Math in Focus

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Review Test Answer Key

Concepts and Skills

Solve.

Question 1.
Name the triangle congruent to ∆ABC.

Answer:
∆XYZ
Explanation:
Given,
∆ABC is congruent to ∆XYZ

Question 2.
ABCDEF is a regular hexagon. Name a quadrilateral congruent to ADCB.

Answer:
ADEF
Explanation:
Given,
ABCDEF is a regular hexagon,
ADCB is congruent to ADEF

Question 3.
\(\overline{S Q}\) and \(\overline{T P}\) are straight lines. Name the triangle similar to ∆PQR.

Answer:
∆TRS
Explanation:
Given,
∆PQR is congruent to ∆TRS

Solve. Show your work.

Question 4.
The two figures are congruent. Find the value of each variable,
a)

b)

Answer:
a) x = 8, y = 6, z = 104 degrees, b) x = 6, y = 4, z = 2
Explanation:
a) Since the triangles are congruent to each other, the longest side of the triangle measures 8 cm which is corresponding to the longest side of the second triangle which is x, therefore x = 8. While y is corresponding to the segment measuring 6 cm, therefore y = 6.
Solving z, the angle of 47 degrees corresponds to the other angle of the second triangle which is already having 29 degress, therefore the other triangle has 47 and 29 degress.
Sum of all the angles = 180 degrees
180 = 47 + 29 + z
180 = 76 + z
180 – 76 = z
104 = z
Therefore, x = 8, y = 6 and z = 104 degrees.
b)Since both the triangles are congruent to each other, the corresponding side of the second triangle is measuring 6 cm which is congruent to the first triangle of x, therefore x is measured as 6 cm. While y on the second triangle which is congruent to the first triangle which is measuring 4 cm, therefore y is measured as 4 cm. While z on the first triangle which is congruent to the first triangle’s measure of 2 cm, therefore, z is measured as 2 cm.
Therefore, x = 6, y = 4 and z = 2.

Question 5.
The two figures are similar. Find the value of each variable.
a)

b)

Answer:
a) x = 2.5 inches, y = 10 inches and z = 30 degrees, b) w = 1.6, x = 13.3 inches and y = 1.8 inches.
Explanation:
a) Since the two triangles are similar the ratio of the corresponding lengths are equal, the side of the bigger triangle corresponds to x of smaller triangle.The side measuring 8 inches in the bigger triangle corresponds to the side of 4 inches in the smaller triangle.
Ratio of corresponding lengths are equal
5/x = 8/4
cross multiplying,
5 x 4 = 8 x X
20 = 8x
20/8 = x
2.5 = x
The side which measures y in the bigger triangle corresponds to the measure of 5 inches in the smaller triangle and the side of 8 inches in the bigger triangle corresponds to the side of 4 inches in the smaller triangle.
Ratio of corresponding lengths are equal
y/5 = 8/4
cross multiplying,
y x 4 = 8 x 5
4y = 40
y = 40/4
y = 10
Since the sum of all the angle in the triangle = 180 degrees
z + 52 degrees + p = 180 degrees
The other angle p whose measure is unknown corresponds to the angle in the larger triangle whose measure is 98 degrees, two triangles are similar, thus their angles are equal. So p is 98 degrees then,
Sum of the angles in the triangle = 180 degrees
180 = z + 52 + p
Substituting p = 98
180 = z + 52 + 98
180 = z + 150
z = 180 – 150
z = 30
Therefore, x = 2.5, y = 10, z = 30
b) Since both the triangles are similar, the ratio of their corresponding lengths are equal. The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle.The side which measures 8 1/3 in the smaller triangle corresponds to the w in the bigger triangle.
Ratio of corresponding lengths are equal
6/10 = w/ 8 1/3
cross multiplying,
6 x 8 1/3 = w x 10
16 = 10 w
w = 16/10
w = 1.6
The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle. The side which measures x inches in the bigger triangle corresponds to the side which measures 8 inches in the smaller triangle.
Ratio of corresponding lengths are equal
6/10 = 8/x
cross multiplying,
6 x X = 8 x 10
6x = 80
x = 80/6
x = 13.33
The side which measures 10 inches in the bigger triangle corresponds to the side which measures 6 inches in the smaller triangle. The side which measures y inches in the bigger triangle corresponds to the side which measures 3 inches in the smaller triangle.
Ratio of corresponding lengths are equal
6/10 = y/3
cross multiplying,
6 x 3 = y x 10
18 = 10y
18/10 = y
y = 1.8
Therefore, x = 13.33, y = 1.8 and w = 1.6

Question 6.
State whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used.
a)

b)

Answer:

a) The triangles are not congruent
Explanation:
The side BC and DA are congruent, however, the sides are not corresponding.
The side BA corresponds to DA but lines are not congruent,
BC corresponds to DC but lines are not congruent, therefore triangles are not congruent.

b) ∠C from ∆TRS corresponds to ∠A from ∆BCA and the two angles are congruent.
∠A from ∆DAC corresponds to ∠C from ∆BCA and the two angles are congruent.
Side AC from ∆DAC is congruent to side CA from ∆BCA.
By the Angle – Side – Angle Congruence Theorem, triangles are congruent.
This theorem states that two triangles are congruent and the sides in between these angles are congruent and corresponds to each other, then the triangle is congruent.

Question 7.
State whether the triangles are similar. Explain with a test for similar triangles.
a)

b)

Answer:

a) Triangles are similar
Explanation:
Both the triangles are similar triangles by Angle-Angle postulate, that states that triangles are similar if two pairs of their corresponding angles are congruent. For example, 60 degree angle of first triangle is corresponding the 60 degree angle of second triangle and also 50 degree angle of first triangle is corresponding the 50 degree angle of second triangle, then the two triangles are similar.
Therefore, triangles are similar.

b) Triangles are not similar
Explanation:
If the triangles are similar triangles by Side-Angle-Side postulate, then their two corresponding sides are proportional and their included angle is congruent. The included angles of two corresponding sides are congruent. However, the corresponding are not proportional, then the side measuring 5 cm corresponds to side measuring 6 cm, while side measuring 8 cm corresponds to side measuring 9 cm.
5/6 not equal to 8/9
0.83 not equal to 0.88
Therefore, Triangles are not similar.

Question 8.
∆ABC undergoes two reflections to be mapped onto ∆A”B”C”.

a) Describe one such pair of reflections.
Answer:
Since there are two reflections showing that A(-4,4) has been reflected into the x-axis and was mapped on to A'(-4,-4). The image of the first reflection is reflected in to the y-axis and shows A”(4, -4)

b) Describe a single transformation that would also map ∆ABC onto ∆A”B”C”.
Answer:
180 degree rotation changes the coordinates from (x,y) to (-x,-y).
Then A(-4, 4) mapped to (4, -4) B(-4, 1) mapped to (4, -1) and C(-2, 1) mapped to C(2, -1). The coordinates of the image of ∆ABC under the rotation of 180 degrees is actually ∆A”B”C”. Therefore, 180 degrees rotation maps the original triangle to ∆A”B”C” in a single transformation.

Describe the single transformations that maps ∆ABC onto ∆A’B’C’, ∆ A’B’C’ onto ∆A”B”C”, and ∆ABC onto ∆A”B”C”.

Question 9.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(-5,4), B(-4,6) and C(-3,4). All the points move 3 units to the right. Therefore, transformation took place is translation.
ii) In ∆ABC mapped to ∆A”B”C”, the original points A(-5,4), B(-4,6) and C(-3,4) mapped to A”(-1, 2), B”(-0.5, 3) and C”(0, 2). ∆A”B”C” is smaller version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (2,0). Therefore, this transformation is dilation.
iii) For ∆A’B’C’ mapped to ∆A”B”C” original points A'(-2,4), B'(-1,6) and C(0,4) are mapped to A”(-1,2),B”(-0.5,3) and C”(0,2). ∆A”B”C” is a smaller version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.

Question 10.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(1,2), B(2,1) and C(3,1) are mapped to A'(3,6), B'(6,3) and C'(9,3).∆A’B’C’ is bigger version of ∆ABC If points A is connected on A”, B to B” and C on C” ” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.
ii) In ∆A’B’C’ mapped to ∆A”B”C”, the original points A'(3,6), B'(6,3) and C'(9,3) mapped to A”(2,4), B”(4,2) and C”(6, 2). ∆A”B”C” is larger version of ∆A’B’C’. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet at the origin. Therefore, this transformation is dilation.
iii) For ∆ABC mapped to ∆A”B”C” original points A(1,2),B(2,1) and C(3,1) are mapped to A”(2,4),B”(4,2) and C”(6,2). ∆A”B”C” is a larger version of ∆ABC. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet on (0,0). Therefore, this transformation is dilation.

Question 11.

Answer:
i) For ∆ABC mapped to ∆A’B’C’, that the original points A(-6,2), B(-6,1) and C(-4,1) are mapped to A'(2,2), B'(2,1) and C'(4,1). All the points move 8 units to the right. Therefore, transformation took place is translation.
ii) In ∆A’B’C’ mapped to ∆A”B”C”, the original points A'(2,2), B'(2,1) and C'(4,1) mapped to A”(-2,2), B”(-2,1) and C”(-4,1). ∆A”B”C” is larger version of ∆A’B’C’. If points A is connected on A”, B to B” and C on C” by the line, then the line should be extended until it makes an intersection, then this will meet at the origin. Therefore, this transformation is dilation.
iii) For ∆ABC mapped to ∆A”B”C” original points A(-6,2),B(-6,1) and C(-4,1) are mapped to A”(-2,2),B”(-2,1) and C”(-4,1). The transformation is a reflection at x = -4. Both C and C” lies at -4 because 0 unit away, then so it is its image. A and B is 2 units to the left of -4 so A” and B” is 2 units to the right of it.

Problem Solving

Solve. Show your work.

Question 12.
Two geometrically similar buckets have bases with radii in the ratio 2 : 5. The top of the small bucket has radius 7 inches.

a) Find the radius of the top of the big bucket.
Answer:
The radius of the top of the big bucket
Explanation:
Given,
2 : 5 = 2/5
Let the radius be p,
7/p = 2/5
cross multiplying, 7 x 5 = 2 x p
35 = 2p
p = 35/2
p = 17.5

b) Write the ratio of the height of the small bucket to the height of the big bucket.
Answer:
Height of the small bucket/ Height of the big bucket
2 : 5

Question 13.
A 5-foot tall vertical rod and a 25-foot tall vertical flagpole stand on horizontal ground. At a certain time of the day, the flagpole casts a shadow 30 feet long.

a) Write a ratio comparing the height of the rod to the height of the flagpole.
Answer:
Height of rod/ Height of the flagpole
5/25 = 1/5

b) Calculate the length of the shadow cast by the rod.
Answer:
Let the length of the shadow of the rod be a,
Height of the rod/Height of flapole = Shadow length of the rod/Shadow length of the flagpole
5/25 = a/30
Cross multiplying, 5 x 30 = a x 25
125 = 25a
a = 125/25
a = 6

Question 14.
Two barrels are geometrically similar. Their heights are 16 inches and 20 inches. The diameter of the base of the larger barrel is 8 inches. Calculate the diameter of the base of the smaller barrel.
Answer:
Let t be the diameter of the base of smaller barrel,
Height of small barrel/ Height of large barrel = Diameter of small barrel/Diameter of large barrel
16 / 20 = t / 8
Cross multiplying, 16 x 8 = t x 20
128 = 20 t
128/20 = t
6.4 = t
Therefore, Diameter of the small barrel = 6.4 inches.

