollars on x – axis and savings r dollars on y – axis as shown above.

d) The point (287.5, 115) is on the line you drew in c). Does this point make sense in the situation? Explain.
Answer:
Yes,

Explanation:
Given the point (287.5, 115) is on the line I drew in c) checking in r = n – 0.6n, if n = 287.5 then r =  287.5 – 0.6 X 287.5 = 287.5 – 172.5 = 115 which matches so this point make sense in the situation.

Question 9.
The side length of a square is t inches. The perimeter of the square is z inches.
a) Express z in terms of t.
Answer:
z = 4t,

Explanation:
Given the side length of a square is t inches. The perimeter of the square is z inches. Expressing z in terms of t as perimeter of sqaure is 4 X side length so it is z = 4t.

b) Make a table to show the relationship between z and t. Use whole number values of t from 1 to 10.
Answer:

Explanation:
Made a table to show the relationship between z and t. Used whole number values of t from 1 to 10 in z – 4t as
if t = 1 then z = 4 X 1 = 4,
if t = 2 then z = 4 X 2 = 8,
if t = 3 then z = 4 X 3 = 12,
if t = 4 then z = 4 X 4 = 16,
if t = 5 then z = 4 X 5 = 20,
if t = 6 then z = 4 X 6 = 24,
if t = 7 then z = 4 X 7 = 28,
if t = 8 then z = 4 X 8 = 32,
if t = 9 then z = 4 X 9 = 36 and
if t = 10 then z = 4 X 10 = 40 shown above.

c) Graph the relationship between z and t in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between z and t in a coordinate plane as the side length of a square is t inches on x – axis and the perimeter of the square is z inches on y – axis as shown above.

d) Use your graph to find the perimeter of the square when the length is 3.5 inches and 7.5 inches.
Answer:
1) 14 inches,
2) 30 inches,

Explanation:
Using my graph the perimeter of the square when the length is 3.5 inches is 4 X 3.5 = 14 inches and when the length is 7.5 inches the perimeter of the square is 4 X 7.5 inches = 30 inches respectively.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.1 Nets of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.1 Answer Key Nets of Solids

Math in Focus Grade 6 Chapter 12 Lesson 12.1 Guided Practice Answer Key

Hands-On Activity

IDENTIFYING A CUBE FROM A NET

Work in pairs.
Step 1: Trace and cut out each figure.

Step 2: Try and fold them into cubes.

Math Journal Can you fold all the above figures into cubes? Discuss with your partner and explain your thinking.
Answer:
The figure (a) can be folded into cube
The figure (b) can also be folded into cube.
The figure (c) cannot be folded into cube.

Hands-On Activity

IDENTIFYING A PRISM FROM A NET

Work in pairs.
Step 1: Trace and cut out each net along the solid lines, Predict what figure can be formed from the net. Then fold the net to make the figure.

Step 2: Name the solid that each net forms
Answer:
If the figure (a) is folded, it will form a square cube.
If the figure (b) is folded, it will form into a rectangular cuboid.
If the figure (c) is folded, it will form into a triangular prism.
If the figure (d) is folded, it will form into a triangular prism.

Match each solid with its net(s). There may be more than one net of each solid.

Question 1.

Answer:
The first solid can have similar of b,f nets. Hence it can be matched with b,f nets.
The second solid have similar of a,e nets. Hence it can be matched with a,e nets.
The third solid have similar of c,d nets. Hence it can be matched with c,d nets.

Name the solid that each net forms.

Question 2.

Answer:
A cuboid can be formed when the above solid is folded.
Therefore, a cuboid is formed from the given net form.

Question 3.

Answer:
A cube can be formed when the above solid is folded.
Therefore, a cube is formed from the given net form.

Question 4.

Answer:
A triangular prism can be formed when the above solid is folded.
Therefore, a triangular prism is formed from the given net form.

Hand-On Activity

CLASSIFYING PYRAMIDS

Work in pairs
Step 1: Trace, cut out, and fold the nets.

Step 2: Name the solid that each net forms.
Answer:
When the nets in figure(a) is folded, it will form a triangular prism.
When the nets in figure(b) is folded, it will form a square prism.

Match each solid with its net(s). There may be more than one net of each solid.

Question 5.

Answer:
The first solid can have similar of a,c nets. Hence it can be matched with a,c nets.
The second solid can have similar of b nets. Hence it can be matched with b nets.

Math in Focus Course 1B Practice 12.1 Answer Key

Name each solid. In each solid, identify a base and a face that is not a base.

Question 1.

Answer:
The above figure is a pyramid since the sides meet at one point.
There is one base for the given figure, which has four equal sides.A base with four equal sides means the base is a square which is BECD.
A face with three sides means the base is a triangle. There are four faces which are not bases. The triangular faces for the given figure are BAE,ACD,AED and BEA.

Question 2.

