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

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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 1
Height of water needed = \(\frac{3}{4}\) • Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 in.
Volume of water needed = Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 in.3
The aquarium needs to have Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 3
a) Find the volume of the metal bar.
Area of parallelogram
= base of parallelogram • height of parallelogram
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 cm2
Volume of metal bar = base of prism • height of prism
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 cm3
The volume of the metal bar is Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 4
The surface area of the metal bar is Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 5
a) Find the length of each side of the square base.
V = Bh
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 = B • Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 ÷ Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 = Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 • B ÷ Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 = B
Length of each side of base
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 cm
The length of each side of the square base is Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 6
The surface area of the candle is Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 7
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
= \(\frac{1}{2}\) • Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 • (Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 + Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2) + Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 + Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 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
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 = Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 • h
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 ÷ Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 = Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 • h ÷ Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2
= Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 ≈ h
The height of the prism is approximately Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 8
The surface area of the prism is approximately Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 2 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?
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 9
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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 10
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?
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 11
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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 12
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.
Math in Focus Grade 6 Chapter 12 Lesson 12.4 Answer Key Real-World Problem Surface Area and Volume 13
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.

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