Question 15.
A photograph is 4 inches wide and 6 inches long. It is enlarged to a width 6 inches.

a) What is the length of the enlarged photo?
Answer:
The length of the enlarged photo is 9 inches
Explanation:
Let the length of the enlarged photo is r,
Width of the original photo/ Length of original photo = Width of enlarged photo/ Length of enlarged photo
4/6 = 6/r
cross multiplying, 4r = 6 x 6
4r = 36
r = 36/4
r = 9

b) If the original photo is enlarged to a length 12 inches, find the new width.
Answer:
New width = 8 inches
Explanation:
Let new width of the photo is i,
Width of the original photo/ length of the original photo = new width of the photo/ length of the photo
4/6 = i/12
Cross multiplying, 4 x 12 = i x 6
48 = 6i
i = 48/6
i = 8
Therefore, new width of the photo = 8 inches.

Question 16.
Anthony made an exact copy of the shape ABCDE using a photocopier.

a) Name the shape congruent to ABCDE on the photocopy.
Answer:
GHIJF
Explanation:
Point A is congruent to point to G, Point B is congruent to point H, Point C is congruent to point I, Point D is congruent to point J, Point E is congruent to point F. Therefore, ABCDE is congruent to GHIJF.

b) Find the values for w, x, y, and z.
Answer:
w = 8, x = 5, y = 3, z = 6
Explanation:-
Solving x, IJ from GHIJF measuring x inches is congruent to CD of ABCDE which measures 5 inches.
IJ = CD
x = 5
Solving y, GF from GHIJF measuring 3y inches corresponds to AE from ABCDE which is measuring 9 inches.
GF = AE
3y = 9
divide 3 on both the sides
3y/3 = 9/3
y = 3
Solving z, HG from GHIJF measuring 4 inches corresponds to BA of ABCDE which is measuring (z – 2).
HG = BA
4 = (z – 2)
Adding 2 on both the sides
4 + 2 = z – 2 + 2
6 = z
Solving w, HI from GHIJH measuring 11 inches corresponds to BC of ABCDE which is measuring (w + y) inches.
HI = BC
11 = w + y
substituting y = 3,
11 = w + 3
Subtract 3 on both the sides,
11 – 3 = w + 3 – 3
8 = w.

Question 17.
A framed poster has a length of 28 inches and a perimeter of 88 inches. A similar poster has a perimeter of 66 inches. Find the area of the similar poster.
Answer:
Area of the similar poster = 252 sq inches
Explanation:
Let the length of the similar poster = p
length of similar poster/ length of framed poster = perimeter of similar poster/ perimeter of framed poster
substituting the values, p/28 = 66/88
cross multiplying, 88 x p = 66 x 28
88p = 1848
dividing 88 on both the sides, 88p/88 = 1848/88
p = 21
Therefore, length of the similar poster = 21 inches.
Perimeter of rectangle = 2 (l + w)
66 = 2 (21 + w)
66 = 42 + 2w
Subtracting 42 from both the sides,
66 – 42 = 42 + 2w – 42
24 = 2w
Divide 2 on both the sides,
24/2 = 2w/2
12 = w
Therefore, Width = 12 inches.
Area of rectangle = length x width
= 21 x 12
= 252 sq inches.

Question 18.
A robotic arm with two circular ends moves through a sequence of translations and rotations. The diagram shows two positions of the arm. All dimensions are in centimeters.

a) Find the values of x and y.
Answer:
x = 6 cm and y = 12 cm
Explanation:
The original image is exactly the same as transformed image, then the measure of the side of the original robotic arm is congruent to corresponding side of its image.
Solving for y:-
Side of original arm = Side of the image
substituting the expressions, y – 1 = 11
Adding 1 on both the sides, y – 1 + 1 = 11 + 1
y = 12
Therefore, y = 12 cm.
Solving for x:-
Other side of the original arm = other side of the image
Substituting the expressions, 2x = y
Substitute y = 12, 2x = 12
Divide 2 on both the sides, 2x/2 = 12/2
x = 6
Therefore, x = 6 cm

b) Calculate the area of the robotic arm. Use \(\frac{22}{7}\) as an approximation for π.
Answer:
Area of the robotic arm = 678.9 sq cm
Explanation:
Area of the arm = pi x diameter (diameter + length)
= 22/7 x 12 (12 + 6)
= 22/7 x 12 (18)
= 22/7 x 216
= 678.9 sq cm.

Question 19.
A table, which is in front of a mirror, has the side view ABCDEFGH shown below (all dimensions are in inches). ABCH and DEFG are rectangles. The side view is reflected in the mirror to form the congruent image STUVWXYZ.

a) State the corresponding angle of ∠CDE and ∠AHG.
Answer: ∠V corresponds to ∠D and ∠Z corresponds to ∠H

Explanation:
Since ABCDEFGH is congruent to STUVWXYZ, the points A,B,C,D,E,F,G and H corresponds to S,T,U,V,W,X,Y and Z respectively.
Therefore, ∠V corresponds to ∠D and ∠Z corresponds to ∠H.

b) Write the length of \(\overline{T U}\), \(\overline{W X}\), and \(\overline{Y Z}\).
Answer:
TU = 3 inches, WX = 16 inches and YZ = 4.5 inches
Explanation:
i) As TU is congruent to BC of table ABCDEFGH, since BC measures 3 inches then TU also has 3 inches.
ii) WX is congruent to EF of table ABCDEFGH, to know the value of EF subtract the two measures of CD, GH to AB,Let the measure of AB = 25 inches and CD and GH = 4.5 inches then,
EF = AB – (CD + GH)
= 25 – ( 4.5 + 4.5 )
= 25 – 9
= 16
Since, EF measures 16 inches therefore, WX also measures 16 inches.
iii) YZ is congruent to GH of table ABCDEFGH, Since GH measures 4.5 inches therefore, YZ also measures 4.5 inches.

c) Find the perimeter of the image.
Answer:
Perimeter of STUVWXYZ = 106 inches.
Explanation:
Perimeter of ABCDEFGH = sum of all the lengths
Substituting all the sides of the table
Perimeter = AB + BC + CD + DF + EF + FG + GH + HA
Substituting all the values
Perimeter = 25 + 3 + 4.5 + 25 + 16 + 25 + 4.5 + 3
= 106
Therefore, the perimeter of ABCDEFGH is 106 inches and ABCDEFGH is congruent to STUVWXYZ then perimeter of STUVWXYZ is 106 inches

Question 20.
Gisele reduced the size of ∆PQR with area 14.4 square centimeters using a photocopier. ∆PQR ~ ∆STU.

a) What is the scale factor of dilation? Express your answer as a percent.
Answer:
The scale factor of dilation is defined as the ratio of the size of the new image to the size of the old image.
Scale factor is 62.5%
Explanation:
PQ measuring 8 cm corresponds to side ST measuring 5 cm, to know the scale factor need to find the ratio of length of ST to length of PQ
Scale factor = length of ST/ length of PQ
= 5/8
= 0.625
Therefore, scale factor is 0.625
To convert it into percentage, multiply it with 100
Scale factor = 0.625 x 100
= 62.5%

b) Find the area of ∆STU.
Answer:
Area of ∆STU = 9 sq cm.
Explanation:
Two triangle are similar to each other,
Scale factor is 0.625,
Area of ∆PQR = 14.4 sq cm,
The area of ∆STU is given by the area of ∆PQR multiplied with scale factor.
Area of ∆STU = ∆PQR x scale factor
= 14.4 x 0.625
= 9

This handy Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 7-9 detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Cumulative Review Chapters 7-9 Answer Key

Concepts and Skills

Find the value of each variable. (Lesson 7.1)

Question 1.

Answer:
Pythagorean Theorem states that the sum of the squares of the legs is the square of hypotenuse of a right angle. Let x be the unknown length in the right angle so,
x² + 15² = 17²
x² + 225 = 289
x² = 289 – 225
x² = 64
x = 8 (Square root of 64)
Therefore, the leg is 8 inches.

Question 2.

Answer:
(3.6)² + (2.7)² = x²
12.96 + 7.29 = x²
20.25 = x²
4.5 = x (Square root of 20.25)
Therefore, x is 4.5 inches.

The side lengths of a triangle are given. Decide whether each triangle is a right triangle. (Lesson 7.1)

Question 3.
6 cm, 8 cm, 12 cm
Answer:
6² + 8² = 12²
36 + 64 = 144
100 = 144 (100 is not equal to 144)
Since the sum of the square of the legs is not equal to the square of hypotenuse, h=the triangle is not a right triangle.

Question 4.
9 in., 7.2 in., 5.4 in.
Answer:
(7.2)² + (5.4)² = (9)²
51.84 + 29.16 = 81
81 = 81
Therefore, the triangle is a right triangle.

Find the distance between each pair of points. If necessary, round your answer to the nearest tenth of a unit. (Lesson 7.2)

Question 5.
P(2, 5), Q(4, 13)
Answer:
d = √(x2 – x1)² + (y2 – y1)²
d = √(4 – 2)² + (13 – 5)²
d = √(2)² + (8)²
d = √4+64
d = √68
d = ±8.2

Question 6.
X(3, -1), Y(4, 2)
Answer:
d = √(x2 – x1)² + (y2 – y1)²
d = √(4 – 2)² + (2 – (-1))²
d = √(2)² + (3)²
d = √4+9
d = √13

For each solid, find the unknown dimension. If necessary, round your answer to the nearest tenth of a unit. (Lesson 7.3)

Question 7.

Answer:
Using pythagorean theorem
18² + w² = 30²
324 + w² = 900
w² = 900 – 324
w² = 536
w ~ 23.2 (Square root of 536)
Therefore, the height w of the cone measures 23.2 cm.

Question 8.

Answer:
Using pythagorean theorem,
z² + 7² = 21²
z² + 49 = 441
z² = 441 – 49
z² = 392
z = 19.8 (Square root of 392)
Therefore, the height is 19.8 inches.

Find the volume of each composite solid. Use 3.14 as an approximation for π. If necessary, round your answer to the nearest tenth. (Lesson 7.4)

Question 9.

Answer:
Finding the radius of the cone:
4² + n² = 5²
16 + n² = 25
n² = 25 – 16
n² = 9
n = 3 ( square root of 9 )
Therefore, the radius of the cone is 3 inches. Deducting 2 inches to radius of the cone determines the radius of the cylinder according to the diagram. Then the radius of the cylinder is 1 inch.
Finding volume of the cone:
For the volume of the cone, using 3.14 as the value of π and let h be the height,
Volume of the cone = 1/3x Area of the base x Height of the cone
= 1/3 πr² x h
= 1/3 x 3.14 x (3)² x h
= 37.68
Therefore, the cone has a volume of 37.68 cubic meters.
Finding the volume of cylinder:
For the volume of the cylinder, using 3.14 as the value of π and the height j measures 10 inches.
Volume of the cylinder = Area of the base x height of cylinder
= πr² x j
= 3.14 x (1)² x 10
= 31.4
Therefore, the cylinder has a volume of 31.4 cubic inches.
By adding two volumes will get the total volume of the figure,
Total volume = Volume of the cone + Volume of the cylinder
= 37.68 + 31.4
= 69.08
Therefore, the total volume is 69.08 cubic meters.

Question 10.

Answer:
Finding the sides of the cube:
g² + h² = 3²
g² + h² = 9
g² + g² = 9 (since g = h)
2 g² = 9
g² = 9/2
g² = 4.5
g ~ 2.1(Square root of 4.5)
Therefore, the sides of the cube measures 2.1 cm.
Finding the volume of the cube:
Volume of the cube = s³
= (2.1)³
~ 9.3
Therefore, the volume of the cube is 9.3 cubic cm.
Finding the volume of the sphere:
Volume of the sphere = 4/3 πr³
= 4/3 x 3.14 x (1.05)³
= 4/3 x 3.14 x 1.2
= 4.8
Volume of the hemisphere = Volume of the sphere/2
= 4.8/2
= 2.4
Therefore, the hemisphere has a volume of 2.4 cubic cm.

Find the coordinates of the image under each translation. (Lesson 8.1)

Question 11.
A (3, -2) is translated 1 unit to the right and 8 units up.
Answer: Locate the point (3, -2) in the coordinate plane then move it 1 unit to the right and 8 times upwards as described by the translation. Thus the image for A will be on points (4,6) on the coordinate plane. Therefore, A(3, -2) is mapped onto A'(4,6).