Answer:
This figure is a prism since it has two ends that are the same shape.
There are two bases for the given figure, both with three sides. A base with three sides means the base is a triangle. The triangular bases for the given figure will be GFH and JKL.
The faces have four sides or two parallel sides. There are three faces which are not bases. The faces are FJKG, FHLJ and GHLK.

Question 3.

Answer:
The above figure has all equal sides and thus form a cube.
There are two bases for the given figure, both with four equal sides. A base with four equal sides means the base is a square which will be PSRQ and TWVU.
The faces have four equal sides. The square faces will be RSWV, SPTW, PQUT and QRVU.

Question 4.

Answer:
This figure is a prism since it has two ends that are the same shape.
There are two bases for the given figure, both with three sides. A base with three sides means the base is a triangle. The triangular bases for the given figure will be ABC and EFD.
The faces have four sides or two parallel sides. There are three faces which are not bases. The faces are AEDC, AEFB and BFCD.

Name the solid that each net forms.

Question 5.

Answer:
The above given figure has four triangles and a square.
Since all the sides meet a point, it can be a pyramid.
When the above given solid is folded it will form a square pyramid.

Question 6.

Answer:
The above given figure has four rectangles and two squares.
When the above given solid is folded it will form a polyhedron.

Question 7.

Answer:
The above given figure has three rectangles and two triangles.
Since it has two ends that are of the same shape.It can be a prism.
When the above given solid is folded it will form a triangular prism.

Question 8.

Answer:
The above given figure has four triangles.
Since all the sides meet a point, it can be a pyramid.
When the above given solid is folded it will form a triangular pyramid.

Decide if each net will form a cube. Answer Yes or No.

Question 9.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 10.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Question 11.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 12.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Question 13.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 14.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Solve. Use graph paper.

Question 15.
In Exercises 9 to 14, you identified some possible nets for a cube. There are other possible nets. Find all of the other possible nets.
Answer:
Each net has 6 squares that when folded properly form the six faces of a cube.

Decide if each net will form a prism. Answer Yes or No.
Question 16.

Answer:
Since it has two ends that are of the same shape.It can be a prism.
Yes, the above given net can form a prism.

Question 17.

Answer:
Since it has two ends that are of the same shape.It can be a prism.
Yes, the above given net can form a prism.

Question 18.

Answer:
No, the above given net can form a prism.

Copy the net of the rectangular prism shown. Then name the vertices that are not already labeled with a letter. Label the vertices.

Question 19.

Answer:

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Surface Area and Volume of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Answer Key Surface Area and Volume of Solids

Math in Focus Grade 6 Chapter 12 Quick Check Answer Key

Name each, prism. In each prism identify a base, a face, an edge, and a vertex.

Question 1.

Answer:
It is a Rectangular prism.
It has 2 bases, 6 faces, 12 edges and 8 vertices.

Question 2.

Answer:
It is a triangular prism.
It has 2 bases, 5 faces, 9 edges and 6 vertices.

Question 3.

Answer:
It is a square prism.
It has 2 bases, 6 faces, 12 edges and 8 vertices.

Find the area of each figure.

Question 4.

Area = •
= cm²
Answer:
The given rectangle is 4cm wide and 9cm long.
The formula for finding the area of rectangle is length×breadth.
The area for the given rectangle is 4×9=36 square centimetre.

Question 5.

Area = \(\frac{1}{2}\) • •
= m²

Answer:
The given trinagle measures 12m wide and 10m long.
The formula to find the area of the given triangle is \(\frac{1}{2}\)×base×height
Therefore, the area of the given triangle will be \(\frac{1}{2}\)×12×10 = 60 square metre.
Question 6.

Area = \(\frac{1}{2}\) • • ( + )
= \(\frac{1}{2}\) • •
= ft²
Answer:
The given quadrilateral has two sides of 10 ft and 7 ft with 5ft base.
Area of the trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
The area will be \(\frac{1}{2}\)×5×(10+7) = \(\frac{1}{2}\)×5×12
Therefore, the area will be 5×6=30 square ft.

Question 7.

Area = \(\frac{1}{2}\) • • ( + )
= \(\frac{1}{2}\) • •
= in.²
Answer:
The formula to find the area of trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area = \(\frac{1}{2}\) × 6.5 × (10+6)
= \(\frac{1}{2}\) × 6.5 × 16
= 6.5×8
= 52 square in.

Find the volume of each solid.

Question 8.

Volume = • •
= in.³
Answer:
Volume of the cube = length×width×height
The given cube measures 8 inches in length.
The volume of the given cube wil be 8×8×8=64 cubic.inches

Question 9.

Volume = • •
= in.³
Answer:
The given cuboid has length of 11 in, width of 5 in and height of 6 in.
Volume of the cuboid = length×width×height
The volume of the given cuboid is 11×5×6=330 cubic. inches.