Question 12.
B(-1, -6) is translated 5 units to the left and 3 units down.
Answer: Locate the point (-1,-6) in the coordinate plane then move it 5 units to the left and 3 units downwards as stated by the translation. Thus the image for B will be on points (-6,-9) of the coordinate plane.

Copy each diagram on graph paper and draw the image under each translation. (Lesson 8.1)

Question 13.
\(\overline{X Y}\) is translated 3 units to the left and 1 unit up.

Answer:

Question 14.
Square ABCD is translated 2 units to the right and 2 units down.

Answer:

Solve. Show your work.

Question 15.
On a coordinate plane, the coordinates of two points are A (3, 4) and B (3, 0). Point A’ is the image of point A and point 8′ is the image of point 8 under a reflection in the y-axis. (Lesson 8.2)

a) Find the coordinates of A’and B’.
Answer:

b) Draw the image of the line segment OA, where O is the origin, under a reflection in the x-axis. Use 1 grid square on both axes to represent 1 unit for the interval from -4 to 4.
Answer:

Question 16.
Triangle ABC and trapezoid STUV are shown on the coordinate plane. On a copy of the graph, draw the images of triangle ABC under a reflection in the y-axis. Then draw the image of trapezoid STUV under a reflection in the x-axis. (Lesson 8.2)

Answer:

Question 17.
A rotation of point A about the origin maps the point onto A’. State the angle and direction of rotation. (Lesson 8.3)

a)

Answer: The point A is on (1,1). The rotation of 90 degree clockwise, the coordinate should have changed from (x,y) to (y, -x). Since A’ is on (1, -1) from the original point (1,1) then it follows the rule for the 90 degree rotation in the clockwise direction.

b)

Answer: Point A is on (-2,-2). The rotation of 90 degree counterclockwise, the coordinate shifts from (x,y) to (-y,x). Since A’ is on (2,-2) from the original point (-2,-2) then it follows the rule for the 90 degree rotation in the counterclockwise direction.

Solve. Show your work.

Question 18.
Triangle ABC has vertices A (2, 3), 8 (3, 1) and C (4, 2). Draw the triangle ABC and its image under a rotation of 180° about the origin. Use 1 grid square on both axes to represent 1 unit for the interval from -4 to 4. (Lesson 8.3)
Answer:

Tell whether each transformation is a dilation. Explain. (Lesson 8.4)

Solve. Show your work.

Question 19.

Answer: The center of dilation is point C(or C’) since it is the point that did not change evn after the transformation. Also note that a transformation in dilation has a scale factor. The triangle A’B’C’ is 1/3 size compared to the original triangle, so the scale factor is 1/3. Therefore, the transformation is dilation.

Question 20.

Answer: Creating lines to connect points of the original rectangle to their corresponding points in their image to pinpoint the intersection of all points, the line won’t ever intersect. Thus, the center of dilation does not exist. Therefore, the transformation is not a dilation.

Solve. Show your work.

Question 21.
The figure described is mapped onto its image by a dilation with its center at the origin, O. Draw each figure and its image on a coordinate plane. (Lesson 8.4)

a) The vertices of triangle DEF are D (-3, -1), E (-3, -4), and F (-1, -3). Triangle DEF is mapped onto triangle D’E’F’ with scale factor -2.
Answer:
Coordinates of D x scale factor
(-3,-1) x -2
(6,2)
Thus the coordinate of D’ is (6,2). Next E’ corresponds to E, then use the coordinates of E to determine the coordinates E’ after the dilation.
Coordinates of E x scale factor
(-3,-4) x -2
(6,8)
Thus, the coordinate of E’ is (6,8). Lastly, F’ corresponds to F so use the coordinates of F to determine the coordinates of F’ after the dilation.
Coordinates of F x scale factor
(-1,-3) x -2
(2,6)
Thus the F’ coordinate (6,6)

b) The vertices of trapezoid ABCD are A (-2, -2), B (-2, -4), C (4, -4) and D (2, -2). Trapezoid ABCD is mapped onto trapezoid A’B’C’D’ with scale factor \(\frac{1}{2}\).
Answer:
Coordinate of A x scale factor
(-2,-2) x 1/2
= (-1,-1)
The coordinate of A’ is (-1,-1)
Coordinates of B x scale factor
(-2,-4) x 1/2
= (-1,-2)
The coordinate of B’ is (-1,-2)
Coordinates of C x scale factor
(4,-4) x 1/2
= (2,-2)
The coordinate of C’ is (2,-2)
Coordinates of D x scale factor
(2,-2) x 1/2
= (1,-1)
The coordinate of D is (1,-1)

Question 22.
The side lengths of a triangle are 3 inches, 6 inches, and 8 inches. The triangle undergoes a dilation. Find the side lengths of the ¡mage of the triangle for each of the scale factors in a) to d). Tell whether each dilation is an enlargement or a reduction of the original triangle. (Lesson 8.4)
a) 2
b) \(\frac{1}{4}\)
c) 120%
d) 1.6
Answer:

a) Side length of 8 inches x scale factor
8 x 2
16
Side length of 6 inches x scale factor
6 x 2
12
Side length of 3 inches x scale factor
3 x 2
6
Therefore, the 3 inches side of the triangle grows into a 6 inches side.

b) At a scale of 1/4
Side length of 8 inches x scale factor
8 x 1/4
2
Side length of 6 inches x scale factor
6 x 1/4
1.5
Side length of 3 inches x scale factor
3 x 1/4
0.75
Therefore, the 3 inches side of the triangle shrinks into a 0.75 inch side.

c) The scale factor is in a percentage form, divide the give percent into 100 to convert it into decimal.
120/100 = 1.2
Thus, the scale factor is 1.3. So, at a scale of 1.2
Side length of 8 inches x scale factor
8 x 1.2
9.6
Therefore, the 8 inches side of the triangle grows into a 9.6 inches side.
Side length of 6 inches x scale factor
6 x 1/2
7.2
Therefore, the 6 inches side of the triangle grows into a 7.2 inches side.
3 x 1.2
3.6
Therefore, the 3 inches side of the triangle grows into a 3.6 inches side.

d) Scale factor is 1.6
Side length of 8 inches x scale factor
8 x 1.6
12.8
Therefore, the 8 inches side of the triangle grows into 12.8 side.
Side length of 6 inches x scale factor
6 x 1.6
9.6
Therefore, the 6 inches side of the triangle grows into 9.6 inches side
Side length of 3 inches x scale factor
3 x 1.6
4.8
Therefore, the 3 inches side of the triangle grows into a 4.8 inches side.

Question 23.
Triangle A with vertices (-1, 3), (-1, 4) and (-3, 3) is mapped onto triangle B. Then triangle B is mapped onto triangle C as shown on the coordinate plane. (Lesson 8.5)

a) Describe the transformation that maps triangle A onto triangle B.
Answer: The point (-3,3) of triangle A is mapped onto (4,4) by moving 1 unit up and 7 units to the right. For the points of triangle A, it also ascends once and moves 7 times to the right. This transformation that moves the original figure up or down or left or right is translation.

b) Describe the transformation that maps triangle B onto triangle C.
Answer: Points (4,4), (6,4) and (6,5) mapped onto (-4,4),(-4,6) and (-5,6) changes the coordinates from its original for (x,y) into (-y,x). The transformation that changes the form of the coordinates from (x,y) into (-y,x) is a 90-degree rotation at the origin in a counterclockwise direction.

Solve. Show your work.

Question 24.
State whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used. (Lesson 9.1)

Answer: AB corresponds to XY and both the measures 10 cm, Triangle A corresponds to Triangle Y while Triangle C corresponds to Triangle Z and this two corresponding angles are congruent by Angle Angle Side Test which states that if the two corresponding angles of two triangles are congruent and any of their side are also congruent, then the triangles are congruent

Question 25.
∆AMS is congruent to ∆ERN. Find the values of x and y. (Lesson 9.1)

Answer:
To angle A of Triangle AMS and angle E of Triangle ERN are equal.
Angle A = Angle E
The measure of angle A is equals to 112 degree while measure of angle E is 16x degree
112 = 16x
112/16 = x
7 = x
Therefore, x is equal to 7
SM of triangle ASM and NR of triangle ERN are equal.
SM = NR
Measure of SM equals to 41 inches while measure of NR is (3x+5y)inches.
41 = (3x+5y)
using x = 7,
41 = 3(7) + 5y
41 = 21 + 5y
5y = 41 – 21
5y = 20
y = 20/5
y = 4

Triangle ABC is similar to triangle POR. Find the value of a. (Lesson 9.2)

Question 26.

Answer:
AC measuring 3 cm corresponds to PQ whose measure is 4.5 cm and CB measuring 5.2 corresponds to QR which measure a. Since the triangles are similar then their corresponding sides are proportional thus,
AC/PQ = CB/QR
3/4.5 = 5.2/a
3a = 23.4
a = 23.4/3
a = 7.8
Therefore, a is equal to 7.8

Name the test you can use to determine whether the two triangles are similar. Then find the value of x. (Lesson 9.2)

Question 27.

Answer:
Triangle ROS and POQ are vertical angles, and pair of opposite angles are formed due to an intersection of line is equal. Observe that angle R is congruent to angle P. Third corresponding angle is also equal. As there are two pairs of corresponding angles in Triangle ROS and POQ, third corresponding angle is also equal. By the Angle-Angle-Angle similarity theorem, which states ate if all the corresponding angles of two triangles are similar, that two triangles are similar.
Similarity of the triangles are proven, to solve x, RS which measures 12 km corresponds to PQ measuring x km and RO whose length is 13 km corresponds to PO measuring 27.3 km
RS/PQ = RO/PO
12/x = 13/27.3
12 x 27.3 = 13x
327.6 = 13x
x = 327.6/13
x = 25.2
Therefore, x = 25.2 km

Solve on a coordinate grid. (Lesson 9.3)

Question 28.
∆ABC undergoes two transformations to form the image ∆A”B”C”.
a) What are the coordinates of ∆A”B”C” if ∆ABC is first translated 2 units to the right and 3 units down, and then reflected in the line x = 2?
Answer:

b) What are the coordinates of ∆A”B”C” if ∆ABC is first reflected in the line x = 2, and then translated 2 units to the right and 3 units down?
Answer:

c) Do the two triangles ∆A”B”C” described in a) and b) have the same coordinates? Are they congruent? Explain.
Answer:
The two traingles are congruent since the only transformation is a reflection and transformation which is known for flipping the original images while translation is known for moving image up and down and left to right. So the size of triangle is still the same and are just plotted at different locations. For their coordinate, the locations are transformed twice, both depends on the first trnasformations that the triangles undergo. Triangle A”B”C” are on Quadrant III while the second one is on Quadrant IV. Therefore, their coordinates are not congruent.

Problem Solving

Solve. Show your work.

Question 29.
A pool table has a width of 5.5 inches. A ball travels a distance of 12.3 inches diagonally from one corner to the other corner of the table. Find the length of the pool table, rounded to the nearest tenth. (Chapter 7)
Answer:
Using pythagorean theorem,
(5.5)² + j² = (12.3)²
30.25 + j² = 151.29
j² = 151.29 – 30.25
j² = 121.04
j ~ 11 (Square root of 121.04)
Therefore, the length of the pool table is 11 inches.

Question 30.
When a drawbridge is open, the bridge forms a right angle with the doorway it leads to. A straight chain connects the far end of the bridge to the top of the doorway. If the doorway is 2 meters high and the chain is 2.5 meters long, find the length of the bridge, rounded to the nearest tenth. (Chapter 7)
Answer:
Using the pythagorean theorem,
y² + z² = x²
(2.5)² + 2² = x²
6.25 + 4 = x²
10.25 = x²
x ~ 3.2 (square root of 10.25)
Therefore, the drawbridge is around 3.2 m in length

Question 31.
Gabriel photocopied an exact copy of the shape ABCDE. The photocopied shape is shown on the right. (Chapter 9)

a) Name the shape congruent to ABCDE.
Answer: Point A corresponds to G , B point to H and point C corresponds to I, D corresponds to J and E corresponds to F. Therefore, the congruent shape to ABCDE is GHIJF.

b) Find the values of w, x, and y.
Answer:
Since ABCDE is congruent to GHIJF, their corresponding angles are equal. Thus, side AB of ABCDE is equal to GH of GHIJF, since these two sides corresponds to each other. So,
AB = GH
The measure of the lengths of AB is 2 feet while GH is 4y feet.
2 = 4y
y = 2/4
y = 0.5
Therefore, y is equal to 0.5
Side EA of ABCDE is equal to FG of GHIJF
EA = FG
The measure of the lengths of EA is (y+2x) feet while FG is 3.5 feet
y + 2x = 3.5
0.5 + 2x = 3.5
2x = 3.5 – 0.5
2x = 3
x = 3/2
x = 1.5
Therefore, x is equal to 1.5
Side BC of ABCDE is equal to HI of GHIJF
BC = HI
The measure of lengths of BC is (x + w) feet while HI is 5.6 feet
x + w = 5.6
1.5 + w = 5.6 (As x = 1.5)
w = 5.6 – 1.5
w = 4.1
Therefore, w is equal to 4.1

Solve. Show your work. (Chapters 7, 9)

Question 32.
Mary wants to pour 130 cm³ of water into the conical cup shown.

a) Find the volume of the conical cup. Use 3.14 as an approximation for π. Round your answer to the nearest tenth.
Answer:

Using pythagorean theorem,
(4.5)² + m² = (7.5)²
20.25 + m² = 56.25
m² = 56.25 – 20.25
m² = 36
m = 6
Therefore, the height of the cone is 6 cm
For the volume of the cone, 3.14 is the value of π and m is the height,
Volume of the cone = 1/3 Area of the base x Height of the cone
= 1/3πr² x m
= 1/3 x (3.14) x (4.5)² x (6)
= 127.17
Therefore, cone has a volume of 127.17 cubic cm.

b) Will the water overflow? Explain.
Answer: Since Mary wants to pour 127.17 cubic centimeters, the water will spill. The conical cup can only accommodate up to 127.17 cubic centimeters of liquid.

Question 33.
Two cones are similar in shape. The ratio of the diameters of their bases is 2 : 7. The radius of the smaller cone is 4.5 inches.

a) Find the radius of the larger cone.
Answer:
Convert the ratio 2:7 into fractions, then the result will be 2/7. Let s be the unknown measure of the radius, so
2/7 = 4.5/s
2 x s = 4.5 x 7
2s = 31.5
s = 31.5/2
s = 15.75
Therefore, the larger cone has a radius measuring 15.75 inches.

b) Write the ratio of the height of the smaller cone to the height of the larger cone.
Answer: Height of the smaller cone / Height of the larger cone

Solve. Show your work. (Chapters 8, 9)

Question 34.
A car is 25 feet long and 6 feet wide. A model of the same car is 9 inches long. What is the width of the model of the car if the model is a dilation of the car?
Answer:
Scale factor = Length of the model car/ Length of the actual car
= 0.75/25
= 0.03
Thus, the scale factor is 0.03
To solve width, multiplying the actual width to the scale factor,
Width of the model car = Width of the car x scale factor
= 6 x 0.03
= 0.18
Therefore, the width of the model car is 0.18 feet,
Converting this into inches, multiplying 0.18 with 12
Thus, the model car has a width of 2.16 inches.

Question 35.
Figure PQRST is dilated with center P and a scale factor of 1.45. Under this dilation, it is mapped onto figure UVWXY. Figure UVWXYis then mapped onto ABCDE by a reflection about segment UV.

a) Tell whether PQRST and ABCDE are congruent or similar figures. Explain.
Answer: The reflection that maps UVWXYZ onto ABCDE does not change the size and shape of the figure. Thus, they are congruent. The dilation that maps PQRST onto UVWXYZ however enlarges the size of PQRST. Therefore, UVWXYZ and PQRST are similar.

b) Find m∠UYX.
Answer: UVWXYZ and PQRST are congruent, then their corresponding angles are equal. The angle P corresponds to angle U, Q to angle V and angle R corresponds to W. The corresponding angle of S is angle X and angle T corresponds to Y . Since angle T corresponds to angle Y, then T is equal to Y. Angle T is equal to 120 degree. Therefore, angle Y is equal to 120 degree.

c) The area of PQRST is 20 square inches. Find the area of ABCDE.
Answer: The area of ABCDE is also a dilated image of PQRST. So, by multiplying the area of the actual image which is 20 square inches to the scale factor, the area of ABCDE
Area of ABCED = Area of PQRST x Scale factor
= 20 x 1.45
= 29
Therefore, ABCDE has an area of 29 square inches.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.3 Relating Congruent and Similar Figures to Geometric Transformations detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.3 Answer Key Relating Congruent and Similar Figures to Geometric Transformations

Math in Focus Grade 8 Chapter 9 Lesson 9.3 Guided Practice Answer Key

Solve.

Question 1.
△ABC and △A’B’C’ are congruent isosceles triangles. Describe a rotation that maps △ABC onto △A’B’C’.

Answer:

Complete each with a value or word and each with +, —, x, or ÷.

Question 2.
△JKL is dilated by a scale factor of 1.2 to form △J’K’L’.

a) Find m∠J’.
△JKL and △J’K’L’ are triangles.
So, m∠J’ = m∠
– 50° – 30° = ∠ sum of triangle
m∠J’ is
Answer:
△JKL and △J’K’L’ are similar triangles.
So, m∠J’ = m∠J
180° – 50° – 30° = 100° ∠ sum of triangle
m∠J’ is 100°

b) Find the length of \(\overline{K^{\prime} L^{\prime}}\).
K’L’ = 2.5 1.2
= cm
Answer:
K’L’ = 2.5 × 1.2
We have to multiply the length with the scale factor to find the scaled length of \(\overline{K^{\prime} L^{\prime}}\).
= 3 cm

Question 3.
△UVW undergoes a geometric transformation to form △XYZ.

a) Identify whether △UVW ~ △XYZ.
m∠V 180° – – ∠ sum of triangle
=
Since m∠W = m∠Z and m∠V = m∠, two pairs of corresponding angles have equal measures. So, △UVW is to △XYZ.
Answer:
m∠V 180° – 110° – 40° ∠ sum of triangle
= 30°
Since m∠W = m∠Z and m∠V = m∠Y, two pairs of corresponding angles have equal measures. So, △UVW is ∼ to △XYZ.

b) If △UVW ~ △XYZ, what is the geometric transformation and scale factor?
Scale factor = \(\frac{Y Z}{?}\)
= \(\frac{6}{?}\)
=
△UVW undergoes a by a scale factor to form △XYZ.
Answer:
Scale factor = \(\frac{Y Z}{WV}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\)
△UVW undergoes a dilation by a scale factor \(\frac{1}{2}\) to form △XYZ.

Copy and complete.

Question 4.
An engineer designs a manufacturing machine to transfer a food package GHIJ onto G’H’I’J’. Describe a sequence of two transformations that maps GHIJ onto G’H’l’J’.
GHIJ is mapped onto G’H’l’J’ by a translation of 2 units up and a reflection in the line .
Think Math
Does the order in which you do the transformations affect the image in Example 11 ? in 4?

Answer:
GHIJ is mapped onto G’H’l’J’ by a translation of 2 units up and a reflection in the line 2.

Technology Activity

Materials
geometry software

EXPLORE SEQUENCES OF TRANSFORMATIONS

STEP 1: Draw a triangle on a coordinate system using a geometry software. Record the position of each vertex.

STEP 2: Translate the triangle 1 unit down. Then reflect it in the x-axis. Select the Translate function and Reflect function, within the Transform menus. Record the position of each vertex of the image triangle.

STEP 3: Change the order of transformations in STEP 2. Record the position of each vertex of this image triangle.

STEP 4: Draw another triangle on a coordinate system using a geometry software. Record the position of each vertex.

STEP 5: Translate the triangle 1 unit to the right. Then dilate it with the center at the origin and scale factor 2. Select the Translate function and Dilate function, within the Transform menus. Record the position of each vertex of the image triangle.

STEP 6: Change the order of transformations in STEP 5. Record the position of each vertex of this image triangle.

Math journal Compare the positions of the original and transformed triangles. Does the order in which you perform two transformations make a difference in the position of the image triangle? Explain.
Answer:

Copy and complete on a graph paper.

Question 5.
DEFG is mapped onto D’E’F’G’ under a transformation. D”E”F”G” is the image of D’E’F’G’ under another transformation.

a) Describe the transformations that map DEFG onto D’E’F’G’ and D’E’F’G’ onto D”E”F”G”. Then describe a single transformation that maps DEFG onto D”E”F”G”.
DEFG is mapped onto D’E’F’G’ by using a reflection in the . D’E’F’G’ is mapped onto D”E”F”G” by using a rotation of about the origin. DEFG is mapped onto D”E”F”G” by using a single transformation, which is .
Answer:
DEFG is mapped onto D’E’F’G’ by using a reflection in the y-axis. D’E’F’G’ is mapped onto D”E”F”G” by using a rotation of 180 degrees about the origin. DEFG is mapped onto D”E”F”G” by using a single transformation, which is reflection in the x-axis.

b) Suppose the order of the transformations is reversed. Draw DEFG, D’E’F’G’, and D”E”F”G” on a coordinate plane. Does the order of the transformations affect the position of D”E”F”G”?
Answer: No, the order of the transformations affects the position of D”E”F”G” if you change the position in the y-axis.

Question 6.
△PQR is mapped onto △P’Q’R’ under a transformation. △P”Q”R” is the image of △P’Q’R’ under another transformation.
a) Describe the transformation that maps △PQR onto △P’Q’R’ and △P’Q’R’ onto △P”Q”R”.

Answer: △PQR is mapped on to △P’Q’R’ by the reflection in the line y = -1 and △P’Q’R’ is mapped onto △P”Q”R” on a coordinate plane by using the anticlockwise rotation of 90 degree about (-3, -3)

b) Suppose the order of the transformations is reversed. Draw △PQR, △P’Q’R’, and △P”Q”R” on a coordinate plane. Does the order of the transformations affect the position of △P”Q”R”?
Answer:

Question 7.
△PQR is mapped onto △P’Q’R’ under a transformation. △P”Q”R” is the image of △P’Q’R’ under another transformation.

a) Describe the transformations that map △PQR onto △P’Q’R’ and △P’Q’R’ onto △P”Q”R”. Then describe a single transformation that maps △PQR onto △P”Q”R”.
△PQR is mapped onto △P’Q’R’ by using a dilation with center (, ) and scale factor . △P’Q’R’ is mapped onto △P”Q” R” by using a rotation of about the point (, ) . △PQR can be mapped onto △P”Q”R” by a single dilation with center (imgg 2, ) and scale factor .
Answer:
△PQR is mapped onto △P’Q’R’ by using a dilation with center (0, 1) and scale factor 1. △P’Q’R’ is mapped onto △P”Q” R” by using a rotation of 180 degrees about the point (0, 1) . △PQR can be mapped onto △P”Q”R” by a single dilation with center (imgg 2, 1) and scale factor 1.

b) If the order of transformations is reversed, draw △PQR, △P’Q’R’, and △P”Q”R” on a coordinate plane.
Answer:

Math in Focus Course 3B Practice 9.3 Answer Key

State whether the figure and image are congruent or similar.

Question 1.
A triangle is rotated 180° about the origin.
Answer: If a triangle is rotated 180° about the origin then the figure is congruent.