Question 10.

Volume = • •
= ft³
Answer:
The given cube measures 12 inches in length.
Volume of the cube = length×width×height
The volume of the given cube wil be 12×12×12=64 cubic.inches

Question 11.

Volume = • •
= ft³
Answer:
The given cuboid measures 22ft, 13ft and 16ft.
Volume of the cuboid = length×width×height
The volume of the given cuboid is 22×13×16=4576 cubic. inches.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Review Test Answer Key

Concepts and Skills

Match each of the solid figures to its net.

Answer:
The net form of Figure 1 is Figure c.
The net form of Figure 2 is Figure a.
The net form of Figure 3 is Figure b.
The net form of Figure 4 is Figure e.
The net form of Figure 5 is Figure d.

Find the surface area of each solid.

Question 6.

Answer:
Given that the length of the cube is 8 in.
The given figure has 6 faces.
Area of 1 face will be 8×8 = 64 sq.in
Surface area of the given solid = 6×base area
The surface area of the given solid will be 6×64 = 384 sq.in

Question 7.

Answer:
Given solid has four triangular faces and one square base.
Area of base will be 8×8 = 64 sq.m
Area of triangle = \(\frac{1}{2}\) × base × width
Area of triangular face will be \(\frac{1}{2}\) × 8 × 10 = 40 sq.m
There are 4 such triangular faces, 4×40 = 160 sq.m
The surface area of the given figure will be 64+160 = 224 sq.m

Find the volume of each prism.

Question 8.

Answer:
Given cube has length of 7 cm.
Area of base = 7×7 = 49 sq.cm
Volume of the prism = base area × height
Volume of the prism will be 49×7 = 343 cu.cm

Question 9.

Answer:
Area of triangular base = \(\frac{1}{2}\) ×base×height
Area of the base will be \(\frac{1}{2}\) ×6×5 = 15 sq.ft
Volume of the given prism will be 3×15 = 45 cu.ft

Solve.

Question 10.
The solid below is made up of cubes, each of which has an edge length of 3 inches.
a) What is the volume of one cube?
Answer:
Given that the length of the cube is 3 inches.
Area of one cube will be = 3×3 = 9 sq.in
Volume of the cube= base area × height
Volume of one cube will be = 9×3 = 27 cu.in

b) What is the volume of the solid figure?

Answer:
The below solid figure has 10 identical cubes.
Therefore, the volume of the solid figure will be 10×27 = 270 cu.in

Problem Solving

Solve.

Question 11.
A fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high. It contains water up to a height of 28 centimeters. How many more cubic centimeters of water are needed to fill the tank to a height of 35 centimeters?
Answer:
Given that a fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high.
Volume of the tank = length × width × height
Volume of water filled up to a height of 28 centimeters is 50×30×28 = 42000 cu.cm
Volume of water to fill up to 35 cm will be 50×30×35 = 52500 cu.cm
More cubic centimeters of water are needed to fill the tank to a height of 35 centimeters will be 52500-42000 = 10500 cu.cm

Question 12.
Find the surface area of a square pyramid given that its base area is 196 square inches and the height of each of its triangular faces is 16 inches.

Answer:
Given that the base area of the given square pyramid is 196 sq.in.
Area of square = 196
196 can also be written as 14 × 14
length × length = 196
Length of one side of the square will be 14 cm.
Area of trinagular face will be \(\frac{1}{2}\) × 14 × 16
= 7 × 16
= 112 cu.in
Surface area of the given solid is 4×112 + 196
= 448 + 196
= 644 sq.in

Question 13.
The volume of a rectangular prism is 441 cubic feet. It has a square base with edges that are 7 feet long.
a) Find the height of the prism.
Answer:
Given that the volume of a rectangular prism is 441 cubic feet and it has a square base with edges that are 7 feet long.
Volume of the prism = Area of the base × height
Area of the base = 7×7 = 49 sq.ft
441 = 49 × height
height = 441÷49
height = 9 ft.
Therefore, the height of the prism is 9 ft.

b) Find the surface area of the prism.

Answer:
Area of the square base = length×length = 49 sq.ft
Area of the rectangular base = length×width= 7×9 = 63 sq.ft
There are 2 square bases and 4 rectangular faces.
Therefore, the surface area of the prism will be 2×49 + 4×63
= 98 + 252
= 350 sq.ft

Question 14.
The volume of a rectangular tank with a square base is 63,908 cubic centimeters. Its height is 64 centimeters. Find the length of an edge of one of the square bases. Round your answer to the nearest tenth of a centimeter.