Question 2.
A pentagon is translated 1 unit to the left and 5 units up.
Answer: Similar figure

Question 3.
A projector dilates a picture by a scale factor of 10, and projects the image on a screen.
Answer: Similar figure

Question 4.
A parallelogram is dilated with center (-2, 4) and scale factor 3.5, and rotated 90° clockwise.
Answer: Similar figure

Question 5.
A cartoon character is reflected in the y-axis and translated to the right.
Answer: Congruent figure

Solve on a coordinate grid.

Question 6.
△ABC undergoes two transformations.

a) What would be the coordinates of △A”B”C” if △ABC is first translated 2 units up and 3 units left, and then reflected in the line x = 1?
Answer:

b) What would be the coordinates of △A”B”C” if △ABC is first reflected in the line x = 1, and then translated 2 units up and 3 units left?
Answer:

c) Do the two triangles △A”B”C” have the same coordinates? Are they congruent? Explain.
Answer:

Solve.

Question 7.
△PQR is mapped onto triangle △P’Q’R’ by using a transformation. △P”Q”R” is the image of △P’Q’R’ by using another transformation. Describe the sequence of transformations from △PQR to △P”Q”R”.
a)

Answer: The above figure has dilation with center (0, 1) and scale factor of 0.5 and also we can see the reflection in the y-axis.

b)

Answer: From the above figure we can see the translation of 1 unit up and then a rotation of 90 degrees counterclockwise about the origin.

c)

Answer: P’Q’R” is the reflection in the x-axis and P”Q”R” is the reflection in the y-axis.

Question 8.
△ABC is mapped onto △A’B’C’, which is then mapped onto △A”B”C”. △ABC and △A”B”C” are shown in each diagram. Describe the sequence of transformations from △ABC to △A”B”C”. Then describe a single transformation from △ABC to △A”B”C”, if any.
a) △ABC is mapped onto △A’B’C’ by a rotation of 180° about (0, 0).

Answer:

b) △ABC is mapped onto △A’B’C’ by a reflection in the line y = 1.

Answer:

Solve. Show your work.

Question 9.
Tom is at position A of a hall whose floor is a square with a bay of window across the front of the room. The length of the square is 30 meters. Describe how he gets to position C by a translation followed by a rotation about the center of the square.

Answer:
Given,
Tom is at position A of a hall whose floor is a square with a bay of window across the front of the room.
The length of the square is 30 meters.
A translation of 30 meters forward followed by a rotation of 90 degrees clockwise direction about the center of the square.

Question 10.
The figure PQRST is dilated with center P and a scale factor 1.2. PQRST is mapped onto UVWXY. The area of PQRST is 24 square inches.

a) Find m∠WVX.
Answer: 75 degrees

b) Find the area of UVWXY.
Answer:
The area of PQRST is 24 square inches.
scale factor = 1.2
24 × 1.2 = 28.8 square inches.

Question 11.
Jack walks into a dark room. The area of the pupils of his eyes dilates to 3 times their normal area to allow more light into his eyes. By what scale factor is the diameter of his pupils enlarged?
Answer:
Given,
Jack walks into a dark room.
The area of the pupils of his eyes dilates to 3 times their normal area to allow more light into his eyes.
Diameter = √3 or 1.7321

Question 12.
Two similar cups are filled with water. The volume of water in the big cup is 80 cubic centimeters and the volume of water in the small cup is 10 cubic centimeters. If the height of the big cup is 8 centimeters, what is the height of the small cup?
Answer:
Given,
The volume of water in the big cup is 80 cubic centimeters and
the volume of water in the small cup is 10 cubic centimeters.
height of the big cup is 8 centimeters
Let the height of the small cup be x.
80/10 = 8/x
8 = 8/x
x = 8/8
x = 1
Therefore the height of the small cup is 1 cm.

Question 13.
Edward was blowing a circular bubble whose volume grew to 27 times its original size. It then drifted 8 units to the right and 11 units up before it burst.
a) Describe the sequence of transformations that the bubble went through.
Answer:
Given,
Edward was blowing a circular bubble whose volume grew to 27 times its original size.
It then drifted 8 units to the right and 11 units up before it burst.
V = 27 = 3³
So, the scale factor is 3
The sequence of transformations that the bubble went through is a dilation with the scale factor 3 and then a translation of 8 units to the right and 11 units up.

b) The original radius of the bubble was 2 centimeters. Find the final radius of the bubble.
Answer:
Given,
The original radius of the bubble was 2 centimeters.
Scale factor = 3
We know that
scale factor = scale/actual
3 = x/2
3 × 2 = x
x = 6 centimeters
Thus the final radius of the bubble is 6 centimeters.

Question 14.
An airplane at an airport terminal is cleared to take off by the control station after it moves from P to Q. The airplane’s path is marked by the thick arrows in the diagram. Describe the sequence of transformations that the airplane undergoes from P to Q.

Answer:

Brain @ Work

Question 1.
A bridge is strung between two big trees on opposite sides of a river at P and Q. The bridge is to be removed and a new bridge is to be built across the river. The new bridge starts at P and spans the shortest distance across the river. Briefly describe how you would find the length of the new bridge using congruent triangles.

Answer:
Given data,
A bridge is strung between two big trees on opposite sides of a river at P and Q.
The bridge is to be removed and a new bridge is to be built across the river.
The new bridge starts at P and spans the shortest distance across the river.

Draw a line from Q to R so that P, Q, R lie on the same straight line. Measure the distance TR such that it is perpendicular to the river bank. And TR is the length of new bridge using congruent triangles.

Question 2.
A souvenir in the shape of a rectangular prism is 20 centimeters by 15.5 centimeters by 13 centimeters. A pattern for a rectangular box, 29.2 centimeters by 18.6 centimeters by 16.9 centimeters, needs to be scaled down to fit the souvenir better. What is the minimum size of the box? Recommend a suitable set of measurements to the nearest 0.1 centimeter.

Answer:

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.2 Understanding and Applying Similar Figures detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.2 Answer Key Understanding and Applying Similar Figures

Math in Focus Grade 8 Chapter 9 Lesson 9.2 Guided Practice Answer Key

Identify the figures that seem similar. Explain why.

Question 1.

Answer: A and E are the similar triangle because all the sides are equal. The shape of A and E are the same but the size is different.

Hands-On Activity

Materials

• ruler
• scissors

EXPLORE ANGLE MEASURES IN SIMILAR TRIANGLES

STEP 1: Draw an acute-angled triangle. Name the angle at each vertex A, B, and C as shown below. Draw a line segment parallel to \(\overline{B C}\). Name the line segment \(\overline{D E}\).

STEP 2: Cut △ABC along its edges. Then cut the triangle along \(\overline{D E}\). You should have one small triangle, △ADE, and a trapezoid, BDEC.

STEP 3: Place ∠B on top of the angles in A ADE, one by one. Which angle has the same measure as ∠B?

STEP 4: Place ∠C on top of the angles of △ADE, one by one. Which angle has the same measure as ∠C?

STEP 5: In the diagram in STEP 1, \(\overline{D E}\) and \(\overline{B C}\) are parallel and intersected by \(\overline{A B}\) and \(\overline{A C}\). What do you know about angle measures formed when a line intersects two parallel lines? How does this fact support what you saw in STEP 3 and STEP 4?

Math Journal Triangles ADE and ABC are similar because they have the same shape but not the same size. In these triangles, △ADE and AABC are called corresponding angles. Two other pairs of corresponding angles are angles AED and ACB and angles DAE and BAC. What did you observe about the measures of each pair of corresponding angles?
Answer: ∠B and ∠D are the corresponding angles.
∠E and ∠C are the corresponding angles.

Solve.

Question 2.
Triangle ABC and triangle DBF are similar triangles. Find the value of x.

Answer:

Question 3.
An engineer wants to make a bridge across a river at \(\overline{P R}\). The diagram shows the known measurements. \(\overline{S P}\) and \(\overline{T R}\) are straight lines and triangle RPQ is similar to triangle TSQ. Find the width x of the river.

The width of the river is feet.
Answer:

The width of the river is 32 feet.

Question 4.
△ABC and △DEF are similar triangles. Find m∠D, m∠E, and m∠F.

The corresponding angles of similar triangles have measures.
So m∠D = m∠ = ,
m∠E = m∠ = ,
and m∠F = 180° – – ∠ sum of triangle
= .
Answer:
The corresponding angles of similar triangles have same or equal measures.
So m∠D = m∠A = 62°,
m∠E = m∠B = 51°,
and m∠F = 180° – 62° – 51° ∠ sum of triangle
m∠F = 67°

Question 5.
The area of △AXY is 12 square centimeters. Find the area of △ABC.
Use the ratio of corresponding lengths to find the ratio of the areas.

Let the area of △ABC be x square centimeters.
Use the ratio of the areas to find x.
\(\frac{x}{12}\) = k² The ratio of the areas equals k².
\(\frac{x}{12}\) = \(\frac{?}{?}\) Substitute for k²,
\(\frac{x}{12}\) • 12 = \(\frac{?}{?}\) • Multiply both sides by .
x = Simplify.
The area of △ABC is square centimeters.
Answer:

\(\frac{x}{12}\) = k² The ratio of the areas equals k².
\(\frac{x}{12}\) = \(\frac{81}{16}\) Substitute 81/16 for k²,
\(\frac{x}{12}\) • 12 = \(\frac{81}{16}\) • 12 Multiply both sides by 12.
x = 60.75 Simplify.
The area of △ABC is 60.75 square centimeters.

Hands-On Activity

Materials:

• two similar triangles
• protractor
• ruler

EXPLORE A MINIMUM CONDITION FOR TWO SIMILAR TRIANGLES

STEP 1: Measure the side lengths of the triangles. Find out if △ABC is similar to △PQR by finding the ratios of the corresponding side lengths.

STEP 2: Measure ∠A, ∠B, ∠P, and ∠Q. Complete the following:
m∠A = m∠ and m∠B = m∠.

Math Journal Without measuring ∠C and ∠R, do you know whether they have the same measure? Explain. State a minimum condition for two triangles to be similar.
Answer:
m∠A = m∠p and m∠B = m∠Q.

Identify whether △ABC is similar to △DEF. Explain with a test for similar triangles.

Question 6.

All three pairs of corresponding side lengths have the ratio.
So, △ABC △DEF.
Answer:

All three pairs of corresponding side lengths have the same ratio.
So, △ABC ≈ △DEF.

Question 7.

Two pairs of corresponding side lengths have ratios.
So, △ABC △DEF.
Answer:

Two pairs of corresponding side lengths have different ratios.
So, △ABC are not similar to △DEF.

Question 8.

m∠BAC ≠ m∠
Two pairs of corresponding side lengths have the ratio and the included angles have measures.
So, △ABC △DEF.
Answer:

m∠BAC ≠ m∠DEF
Two pairs of corresponding side lengths have the different ratio and the included angles have different measures.
So, △ABC is not similar to △DEF.

Math in Focus Course 3B Practice 9.2 Answer Key

Identify the figures that seem similar. Explain why.

Question 1.

Answer:
A and D are similar figures with the same shape.

Question 2.

Answer: B, C and D are similar figures with a similar shape.

Triangle ABC is similar to triangle POR. Find the scale factor by which △ABC is enlarged to △PQR.

Question 3.

Answer:
Scale factor = scale/actual length
Scale factor = 4.5/3 = 1.5

Question 4.

Answer:
Scale factor = scale/actual length
Scale factor = 12/4 = 3

Question 5.

Answer:
Scale factor = scale/actual length
Scale factor = 20/15 = 4/3 = 1 1/3

Each pair of figures are similar. Find the value of each variable.

Question 6.
ABCD ~ EFGH

Answer:
By seeing the above figures we can say that it is a parallelogram.
We know that the opposite sides of the parallelogram are equal.
AB = 3EF
AD = 3EH
AB = 6 in.
EF = 3 × 6 in = 18 in.
AD = 9 in.
EF = 3x = 3 × 9 = 27 in.
Thus the value of x is 6 in.

Question 7.
△ABC ~ △DEF

Answer:
As the given triangles are similar the angles will be same.
∠B = 64°
∠C = 78°
Sum of angles = 180°
∠x° + 64° + 78° = 180°
∠x° = 180° – 64° – 78°
x° = 38°
x° = y° = 38°

Question 8.
ABCD ~ EFGH

Answer:
90/6 = 10/x
9/6 = 1/x
x × 9 = 6
x = 6/9
x = 2/3

Question 9.
△ABC ~ △DEF

Answer:
AB = ED
BC = EF
AC = DF
BC/EF = 4.5/3 = 1.5
Scale factor = 1.5
AD = 5.2 cm
AB = 5.2 × 1.5 = 7.8
z° = 35°

Explain, with a test, why the two triangles in each figure are similar. Find the unknown lengths.

Question 10.

Answer:
By using the SSS test we can say that the two triangles are similar because the shape of two triangles are same.
AB = DE
AC = CE
BC = CD
AB = 2 cm
DE = 4 cm
scale factor = 4/2 = 2
BC = 1 cm
CD = 1 × 2 = 2 cm
CD = y = 2 cm
CE = 3 cm
CE = 3/2 = 1.5 cm
x = 1.5 cm

Question 11.

Answer:
AB = AE
2 + 1 = 3 m
So, AE = x = 3 m
AD + DE = BC
2 + 4 = 6m
Thus x = 3 and y = 6

Solve. Show your work.

Question 12.
Mia made some copies of a drawing using a photocopier. The drawing was either enlarged or reduced. Find the value of x.
a)

Answer:
7.5/6 = 1.25
3 × 1.25 = 3.75 in.
Thus the value of x is 3.75 in.

b)

Answer:
3/2.25 = 1.3 in
x = 2.4 × 1.3 = 3.12 in
Thus the value of x is 3.12 in.

Question 13.
The slope of a wheelchair ramp is \(\frac{1}{15}\).

a) Suppose a wheelchair ramp has to rise 3 feet. Find the horizontal distance it covers.
Answer:
3/x = 1/15
3 × 15 = x
x = 45 feet
Thus the horizontal distance it covers is 45 feet.

b) Suppose there is space for a wheelchair ramp to cover at most 30 feet horizontally. How high can it rise then?
Answer:
x/30 = 1/15
x = 30/15
x = 2
Thus it rise 2 feet.

Question 14.
A circle has 9 times the area of another circle. If the radius of the larger circle is 27 meters, find the radius of the smaller circle.
Answer:
Given,
A circle has 9 times the area of another circle.
Radius of the larger circle is 27 meters.
Area of circle (A1) = 9 Area of circle2(A2)
A1 = 9A2
We know that,
Area of the circle = πr²
where r is the radius of the circle
r1 = 27 meters
r2 = ?
A1 = πr²
A1 = 3.14 × r²
A1 = 9A2
π(27)² = 9πr²
r² = (27)² /9
r = 27/3
r2 = 9 meters
Thus the radius of the smaller circle = 9 meters

Question 15.
The two ellipses shown on the right are similar. The area of the inner ellipse is 18 square inches and the shaded area is 32 square inches. Find the values for x and y.

Answer:
Given,
The area of the inner ellipse is 18 square inches.
a = 12/2 = 6 in.
18 = 6 × b
18/6 = b
b = 3
x = 3 + 3
x = 6 in.

Question 16.
Two trees on a street have heights 10 feet and 28 feet. At a certain time of a day, the shorter tree casts a shadow of length 15 feet on the ground. How far apart are the trees?

Answer:
Given,
Two trees on a street have heights 10 feet and 28 feet.
At a certain time of the day, the shorter tree casts a shadow of length 15 feet on the ground.
Let x be the distance between the two trees.
x + 15
10 : 28 = x : x + 15
10/28 = x/x + 15
10(x + 15) = x × 28
10x + 150 = 28x
150 = 28x – 10
18x = 150
x = 150/18
x = 8.3
x + 15 = 8.3 + 15 = 23.3 feet

Question 17.
A bag in the shape of two regular pentagons is shown on the right. The ratio of the shaded area to the area of the larger pentagon is 27 : 36. If the larger pentagon has sides of length 4 inches, find the length of the sides of the smaller pentagon.

Answer:
Given,
A bag in the shape of two regular pentagons is shown on the right.
The ratio of the shaded area to the area of the larger pentagon is 27 : 36.
Larger pentagon has sides of length 4 inches
Let the length of the sides of the smaller pentagon be x.
27/36 = x/4
3/4 = x/4
(3 × 4)/4 = x
x = 12/4
x = 3
Thus the length of the sides of the smaller pentagon is 3 inches.

Question 18.
A cable car travels from a village up to a resort on the top of a mountain. When the cable car has traveled 150 feet along the cable, it is 100 feet above the ground. The total distance the cable car must travel to reach the resort is 12,000 feet. How high is the mountain?

Answer:
Given that,
A cable car travels from a village up to a resort on the top of a mountain. When the cable car has traveled 150 feet along the cable, it is 100 feet above the ground.
The total distance the cable car must travel to reach the resort is 12,000 feet.
12000 ÷ 150 = 80
100 × 80 = 8000 feet
Thus the height of the mountain is 8000 feet.

Question 19.
Math journal Are similar figures ever congruent? If so, under which conditions are they congruent? Use two congruent triangles and two similar triangles in your explanation.
Answer:
Similar figures are not congruent until the scale factor is 1 or -1. The two triangles are said to be similar when all the three pairs of the figure are corresponding sides have the same ratio. If the ratio of the scale factor is 1 or -1 then the corresponding lengths are the same. When all the three pairs of corresponding lengths are the same then the triangle is said to be congruent.

Question 20.
Math journal Do you know all circles are similar? What other geometric shapes are similar?
Answer:
All circles are said to be similar because the radii are equidistant from their center.
Two shapes are said to be similar if and if it has the same shape but not the same size.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.1 Understanding and Applying Congruent Figures detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.1 Answer Key Understanding and Applying Congruent Figures

Math in Focus Grade 8 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Identify the figures that seem congruent. Explain why they seem so.

Question 1.

Answer: Figures A and F seems congruent because they have the same shape and same size.

Question 2.

Answer: C and D seems congruent because they have the same shape and same size.

Complete.

Question 3.
The quadrilaterals below are congruent. Write the statement of congruence.

So, the statement of congruence is PQRS ≅ ____________.
Answer:
The corresponding angles from the above figure is
∠P = ∠U
∠S = ∠T
∠R = ∠W
∠S = ∠V
So, the statement of congruence is PQRS ≅ UTWV.

Question 4.
To make two congruent quadrilaterals, Wendy cut two pieces of cardboard, one on top of the other. She measured the lengths of some sides of the quadrilaterals.
a) Find the values of the variables x, y, and z. All lengths are in inches.

Solve for x:
x = LM =
Solve for y.
2y – 5 = IJ =
2y – 5 + 5 = Add 5 to both sides.
2y = Simplify.
2y – y = Subtract y from both sides.
y = Simplify.

Solve for z:
x – 6 = JG =
– 6 = Substitute x = .
= z Simplify.
So, the value of x is , y is , and z is .
Answer:

Solve for x:
x = 12 LM = 12
Solve for y.
2y – 5 = y + 2 IJ = NK
2y – 5 + 5 = y + 2 + 5 Add 5 to both sides.
2y = y + 7 Simplify.
2y – y = 7 Subtract y from both sides.
y = 7 Simplify.

b) What is the length of each side of GHIJ? of NKLM?
GH = in., HI = in., IJ = in., JG = in.
NK = in., KL = in., LM = in., MN = in.
Answer:
GH = 12 in., HI = 10 in., IJ = 7 in.,
JG = in.
x – 6 = z
12 – 6 = z
z = 6
NK = 7 in., KL = 6 in., LM = 12 in.,
MN = 10 in.
2z – 2 = 2 (6) – 2 = 12 – 2 = 10

Technology Activity

Materials:
geometry software

OBSERVE THE CONGRUENCE IN TRIANGLES

STEP 1: Draw △ABC and △DEFso that m∠CAB m∠FDE, AB = DE, and AC = DF. Draw the two triangles such that point A is 10 units to the left of point D.

STEP 2: Translate △ABC 10 units to the right. Select the Translate function within the Transform menus. Do the two triangles overlap? Are the two triangles congruent?

STEP 3: Repeat STEP 1 and STEP 2 for triangles with different dimensions. Which pairs of corresponding parts did you specify to be congruent? Was this sufficient to make the triangles congruent?

STEP 4: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that m∠CAB = m ∠FDE, AB = DE, and m∠CBA = m∠FED.

STEP 5: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that BC = EF, AB = DE, and AC = DF.

STEP 6: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that m∠CAB = m ∠FDE = 90°, BC = EF, and AC = DF.

Math Journal How are △ABC and △DEF related in each case? What are the four sets of the minimum conditions for two triangles to be congruent?
Answer:

Question 5.
Justify whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used.
a)

LN = , m∠ = m∠, and LM = .
By the SAS test, △LMN ≅ .
Caution
In the SAS test, the angle congruent to its corresponding angle must be the included angle, which is the angle between the congruent corresponding sides.
Answer:
LN = XZ, m∠ = NML = m∠YXZ, and LM = XY.
By the SAS test, △LMN ≅ △XYZ.

b)

m∠ABE = m∠ Corr. ∠s
m∠BAE = m∠ Corr.∠s
m∠AEB = m∠ ∠sum of triangle
Three pairs of congruent angles do not ensure that the triangles are .
Answer:
m∠ABE = m∠BCD Corr. ∠s
m∠BAE = m∠CBD Corr.∠s
m∠AEB = m∠BDC ∠sum of triangle
Three pairs of congruent angles do not ensure that the triangles are CONGRUENT.

Math in Focus Course 3B Practice 9.1 Answer Key

Name the figures that are congruent. Name the corresponding congruent line segments and angles.

Question 1.
ABCD is a parallelogram with diagonal \(\overline{B D}\).

Answer: Yes the above parallelogram is congruent (△BAD ≅ △DCB).
AB = CD and AD = BC because opposite sides of a parallelogram are equal.
∠ABD = ∠CDB And ∠BAD = ∠DCB

Question 2.

Answer:
ABCD ≅ PQRS
AB = PR = 5 cm
AD = RS = 9.22 cm
BC = PQ = 9 cm
DC = QS = 7 cm
∠ABC = ∠RPQ
∠DCB = ∠SQP

Solve. Show your work.

Question 3.
WXYZ is a rectangle with diagonal \(\overline{W Y}\). Explain, using the given test for congruent triangles, why △WXY is congruent to △YZW.

a) SSS
Answer: WX = ZY
WZ = XY
WY = WY

b) SAS
Answer:
WX = ZY
m∠ZWX= m∠ZYX
WZ = XY

c) ASA
Answer:
m∠ZWY= m∠XYW

d) HL
Answer:
ΔWXY = ΔWZY (right angled triangle)
WY = WY (Hypotenuse)

Question 4.
ABCDE is a regular pentagon with congruent diagonals \(\overline{B E}\) and \(\overline{B D}\).
a) Justify △ABE ≅ △CBD with a test for congruent triangles.
Answer:
We can test whether the given figures are congruent by testing SSS, ASA, SAS, and HL.
SSS test:
AB = BC
AE = CD
BE = BD

b) Name a pair of congruent quadrilaterals.

Answer: △ABE ≅ △CBD because the shape and size are equal.

Question 5.
△ABC is congruent to △PQR. Find the values of x, y, and z.

Answer:
Given,
△ABC is congruent to △PQR.
AB = PQ
BC = QR
AC = PR
AB = 8 cm
So, PQ = y = 8 cm
BC = x cm
We know that
BC = QR
So, BC = 3.4 cm
m∠BCA= m∠QPR
m∠BCA= 67°
So, m∠QPR = 67°

Question 6.

a) Name the figure that is congruent to trapezoid BADC.
Answer: The figure that is congruent to trapezoid BADC is AECD.

b) Find the length of each side of the trapezoid you named in a).
Answer:
From the figure, we can see that
BC = ED
AC = AD
BE = CD
Now we will find the unknown values of the figure.
BC = 6 in.
So, ED = 6 in.
AC = 10 in.
AC = AD = 10 in.
BE = CD
CD = 16 in
So, BE = CD = 16 in.