Answer:
Given that the volume of a rectangular tank with a square base is 63,908 cubic centimeters and it is 64 cm high.
Volume of a rectangular prism = Area of base × height
63,908 = Area of base × 64
Area of base will be 63908 ÷ 64 = 998.56 sq.cm
Area of the base when rounded to nearest ten will be 1000.
Area of base = length × length
1000 = length × length
10 × 10 = length × length
Length of an edge of one of the square bases will be 10 cm.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.4 Real-World Problem: Surface Area and Volume detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.4 Answer Key Real-World Problem: Surface Area and Volume

Math in Focus Grade 6 Chapter 12 Lesson 12.4 Guided Practice Answer Key

Complete.

Question 1.
Find the volume of water needed to fill three fourths of the aquarium.

Height of water needed = \(\frac{3}{4}\) •
= in.
Volume of water needed = • •
= in.³
The aquarium needs to have cubic inches of water added to it to be \(\frac{3}{4}\) full.
Answer:
Volume = length × width × height
Volume of water needed = 25×12×14
= 300×14
= 4200 cu.in
Height of water needed = \(\frac{3}{4}\) × 4200
= 3 × 150
= 3150 cu.in
The aquarium needs to have 3150 cubic inches of water added to it to be \(\frac{3}{4}\) full.

Question 2.
A metal bar has bases that are parallelograms.

a) Find the volume of the metal bar.
Area of parallelogram
= base of parallelogram • height of parallelogram
= •
= cm²
Volume of metal bar = base of prism • height of prism
= •
= cm³
The volume of the metal bar is cubic centimeters.
Answer:
Area of parallelogram
= base of parallelogram × height of parallelogram
= 8 × 4
= 32 cm²
Volume of metal bar = base of prism × height of prism
= 32 × 24
= 768  cm³
The volume of the metal bar is 480 cubic centimeters.

b) Find the surface area of the metal bar.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases

The surface area of the metal bar is square centimeters.
Answer:
Given parallelogram is 5cm high, 24cm long and 8cm wide.
The perpendicular height of the base is 4cm.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases
= (24+8+8+24) × 5 + (20+20)
= (64×5) + 40
= 320 + 40
= 360 sq.cm

Question 3.
A candle is a square prism. The candle is 15 centimeters high, and its volume is 960 cubic centimeters.

a) Find the length of each side of the square base.
V = Bh
= B •
÷ = • B ÷
= B
Length of each side of base
=
= cm
The length of each side of the square base is centimeters.
Answer:
Given that the square prism is 15cm high and its volume is 960 cubic centimeters.
V = Bh
960 = B × 15
960 ÷ 15 = 15 × B ÷ 15
64 = B
Base is side × side = 8 × 8
The length of each side of the square base is 8 centimeters.

b) Find the surface area of the candle.
Surface area of candle
= perimeter of base • height + area of two bases

The surface area of the candle is square centimeters.
Answer:
= perimeter of base × height + area of two bases
= (8 + 8 +8 + 8) × 15 + (2×64)
= 32 × 15 + 128
= 480 + 128
= 608 sq.cm

Question 4.
A storage chest is a prism with bases that are pentagons. The diagram shows some of the dimensions of the storage chest. The volume of the storage chest is 855 cubic inches.

a) Find the height AS of the prism. Round your answer to the nearest hundredth.
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) • • ( + ) + •
= +
= in.²

Answer:
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) × 3× (3+7) + 7× 9
= 3×5 + 63
= 15+63
= 78 sq.in

V = Bh
= • h
÷ = • h ÷
= ≈ h
The height of the prism is approximately inches.
Answer:
855 = 78×h
855 ÷ 78 = 78 × h ÷ 78
10.96 = h
Height of the prism will be 10.96 in.

b) Find the surface area of the prism. Round your answer to the nearest hundredth.
Surface area of prism
= perimeter of base • height + area of two bases

The surface area of the prism is approximately square inches.
Answer:
Area of two bases = 78 + 78 = 156 sq.in
= perimeter of base • height + area of two bases
= (5+3+3+9+7)×10.96 + 156
= 27×10.96 + 156
= 295.92 + 156
= 451.92 sq.in
The surface area of the prism is approximately 500 sq.inf

Math in Focus Course 1B Practice 12.4 Answer Key

Solve.

Question 1.
Savannah has a water bottle that is a rectangular prism. The bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water. Then, she drank \(\frac{1}{4}\) of the volume of water in her water bottle. How many cubic centimeters of water were left in the water bottle?

Answer:
Given that the bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water.
Therefore, the volume of the rectangular prism = length×width×height
= 7×5×18
= 630 cu.cm
Volume of water in the bottle is 630 cu.cm
Out of which she drank \(\frac{1}{4}\), so the left over water will be \(\frac{1}{4}\) of total volume
= \(\frac{1}{4}\) × 630
= 157.5 cu.cm
Therefore, 157.5 cu.cm of water will be left in the water bottle.

Question 2.
A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches.
a) Find the height of the prism.
Answer:
Given that a rectangular prism has a square base with edges measuring 8 inches each and its volume is 768 cubic inches.
Volume of rectangular prism = 768
Area of base × height = 768
8×8× height = 768
height = 768÷64
Height of the prism is 12 in

b) Find the surface area of the prism.
Answer:
Surface area of the base = 8×8 = 64sq.in
Total area of 2 bases = 2×64 = 128 sq.in
Surface area of the prism = perimeter of base × height + total area of 2 bases
= 4×8×12 + 128
= 144 +128
= 272 sq.in

Question 3.
A triangular prism has the measurements shown.
a) Find the volume of the prism.
Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
`Volume of triangular prism = 49×16 = 784 cu.ft

b) Find the surface area of the prism.

Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
Area of two triangular faces will be 2×49 = 98 sq.ft
Area of rectangular face = length×width
Area of first rectangular face = 16×10 = 160 sq.ft
Area of second rectangular face = 14×16 = 192 sq.ft
Area of third rectangular face = 16×19.6 = 313.6 sq.ft
Surface area of prism = 98+160+192+313.6
= 763.6 sq.ft

Question 4.
The volume of Box A is \(\frac{2}{5}\) the volume of Box B. What is the height of Box A if it has a base area of 32 square centimeters?

Answer:
Box B:
Box B is 16cm wide, 8cm long and 10cm high.
Volume of Box B will be = 16×8×10 = 1280 cu.cm
Given that volume of Box A is \(\frac{2}{5}\) the volume of Box B.
Therefore, volume of Box A = \(\frac{2}{5}\)×1280
= 2×256
= 512
Volume of Box A = Area of Base × height
512 = 32 × height
Height = 16 cm.
Therefore height of Box A will be 16cm.

Question 5.
The ratio of the length to the width to the height of an open rectangular tank is 10 : 5 : 8. The height of the tank is 18 feet longer than the width.
a) Find the volume of the tank.
Answer:
Let the width of the tank be ‘a’ ft.
Height of the tank is 18 feet longer than the width, therefore, it will be 18+a
If height equals to 8 parts, width equals 5 parts, then 8 – 5 parts = 18 ft
3 parts = 18 ft.
1 part = 6ft
Therefore, now the length will be 10×6=60 ft, width will be 5×6=30ft and height will be 8×6=48ft.
Volume of the tank will be 60×30×48=86400 cu.ft

b) Find the surface area of the open tank.
Answer:
Since the rectangular tank resembles rectangular prism.
Surface area of rectangular prism = 2×(length×width + width×height + height×length)
The surface area of the open tank = 2×(60×30 + 30×48 + 48×60)
= 12240 sq.ft

Question 6.
Janice is making a gift box. The gift box is a prism with bases that are regular hexagons, and has the dimensions shown in the diagram.

a) Find the height PQ of the prism.
Answer:
Given that the volume of prism is 2835 cu.cm
Volume of prism = Area of hexagonal base × height
6 equilateral triangles form a regular hexagon.
Therefore, Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
= 6×(7×3)
= 6×21
= 126 sq.cm
2835 = 126 × height
Height = 2835÷126
height = 22.5cm
The height PQ of the prism will be 22.5cm

b) Find the surface area of the prism.
Answer:
Surface area of prism will be = Perimeter × height + Area of base
= 7×6 ×22.5 + 126 + 126
= 42×22.5 + 252
= 945 + 252
= 1197 sq.cm
The surface area of the prism will be 1197 sq.cm

Question 7.
Container A was filled with water to the brim. Then, some of the water was poured into an empty Container B until the height of the water in both containers became the same. Find tT.

Answer:
Container A:
Container A is a rectangular prism of 40cm high, 25cm long and 30 cm wide.
Volume of the rectangular prism is = length × width × height
Volume of water in the container will be 40×25×30 = 30000 cu.cm
The volumes of two containers will be same.
Container B:
Therefore the volume of Container B will be 30000 cu.cm
Given that the container B is 18cm long and 25cm wide.
Volume of the rectangular prism is = length × width × height
30000 = length × width × height
30000 = 18 × 25× height
Height = 30000÷450
Height = 66.6cm.
The new height of the water in both containers will be 66.6cm

Brain @ Work

Question 1.
The volume of a cube is 1000 cubic inches. If each of the edges is doubled in length, what will be the volume of the cube?
Answer:
Given that the volume of a cube is 1000 cubic inches.
Volume of a cube = Area of base × height
1000 =length × width × height
1000 = 10 × 10 × 10
Therefore, we can say length of the cube will be 10 in.
If the length of edge is doubled, then it will be 2×10 = 20 in.
Volume of the new cube will be 20×20×20 = 8000 cu.in

Question 2.
The volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number. Find the length of each edge of the cube.
Answer:
Given that the volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number.
Volume of the cube = Area of base × height
Surface area will be 6×Area of base
Given surface area is x sq.ft
x = 6 × Area of base
Area of base = \(\frac{x}{6}\)
Volume of the cube = \(\frac{x}{6}\) × height
x = \(\frac{x}{6}\) × height
Height = \(\frac{6x}{x}\)
Height will be 6 ft.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.3 Volume of Prisms detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.3 Answer Key Volume of Prisms

Math in Focus Grade 6 Chapter 12 Lesson 12.3 Guided Practice Answer Key

Find the volume of each rectangular prism.