Question 7.
A piece of fabric has many printed shapes of congruent triangles and congruent quadrilaterals on it.
a) The pieces of fabric shown are congruent quadrilaterals. Find the value of x, y, and z.

Answer:
z + 3 = 6 cm
z = 6 – 3
z = 3 cm
9x° = 63°
x° = 63/9
x = 7
y° = 49.5

b) The pieces of fabric shown are congruent triangles. Find the value of p, q, and r.

Answer:
2p + q = 18 cm—- eq. 1
p – 2q = 16 cm—-eq. 2
2p + q = 18 cm—- eq. 1 × 2
4p + 2q = 36 cm
p – 2q = 16 cm
5p = 52 cm
p = 52/5
p = 10.4 cm
substitute the value of p in the eq.2
10.4 – 2q = 16 cm
10.4 – 16 = 2q
-5.6 = 2q
q = -5.6/2
q = -2.8 cm
The sum of all three angles = 180°
55° + 58° + r° = 180°
113° + r° = 180°
r° = 180° – 113°
r° = 67°

Question 8.
The origami cornflower is formed by making symmetrical folds in a paper square. In the diagram, AB = BC = DE = EF and OA = OC = OD = OF. Write two possible statements of congruence for ABCO.

Answer:
△ABO and △BCO seems congruent
AB = BC.

Question 9.
In making this suspension bridge, engineers connected two cables at the same point P on the concrete piers and at the same distance from the base of the concrete piers at R.
a) Which congruence test ensures that △PQR is congruent to △PSR? Explain.
Answer: SAS (Side Angle Side). Each triangle in the below figure is a right angled triangle at R and congruent with the same size and shape.

b) How many other pairs of congruent triangles are attached to each concrete pier? Explain.

Answer: There are 10 pairs of congruent triangles that are attached to each concrete pier.

Question 10.
This skyscraper has several symmetric triangles overlaid on a grid of horizontal and vertical lines. In each direction, these lines are parallel.

a) This face of the building is symmetric about \(\overline{G K}\), so AB = CD, AX = CX, and BX = DX. Give the statement of congruence for △ABX and tell which congruence test you used.
Answer: SSS test is used to tell whether the figure is congruent. △ABX ≅ △CDX

b) Is △EFG congruent to △HJK? Explain.
Answer: No because the shape of the triangles is same but the size of the triangles are different.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 13 Practice 4 Subtraction with Regrouping to score better marks in the exam.

Math in Focus Grade 1 Chapter 13 Practice 4 Answer Key Subtraction with Regrouping

Regroup.

Question 1.
27 = 1 ten ________ ones
Answer:

Question 2.
15 = 0 tens ________ ones
Answer:

Question 3.
30 = 2 tens ________ ones
Answer:

Subtract.

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Fill in the missing numbers.

Question 10.
21 – 5 = ___

Answer:

Question 11.
36 – 7 = ___

Answer:

Question 12.
25 – 18 = ___

Answer:

Question 13.
31 – 18 = ___

Answer:

Question 14.
32 – 14 = ___

Answer:

Question 15.
30 – 17 = ___

Answer:

Subtract.

Then solve the riddle.

Question 16.
38 – 9 = 29
Answer:

Question 17.
30 – 18 = _____
Answer:

Question 18.
32 – 5 = ___
Answer:

Question 19.
35 – 8 = ___
Answer:

Question 20.
27 – 7 = _____
Answer:

Question 21.
34 – 19 = _____
Answer:

How do you cut the sea?

Match the letters to the answers below to find out.

Nafasha drops a ball into each number machine. Write the missing numbers in the blanks to show what happens to each ball.
Answer:

Question 22.

Answer:

Question 23.

Answer:

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 13 Practice 5 Adding Three Numbers to score better marks in the exam.

Math in Focus Grade 1 Chapter 13 Practice 5 Answer Key Adding Three Numbers

Add.

Question 1.

4 + 5 + 6 = ____
Answer: 4+5+6=15
Explanation:
By adding 4 with 5 we get 9,
Then by adding 9 with 6 we get 15.

Question 2.

8 + 7 + 7 = ____
Answer:
Explanation:8+7+7=22
By adding 8 with 7 we get 15,
Then by adding 15 with 7 we get 22.

Question 3.

6 + 9 + 8 = ____
Answer:
Explanation:6+9+8=23
By adding 6 with 9 we get 15,
Then by adding 15 we get 23

Question 4.

5 + 4 + 8 = ____
Answer:
Explanation:5+4+8=17
By adding 5 with 4 we get 9,
Then by adding 9 with 8 we get 17.

Make ten. Then add.

Example

Question 5.
5 + 8 + 5 = ____
Answer:

Explanation:
By adding 5 with 5 we get 10,
Then by adding 8 with 10 we get 18.

Question 6.
8 + 9 + 2 =___
Answer:

Explanation:
By adding 8 with 2 we get 10,
Then by adding 9 with 10 we get 19.

Question 7.
9 + 7 + 2 = ___
Answer:

Explanation:
By adding 9 with 2 we get 11,
Then by adding 11 with 7 we get 18

Question 8.
9 + 4 + 4 = ___
Answer:

Explanation:
By adding 9 with 4 we get 13,
Then by adding 13 with 4 we get 17.

Question 9.
2 + 9 + 5 = ___
Answer:

Explanation:
By adding 2 with 9 we get 11,
Then by adding 11 with 5 we get 16.

This handy Math in Focus Grade 4 Workbook Answer Key Chapter 11 Practice 2 Properties of Squares and Rectangles detailed solutions for the textbook questions.

Math in Focus Grade 4 Chapter 11 Practice 2 Answer Key Properties of Squares and Rectangles

All the figures are rectangles. Find the measures of the unknown angles.

Example

Find the measure of ∠a.

Question 1.
Find the measure of ∠b.

Answer:
Measure of ∠b = 90⁰ – 27⁰
∠b= 63⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 27⁰ from 90⁰the difference is equal to 63⁰.
Question 2.
Find the measure of ∠c.

Answer:
Measure of ∠c = 90⁰ – 45 ⁰
∠c= 45⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 45⁰ from 90⁰the difference is equal to 45⁰.

All the figures are rectangles. Find the measures of the unknown angles.

Question 3.
Find the measure of ∠p.

Answer:
Measure of ∠p = 90⁰ – 36⁰– 18⁰
∠p= 36⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 36⁰ and 18⁰ from 90⁰the difference is equal to 36⁰.

Question 4.
Find the measure of ∠m.

Answer:
Measure of ∠m = 90⁰ – 39⁰– 22⁰
∠m= 29⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 39⁰ and 22⁰ from 90⁰the difference is equal to 29⁰.

The figure is a rectangle. Find the measure of the unknown angle.

Question 5.
Find the measure of ∠s.

Answer:
Measure of ∠s = 90⁰ – 25⁰– 12⁰
∠s= 53⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 25⁰ and 12⁰ from 90⁰the difference is equal to 53⁰.

Find the lengths of the unknown sides.

Question 6.
The figure is made up of a rectangle and a square. Find BC and GE.

Answer:
In the above image we can observe rectangle and a square.
From the rectangle ABFG we know that
GA = BF = 12 cm
AB = FG = 15cm
From the square CDEF we know that
CD = DE = EF = FC = 8 cm
BC = GA – CF
BC = 12 – 8
BC = 4 cm
GE = AB + EF
GE = 15 + 8
GE = 23 cm

Find the lengths of the unknown sides.

Question 7.
The figure is made up of two rectangles. Find BD and FG.

Answer:
In the above image we can observe two rectangles.
From the rectangle ABCG we know that
AB = CG = 20 feet
BC = GA = 17 feet
From the rectangle CDEF we know that
CD = EF = 5 feet
DE = FC = 14 feet
BD = BC + CD
BD = 17 + 5
BD = 22 feet
FG = AB – FC
FG = 20 – 14
FG = 6 feet

Question 8.
The figure is made up of two rectangles. Find QR and RT.

Answer:
In the above image we can observe two rectangles.
From the rectangle PQRS we know that
PQ = RS = 5 yards
QR = SP
From the rectangle STUV we know that
ST = UV = 6 yards
TU = VS = 4 yards
QR = PV + VS
QR = 3 + 4
QR = 7 yards
RT = RS + ST
RT = 5 + 6
RT = 11 yards

Find the lengths of the unknown sides.

Question 9.
The figure is made up of two rectangles. Find FG.

Answer:
In the above image we can observe two rectangles.
From the rectangle ABGH we know that
BG = HA = 5 miles
AB = GH
From the rectangle CDEF we know that
DE = FC = 8 miles
CD = EF
CG = HA – BC
CG = 5 – 3
CG = 2 miles
FG = DE – CG
FG = 8 – 2
FG = 6 miles

Question 10.
The figure is made up of a square and a rectangle. Find BC.

Answer:
In the above image we can observe rectangle and a square.
From the rectangle ABGH we know that
BG = HA = 10 feet
AB = GH
From the square CDEF we know that
CD = DE = EF = FC = 4 feet
BC = HA – FC – FG
BC = 10 – 4 – 2
BC = 4 feet

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.4 Sketching Graphs of Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.4 Answer Key Sketching Graphs of Linear Equations

Math in Focus Grade 8 Chapter 4 Lesson 4.4 Guided Practice Answer Key

Use graph paper. Use 1 grid square to represent 1 unit for the x interval from -2 to 2, and the y interval from -2 to 4.

Question 1.
Graph the equation y = \(\frac{3}{2}\)x + 1.
Answer:

Use graph paper. Use 1 grid square to represent 1 unit on both axes for the x interval from 0 to 3, and the y Interval from -3 to 7.

Question 2.
Graph the equation y = 2x + 1.
Answer:

Question 3.
Graph the equation y = –\(\frac{1}{3}\)x – 2.
Answer:

Use graph paper. Use 1 grid square to represent 1 unit on both axes for the x interval from -2 to 2, and the y interval from 0 to 10.

Question 4.
Graph a line with slope -2 that passes through the point (2, 2).
Answer:

Question 5.
Graph a line with slope 2 that passes through the point (-2, 1).
Answer:

Math in Focus Course 3A Practice 4.4 Answer Key

For this practice, use 1 grid square to represent 1 unit on both axes for the interval from -6 to 6.

Graph each linear equation.

Question 1.
y = \(\frac{1}{3}\)x + 1
Answer:

Question 2.
y = \(\frac{1}{6}\)x + 3
Answer:

Question 3.
y = \(\frac{1}{2}\)x + 2
Answer:

Question 4.
y = \(\frac{2}{3}\)x – 1
Answer:

Question 5.
y = -x + 5
Answer:

Question 6.
y = 3 – \(\frac{1}{4}\)x
Answer:

Question 7.
y = 1 – \(\frac{1}{2}\)x
Answer:

Question 8.
y = –\(\frac{1}{5}\)x – 2
Answer:

Question 9.
Math Journal Graph the equation y = 2 – \(\frac{2}{3}\)x. Explain how to use the graph to find other solutions of the equation.
Answer:

First select a value of the x-coordinate on the x-axis. Then trace it along the vertical grid lines until the gridline intersects with the graph.
Then trace horizontally from the graph to the y-axis to obtain the corresponding y-coordinate. This pair of x-coordinate and y-coordinate is one solution to the equation.

Question 10.
Math Journal Martha says that the point (4, -2) lies on the graph of the equation y = –\(\frac{1}{4}\)x – 1. Explain how you can find out if she is right without actually graphing the equation.
Answer:

Graph each line with the given slope that passes through the given point.