Question 1.

Length = 5\(\frac{1}{4}\) in.
Width = 6 in.
Height = 12 in.
V = lwh
= • •
= in.³
Answer:
V = lwh
Length = 5\(\frac{1}{4}\) in.
(5×\(\frac{4}{4}\)+\(\frac{1}{4}\)) = \(\frac{21}{4}\)
V = \(\frac{21}{4}\) × 6 × 12 = 378 cu.in

Question 2.

Length = 8 cm
Width = 7.2 cm
Height = 3 cm
V = lwh
= • •
= cm³
Answer:
V = 8 × 7.2 × 3
172.8 cu.cm

Question 3.

Length = 4 ft
Width = 3 ft
Height = 8 \(\frac{1}{3}\) ft
Volume =
= • •
= ft³
Answer:
V = lwh
Height = 8 \(\frac{1}{3}\) ft
(8 × \(\frac{3}{3}\))+\(\frac{1}{3}\) = \(\frac{25}{3}\)
Volume = 4×3×\(\frac{25}{3}\) = 100 cu.ft

Tell whether slices parallel to each given slice will form uniform cross sections. If not, explain why not.

Question 4.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is larger than the area’s of the cube’s face.

Question 5.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 6.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Find the volume of each prism.

Question 7.

Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = •
= cm²
Volume of prism = •
= cm²
The volume of the prism is cubic centimeters.
Answer:
Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = 6×5.5 = 33 sq.cm
Volume of prism = Base area of prism × height
Volume of prism = 33×9 = 297 cu.cm
The volume of the prism is 297 cubic centimeters.

Question 8.

Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
Height of prism = 14 in.
Area of base = \(\frac{1}{2}\) • •
= in.²
Volume of prism = •
= in.³
The volume of the prism is cubic inches.
Answer:
Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
(3×\(\frac{2}{2}\))+\(\frac{1}{2}\) = \(\frac{7}{2}\)
Height of prism = 14 in.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of base = \(\frac{1}{2}\) × 10 × \(\frac{7}{2}\) = \(\frac{35}{2}\) sq.in
Volume of prism = Base area of prism × height
Volume of prism = \(\frac{35}{2}\) × 14 = 245 cu.in

Question 9.

Length of shorter base of trapezoid = 4 ft
Length of longer base of trapezoid = 10 ft
Height of trapezoid = 2 ft
Height of prism = 12 ft
Area of base = \(\frac{1}{2}\) • ( + )
= \(\frac{1}{2}\) • •
= ft²
Volume of prism = •
= ft³
The volume of the prism is cubic feet.
Answer:
The formula to find the base area = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (4+10)
= \(\frac{1}{2}\) × 2 × 14
= 14 sq.ft
Volume of prism = Base area of prism × height
Volume of prism = 14×12
= 168 cu.ft

Hands-On Activity

DETERMINING THE RELATIONSHIP BETWEEN VOLUME AND SURFACE AREA OF PRISMS

Materials:

• 27 unit cubes.

Work in pairs.
Step 1: Build the cube and the rectangular prism using unit cubes.

Step 2: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about the volumes of the cube and the rectangular prism?
Answer:
Volume of cube = lwh
Volume of the cube = 2×2×2 = 8 cu.cm
Volume of the rectangular prism = 2×4×1 = 8 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 3: Find the surface area of the cube. Draw its net if it helps you. Find the surface area of the rectangular prism. Draw its net if it helps you. What can you say about the surface areas of the cube and the rectangular prism?
Answer:
Area of one square will be 2×2 = 4 sq.cm
Volume of cuboid = 6 × Area of base
There are 6 faces for a cube, therefore the surface area will be 6×4 = 24 sq.cm
The surface areas of the rectangular prism will be 2(2×4 + 2×1 + 1×4) = 28 sq.cm
The surface areas of the cube and the rectangular prism are different.

Step 4: Now build these rectangular prisms using unit cubes.

Step 5: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about their volumes?
Answer:
Volume of the cube = 3×3×3 = 27 cu.cm
Volume of the rectangular prism = 3×9×1 = 27 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 6: Find the surface area of the cube. Find the surface area of the . rectangular prism. Draw their nets if it helps you. What can you say about their surface areas?
Area of one square will be 3×3 = 9 sq.cm
There are 6 faces for a cube, therefore the surface area will be 6×9 = 54 sq.cm
The surface areas of the rectangular prism will be 2(3×9 + 9×1 + 1×3) = 78 sq.cm
The surface areas of the cube and the rectangular prism are different.

Math Journal Based on the activity, what can you conclude about prisms with the same volume? Discuss with your partner and explain your thinking.
Answer:
Two prisms of different measurements might have the same volume, they might not have the same surface area.
For example: A rectangular prism with side lengths of 1 cm, 2 cm, and 2 cm has a volume of 4 cu cm and a surface area of 16 sq cm. A rectangular prism with side lengths of 1 cm, 1 cm, and 4 cm has the same volume but a surface area of 18 sq cm

Math in Focus Course 1B Practice 12.3 Answer Key

Solve.

Question 1.
A cube has edges measuring 9 inches each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 9 inches each.
The base area will be equal to 9×9=81 sq.in
Volume of cube = Area of base × height
Volume of the cube will be 81×9=729 cu.in

Question 2.
A cube has edges measuring 6.5 centimeters each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 6.5 cm each.
The base area will be equal to 6.5×6.5=42.25 sq.cm
Volume of cube = Area of base × height
Volume of the cube will be 81×9=274.625 cu.cm

Question 3.
A storage container is shaped like a rectangular prism. The container is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high. Find the volume of the storage container.
Answer:
Given that: A rectangular prism is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high.
Area of base will be 20×10=200 sq.ft
Volume of prism = Area of base × height
Volume of the storage container will be 200×5\(\frac{1}{2}\)
= 200×(5×\(\frac{2}{2}\) + \(\frac{1}{2}\))
= 200×(\(\frac{10}{2}\) + \(\frac{1}{2}\))
= 200×\(\frac{11}{2}\)
= 100×11
= 1100 cu.ft

Question 4.
Find the volume of the peppermint tea box on the right.

Answer:
Given that: The peppermint tea box is 12.6 cm long, 6.7 cm wide, and 7.8 cm high.
Area of  base will be 12.6×6.7 = 84.42 sq.cm
Volume of prism = Area of base × height
Volume of the box will be 7.8×84.42 = 658.476 cu.cm

Question 5.
The solid below is made of idential cubes. Each cube has an edge length of 2 inches. Find the volume of the solid.

Answer:
Given that: Each cube has an edge length of 2 inches.
Let us first find the volume of a single cube
Area of the base of a single cube will be 2×2=4 sq.in
Volume of cube = Area of base × height
Volume of a single cube will be  4×2=8 cu.in
The given solid is made of idential cubes. There are 9 such cubes.
Therefore, the volume of all the 9 identical cubes or the given solid figure will be 9×8=72 cu.in

Find the volume of the triangular prism.

Question 6.

Answer:
Given that: A triangular prism is 15ft long, 10ft wide and 6ft high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 10 × 6
= 10 × 3
= 30 sq.ft
Volume of prism = Area of base × height
Volume of the given triangular prism will be 30×15=450 cu.ft

Question 7.

Answer:
Given that: A triangular prism is 12cm long, 6.7cm wide and 3cm high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 6.7 × 3
= \(\frac{1}{2}\) × 20.1
= 10.05 sq.cm
Volume of prism = Area of base × height
Volume of the given triangular prism will be 10.05×12=120.6 cu.cm

Tell whether slices parallel to each given slice will form uniform cross-sections. If not, explain why not.

Question 8.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 9.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the circular face.

Question 10.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Copy the solid. Draw a slice that has the same cross section as the bases In each prism.

Question 11.

Answer:

The bases for the above given figure is square.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 12.

Answer:

The bases for the above given figure is hexagon.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 13.

Answer:

The bases for the above given figure is triangle.
Therefore, a slice that has the same cross section is drawn for the given prism.

Solve.

Question 14.
The bases of the prism shown are trapezoids. Find the volume of the prism.

Answer:
Length of shorter base of trapezoid = 2 m
Length of longer base of trapezoid = 6 m
Height of trapezoid = 2 m
Height of prism = 10 m
The area of the given figure = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (2+6)
= \(\frac{1}{2}\) × 2 × 8
= 8 sq.m
Volume of prism = base area × height
The volume of prism = 10×8
= 80 cu.m

Question 15.
A cube has a volume of 125 cubic inches. Find the length of its edge.
Answer:
Given that: A cube has a volume of 125 cu.in
Let us assume the length of a cube be ‘a’ unit.
As cube has all equal sides, the base area will be a×a = a² sq.units
Now, the volume will be a²×a = a³ cubic units.
Therefore we can say that the volume of a cube is the length of an edge taken to the third power.
a³ = 125 and 125 in cube can be written as 5³
a³ = 5³
a = 5
Therefore, the length of its edge will be 5 inches.

Question 16.
The volume of a triangular prism is 400 cubic centimeters. Two of its dimensions are given in the diagram. Find the height of a triangular base.

Answer:
Given that: The volume of a triangular prism is 400 cubic centimeters.
The three rectangular faces measures 10cm long and 8cm wide.