Question 11.
Slope = \(\frac{2}{5}\); (5, 4)
Answer:
y = \(\frac{2}{5}\)x + 4

Question 12.
Slope = \(\frac{2}{3}\); (6, 1)
Answer:

Question 13.
Slope = -3; (1, 0)
Answer:

Question 14.
Slope = -2; (-1, -2)
Answer:

Question 15.
Math Journal Suppose that Emily shows you some of her homework:

Graph the equation y = -2x + \(\frac{1}{2}\).
Describe Emily’s mistake. Graph the equation correctly.
Answer:
Emily made three mistakes. She labeled the coordinates (6, 3) incorrectly, drew the slope of the line incorrectly, and mistook the value for the slope as the y-intercept.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.3 Writing Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.3 Answer Key Writing Linear Equations

Math in Focus Grade 8 Chapter 4 Lesson 4.3 Guided Practice Answer Key

For each line, state its slope and its y-intercept.

Question 1.
5x + 4y = 8
First, write the equation in the slope-intercept form.
5x + 4y = 8
5x + 4y – = 8 –      Subtract 5x from both sides.
=     Simplify.
=     Divide both sides by .
y =       Write in slope-intercept form.
Comparing the equation y = with y = mx + b:
Slope: m =
y-intercept: b =
Answer:
First, write the equation in the slope-intercept form.
5x + 4y = 8
5x + 4y – 5x = 8 – 5x    Subtract 5x from both sides.
4y = 8 – 5x    Simplify.
y = 2 – 5/4 x    Divide both sides by 4.
y = -5/4 x + 2      Write in slope-intercept form.
Comparing the equation y = -5/4 x + 2  with y = mx + b:
Slope: m = -5/4
y-intercept: b = 2

Question 2.
2x – 3y = 7
First write the equation in the slope-intercept form.
2x – 3y = 7
2x – 3y + = 7 +    Add 3y to both sides.
=         Simplify.
=      Subtract from both sides.
=      Simplify.
=     Divide both sides by .
y =       Write in slope-intercept form.
Comparing the equation y = with y = mx + b:
Slope: m =
y-intercept: b =
Answer:
First write the equation in the slope-intercept form.
2x – 3y = 7
2x – 3y + 3y = 7 + 3y   Add 3y to both sides.
2x = 7 + 3y        Simplify.
2x – 2x = 7 + 3y – 2x     Subtract 2x from both sides.
0 = 7 + 3y – 2x     Simplify.
3y  = 2x – 7    Divide both sides by 3.
y = 2/3 x – 7/3      Write in slope-intercept form.
Comparing the equation y = 2/3 x – 7/3 with y = mx + b:
Slope: m = 2/3
y-intercept: b = -7/3

Question 3.
5y – x = 15
Answer:
First write the equation in the slope-intercept form.
Add x on both sides
5y – x + x = 15 + x
5y = 15 + x
Subtract 5y on both sides
5y – 5y = 15 + x – 5y
0 = 15 + x – 5y
5y = x + 15
Divide by 5 on both sides
y = 1/5 x + 3

Question 4.
2y – 3x = -4
Answer:
First write the equation in the slope-intercept form.
2y = 3x -4
Divide by 2 on both sides
y = 3/2 x – 2

Question 5.
6y + 5x = 24
Answer:
First write the equation in the slope-intercept form.
Subtract 5x on both sides
6y + 5x – 5x = 24 – 5x
6y = 24 – 5x
Divide by 6 on both sides
y = 4 – 5x
y = -5x + 4

Question 6.
3y + 4x = 3
Answer:
First write the equation in the slope-intercept form.
Subtract 4x on both sides
3y + 4x -4x = 3 – 4x
3y = -4x + 3
Divide by 3 on both sides
y = -4/3 x + 1

Use the given slope and y-intercept of a line to write an equation in slope-intercept form.

Question 7.
Slope, m = –\(\frac{2}{3}\)
y-intercept, b = 4
y = mx + b
y = Substitute the given values for and b.
Answer:
y = mx + b
y = –\(\frac{2}{3}\) Substitute the given values for 4 and b.

Question 8.
Slope, m = 4
y-intercept, b = -7
Answer:
y = mx + b
y = 4 Substitute the given values for -7 and b.

Solve.

Question 9.
A line has the equation 3y + 6 = 10x. Write an equation of a line parallel to this given line that has a y-intercept of 2.
First write the given equation in slope-intercept form.
3y + 6 = 10x
3y + 6 – = 10x –      Subtract both sides by 6.
=     Simplify.
y =     Divide both sides by .
y =      Write in slope-intercept form.
The line has slope m = and y-intercept b = .
Then write an equation for the parallel line with slope m = and y-intercept b = .
y = . Substitute the values of m and b.
So, an equation of the line parallel to 3y = 10x – 6 is .
Answer:
First write the given equation in slope-intercept form.
3y + 6 = 10x
3y + 6 – 6 = 10x – 6     Subtract both sides by 6.
3y = 10x – 6    Simplify.
y = 10/3 x – 2    Divide both sides by 3.
y = 10/3 x – 2      Write in slope-intercept form.
The line has slope m = 10/3 and y-intercept b = -2.
Then write an equation for the parallel line with slope m = 10/3 and y-intercept b = -2
y = 10/3 x + 2. Substitute the values of m and b.
So, an equation of the line parallel to 3y = 10x – 6 is 10/3 x + 2.

Question 10.
A line has slope -3 and passes through the point (-6, 8). Write an equation of the line.
Answer:
Given,
A line has slope -3 and passes through the point (-6, 8)
m = -3
y-intercept = 8
y = -3x + 8

Question 11.
A line has slope \(\frac{1}{3}\) and passes through the point (0, 1). Write an equation of the line.
Answer:
Given,
A line has slope \(\frac{1}{3}\) and passes through the point (0, 1).
m = \(\frac{1}{3}\)
y-intercept = 1
y = mx + b
y = \(\frac{1}{3}\)x + 1

Question 12.
A line has slope 2 and passes through the point (1, 5). Write an equation of the line.
Answer:
Given,
A line has slope 2 and passes through the point (1, 5).
m = 2
y-intercept = 5
y = mx + b
y = 2x + 5

Question 13.
Write an equation of the line that passes through the point (-2, 1) and is parallel to y = 5 – 3x.
First, write the equation in slope-intercept form.
y = 5 – 3x
y =     Write in slope-intercept form.
The line has slope m = .
So, the line parallel to y = 5 – 3x has slope m = .
Then use the slope m = . and the fact that (-2, 1) lies on the parallel line to find the y-intercept.
y = mx + b Write in slope-intercept form.
=      Substitute the values for m, x, and y.
=     Simplify.
=     Subtract from both sides.
=     Simplify.
The y-intercept is .
So, an equation of the line is .
Answer:
First, write the equation in slope-intercept form.
y = 5 – 3x
y = -3x + 5    Write in slope-intercept form.
The line has slope m = -2.
So, the line parallel to y = 5 – 3x has slope m = -3.
Then use the slope m = -2. and the fact that (-2, 1) lies on the parallel line to find the y-intercept.
y = mx + b Write in slope-intercept form.
y = -2x + 1 Substitute the values for m, x, and y.
So, an equation of the line is y = -2x + 1.

Question 14.
Write an equation of the line that passes through the pair of points (-2, -5) and (2, -1).
Answer:
The line passes through the points (-2, -5) and (2, -1).
Slope m = \(\frac{-1-(-5)}{2-(-2)}\)
= \(\frac{4}{4}\)
= 1
The line passes through the y-axis at the point (0, 0).
Thus m = 1 and y-intercept b is 0.
y = x

Math in Focus Course 3A Practice 4.3 Answer Key

Find the slope and the y-intercept of the graph of each equation.

Question 1.
y = -5x + 7
Answer:
Given,
y = -5x + 7
The equation of the line is y = mx + b
Slope, m = -4
y-intercept = 7

Question 2.
y = 2x + 3
Answer:
Given,
y = 2x + 3
The equation of the line is y = mx + b
Slope, m = 2
y-intercept = 3

Question 3.
5x + 2y = 6
Answer:
Given,
5x + 2y = 6
The equation of the line is y = mx + b
2y = -5x + 6
y = -5/2 x + 3
Slope, m = -5/2
y-intercept = 3

Question 4.
2x – 7y = 10
Answer:
Given,
The equation of the line is y = mx + b
2x – 7y = 10
-7y = -2x + 10
7y = 2x – 10
y = 2/7 x – 10/7
Slope, m = 2/7
y-intercept = 10/7

Use the given slope and y-intercept of a line to write an equation in slope-intercept form.

Question 5.
Slope, m =\(\frac{1}{2}\)
y-intercept, b = 3
Answer:
The equation of the line is y = mx + b
y = \(\frac{1}{2}\)x + 3

Question 6.
Slope, m = -2
y-intercept, b = 5
Answer:
The equation of the line is y = mx + b
y = -2x + 5

Solve. Show your work.

Question 7.
A line has the equation 4y = 3x – 8. Find an equation of a line parallel to this line that has a y-intercept of 2.
Answer:
A line has the equation 4y = 3x – 8.
y = 3/4 x – 2
A line parallel to this line that has a y-intercept of 2
y = 3/4 x + 2

Question 8.
A line has the equation 3y = 3 – 2x. Find an equation of a line parallel to this line that has a y-intercept of 5.
Answer:
A line has the equation 3y = 3 – 2x.
y = 1 – 2/3 x
y = -2/3 x + 1
y-intercept = 5
y = -2/3 x + 5

Question 9.
Math Journal Ira says that the graphs of the equations y = -3x + 7 and y = 3x – 7 are parallel lines. Do you agree? Explain.
Answer:
No. The slope of the equation y = -3x + 7 is -3 and the slope of the equation y = 3x – 7 is 3.
So, the graphs of the equations are not parallel lines.

Question 10.
Find an equation of the line that passes through the point (0, 4) and has a slope of –\(\frac{1}{3}\).
Answer:
Slope = –\(\frac{1}{3}\)
y-intercept = 4
The equation of the line is y = mx + b
y = –\(\frac{1}{3}\)x + 4

Question 11.
A line has slope –\(\frac{1}{2}\) and passes through the point (-4, -2). Write an equation of the line.
Answer:
The equation of the line is y = mx + b
slope = –\(\frac{1}{2}\)
y – intercept = -2
y = –\(\frac{1}{2}\)x – 2

Question 12.
Find an equation of the line that passes through the point (-5, 7) and is parallel to y = 4 – 3x.
Answer:
The equation of the line is y = mx + b
y = 4 – 3x
y = -3x + 4
y = -3x + 7

Question 13.
Find an equation of the line that passes through the point (0, 2) and is parallel to 6y = 5x – 24.
Answer:
6y = 5x – 24
The equation of the line is y = mx + b
y = 5/6 x – 4
The line that passes through the point (0, 2)
y = 5/6 x + 2

Question 14.
Find an equation of the line that passes through the pair of points (-5, -1) and (0, 4).
Answer:
The line passes through the points (-5, -1) and (0, 4).
Slope m = \(\frac{4-(-1)}{0-(-5)}\)
= \(\frac{5}{5}\)
= 1
The line passes through the y-axis at the point (0, 0).
Thus m = 1 and y-intercept b is 0.
y = x

Question 15.
Find an equation of the line that passes through the pair of points (-3, 2) and (-2, 5).
Answer:
The line passes through the points (-3, 2) and (-2, 5)
Slope m = \(\frac{5-2}{-2-(-3)}\)
= \(\frac{3}{1}\)
= 3
The line passes through the y-axis at the point (0, 0).
Thus m = 3 and y-intercept b is 0.
y = 3x

Question 16.
Math Journal Can you write a linear equation in the slope-intercept form using the points (3, 4) and (5, 8)? Explain.
Answer:
The line passes through the points (3, 4) and (5, 8)
Slope m = \(\frac{8-4}{5-3}\)
= \(\frac{4}{2}\)
= 2
The line passes through the y-axis at the point (0, 0).
Thus m = 2 and y-intercept b is 0.
y = 2x