The area of three rectangular faces will be 3×10×8 = 240 sq.cm.
Inorder to get the areas of the triangular bases, we need to subtract the area of the rectangular faces from the total volume.
Area of two triangular bases will be 400-240 = 160 sq.cm
Area of one triangular base will be 160÷2 = 80 sq.cm
From the given figure, the base length of the triangle is 10 cm and the height is ‘h’ cm.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of the triangle is \(\frac{1}{2}\)×10×h = 80
5×h = 80
h = 80÷5
h = 16 cm.

Question 17.
A cross-section of the triangular prism shown below is parallel to a base. The area of the cross-section is 24 square feet. The ratio of DM to MA is 3 : 5 and the length of \(\overline{F O}\) is 6 feet. Find the volume of the triangular prism.

Answer:
Given that the area of cross-section is 24 sq.ft and ratio of DM to MA is 3 : 5.
The length of FO is 6 ft.
By observing the FO length and the ratio, we can say that the FO length is twice of the ratio of DM.
Therefore, the length of OC will be twice of the ratio of MA.
Length of OC = 10 ft.
Total length = FO + OC = 6+10 = 16 ft.
Volume of the triangular prism = Area of triangular base × length
Volume of the triangular prism = 24×16
Volume of the triangular prism = 384 cu.ft

Question 18.
The volume of the rectangular prism shown below is 2,880 cubic inches. The cross-section shown is parallel to a base. The area of the cross-section is 180 square inches. The length of \(\overline{A B}\) is x inches, and the length of \(\overline{B C}\) is 4x inches.
a) Find the length of \(\overline{A C}\).
Answer:
Given that: The volume of the rectangular prism is 2,880 cubic inches and the area of the cross-section is 180 square inches.
The length of AB is x inches and BC is 4x inches.
Length of AC will be AB+BC = x+4x = 5x

b) Find the value of x.

Answer:
Volume of the rectangular prism = 2,880 cubic inches
Volume of the rectangular prism = Area of rectangular base × length
2,880 = 180 × 5x
5x = 2880 ÷ 180
5x = 16
x = \(\frac{16}{5}\)
Length of AB will be \(\frac{16}{5}\) in.
Length of BC will be 4×\(\frac{16}{5}\) = \(\frac{64}{5}\) in.
Length of AC will be \(\frac{80}{5}\) = 16 in.

Question 19.
In the diagram of a cube shown below, points A, B, C, and D are vertices. Each of the other points on the cube is a midpoint of one of its sides. Describe a cross-section of the cube that will form each of the following figures.
a) a rectangle
b) an isosceles triangle
c) an equilateral triangle
d) a parallelogram

Answer:
a) a rectangle
By joining the CDLJ points, a rectangular cross section can be formed.
b) an isosceles triangle
By joining the MHL points, an isosceles triangle cross section can be formed.
c) an equilateral triangle
By joining the CDM points,an equilateral triangle cross section can be formed.
d) a parallelogram
By joining the MHGL points, a parallelogram cross section can be formed.

Solve. Use graph paper.

Question 20.
Points A, B, C, and D form a square. The area of the square is 9 square units.
a) Find the side length of square ABCD.
Answer:
Given that the area of the square is 9 square units.
Area of square = length × length
9 = length × length
3 × 3 = length × length
Length of the square will be 3 units.

b) The coordinates of point A are (2, 6). Points B and C are below \(\overline{A D}\). Point B is below point A, and point D is to the right of point A. Plot the points in a coordinate plane. Connect the points in order to draw square ABCD.
Answer:
The coordinates of point A are (2, 6).
Point B and C are below point A, so they will be (0,2) and C will be (0,8).
D will be (8,6)

c) The points E, F, G, and H also form a square that is the same size as square ABCD. Point E is 4 units to the right of point A, and 3 units up. Points F and G are below \(\overline{E H}\). Point F is below point E, and point H is to the right of point E. Plot the points in the coordinate plane. Draw \(\overline{E H}\) and \(\overline{G H}\) with solid lines, and \(\overline{E F}\) and \(\overline{F G}\) with dashed lines.
Answer:
Point E is 4 units to the right of point A, and 3 units up. So, it will be (6,9).
Points F and G are below, so they will be (6,3) and (12,3).
Point H will be (12,9).

d) Draw \(\overline{A E}\), \(\overline{D H}\), and \(\overline{C G}\) with solid lines, and \(\overline{B F}\) with a dashed line, Use the solid and dashed lines to see the figure as a solid. Name the type of prism formed.
Answer:
After joining AE, DH, CG and BF.Below prism will be formed.
Since all the sides are of same length, it will form a cube.

e) If the height of the prism is 7 units, find the volume of the prism.
Answer:
Area of the base 7×7 = 49 sq.units
Volume of the prism = 49×7 = 343 cu.units