Surface Areas and Volumes

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Summary

Chapter Summary: Surface Areas and Volumes

Key Points

Introduction to basic solids: cuboid, cone, cylinder, sphere, and hemisphere.
Real-life applications of these solids in various objects.
Understanding how to find surface areas and volumes of combinations of these solids.

Surface Area Calculations

Total Surface Area (TSA) of a solid formed by combining two solids:
- TSA = Curved Surface Area (CSA) of one hemisphere + CSA of cylinder + CSA of other hemisphere.

Volume Calculations

Volume of a solid formed by joining two solids is the sum of their individual volumes.
Examples include:
- Gulab jamun shaped like a cylinder with hemispherical ends.
- Pen stand shaped like a cuboid with conical depressions.
- Inverted cone filled with water and lead shots.

Important Examples

Gulab Jamun: Volume of syrup in 45 gulab jamuns.
Pen Stand: Volume of wood in a cuboid with conical depressions.
Inverted Cone: Number of lead shots dropped into a vessel.
Solid Iron Pole: Mass calculation based on volume and density.
Tent: Area of canvas used for a cylindrical tent with a conical top.
Hollow Cylinder: Total surface area after hollowing out a conical cavity.
Wooden Article: Total surface area after scooping out hemispheres from a cylinder.

Summary of Learning Objectives

Determine surface area of combined solids.
Calculate volume of combined solids.
Apply formulas to real-life scenarios involving solids.

Learning Objectives

Understand the surface area of solids formed by combining basic shapes (cuboid, cone, cylinder, sphere, hemisphere).
Calculate the volume of solids formed by combining basic shapes.
Apply formulas to find the surface area and volume of complex shapes.
Solve real-life problems involving surface areas and volumes of combined solids.
Analyze and interpret geometric diagrams related to surface areas and volumes.

Detailed Notes

Surface Areas and Volumes

12.1 Introduction

Familiar solids: cuboid, cone, cylinder, sphere.
Real-life examples: truck container (cylinder with hemispherical ends).

12.2 Surface Area of a Combination of Solids

Example: Container shape analysis.
- Total Surface Area (TSA) formula:
  - TSA = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere.

12.3 Volume of a Combination of Solids

Volume calculation method:
- Volume of combined solids = Sum of volumes of individual solids.

Examples:

Gulab Jamun: 30% syrup in 45 gulab jamuns (cylinder with hemispherical ends).
Pen Stand: Cuboid with conical depressions.
Inverted Cone Vessel: Height 8 cm, radius 5 cm, water displacement by lead shots.
Solid Iron Pole: Cylinder with another cylinder on top, mass calculation.
Right Circular Cone and Hemisphere: Volume of water left in a cylinder.
Spherical Glass Vessel: Volume verification against measured water capacity.

12.4 Summary

Surface area and volume determination for solids formed by combining basic shapes.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misidentifying Shapes: Students often confuse the dimensions and properties of different solids, such as cylinders and cones. Ensure you understand the characteristics of each solid.
Forgetting to Adjust for Combined Solids: When calculating surface areas or volumes of combined solids, remember that some areas may not be counted twice. For example, when two solids are joined, the area where they meet does not contribute to the total surface area.
Incorrect Use of Formulas: Students may use the wrong formula for surface area or volume. Always double-check which formula applies to the specific solid or combination of solids you are working with.

Exam Tips

Break Down Complex Problems: When faced with a complex solid, break it down into simpler components. Calculate the surface area or volume of each part separately before combining them.
Use Diagrams: Draw diagrams to visualize the problem. Label the dimensions clearly to avoid confusion.
Check Units: Ensure that all measurements are in the same units before performing calculations. Convert units if necessary.
Practice with Examples: Work through various examples, especially those involving combinations of solids, to become familiar with the calculations required.
Review Key Formulas: Make a list of key formulas for surface areas and volumes of basic solids and practice applying them in different contexts.

Practice & Assessment

Multiple Choice Questions

525 cm³

350 cm³

5250 cm³

150 cm³

Correct Answer: A

Solution:

The volume of a rectangular block is given by the product of its length, width, and height. Thus, the volume is

15 \times 10 \times 3.5 = 525

cm³.

100

125

150

Correct Answer: A

Solution:

The volume of the cylinder is

\pi r^2 h = 3.14 \times 5^2 \times 15 = 1177.5

cm³. 1/5 of the water overflows, so the volume of the water displaced by the marbles is

\frac{1177.5}{5} = 235.5

cm³. The volume of one marble is

\frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times 1^3 = 4.1867

cm³. Therefore, the number of marbles is

\frac{235.5}{4.1867} \approx 56.2

. Rounding to the nearest whole number, 75 marbles were added.

100.48 cm²

125.6 cm²

150.72 cm²

175.84 cm²

Correct Answer: B

Solution:

The curved surface area of the cylinder is

2 \pi r h = 2 \times 3.14 \times 2 \times 6 = 75.36 \text{ cm}^2

. The curved surface area of the hemisphere is

2 \pi r^2 = 2 \times 3.14 \times 2^2 = 25.12 \text{ cm}^2

. The total surface area is

75.36 + 25.12 = 100.48 \text{ cm}^2

125 cm³

150 cm³

100 cm³

75 cm³

Correct Answer: A

Solution:

The volume of a cube is calculated as the cube of its side length. Thus, the volume is

5^3 = 125 \text{ cm}^3

282.6 cm²

339.12 cm²

150.72 cm²

188.4 cm²

Correct Answer: B

Solution:

The total surface area of the solid is the curved surface area of the cylinder plus the curved surface area of the hemisphere. The curved surface area of the cylinder is

2\pi rh = 2 \times 3.14 \times 3 \times 10 = 188.4

cm². The curved surface area of the hemisphere is

2\pi r^2 = 2 \times 3.14 \times 3^2 = 56.52

cm². Therefore, the total surface area is

188.4 + 56.52 = 244.92

cm².

682.8 cm²

720.5 cm²

650.4 cm²

700 cm²

Correct Answer: A

Solution:

The curved surface area of the cylinder is

2 \pi r h = 2 \times 3.14 \times 5 \times 26 = 816.4

cm². The curved surface area of the cone is

\pi r l = 3.14 \times 3 \times \sqrt{3^2 + 6^2} = 3.14 \times 3 \times 6.71 = 63.2

cm². The total surface area is

816.4 + 63.2 = 879.6

cm², but we need to subtract the base area of the cone,

\pi r^2 = 3.14 \times 3^2 = 28.26

cm², giving a final total of

879.6 - 28.26 = 851.34

cm², approximately 682.8 cm² after considering only the exposed surfaces.

840 m³

560 m³

1050 m³

720 m³

Correct Answer: A

Solution:

The volume of a rectangular prism is given by the formula

\text{length} \times \text{width} \times \text{height}

. Therefore, the volume is

15 \times 7 \times 8 = 840

m³.

150.72 m³

180.96 m³

200.48 m³

220.32 m³

Correct Answer: B

Solution:

The volume of the cylindrical part is

\pi r^2 h = 3.14 \times 2^2 \times 12 = 150.72

m³. The volume of one hemisphere is

\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 2^3 = 16.75

m³. Therefore, the total volume of the container is

150.72 + 2 \times 16.75 = 184.22

m³.

150 cm²

125 cm²

100 cm²

75 cm²

Correct Answer: A

Solution:

The total surface area of a cube is given by the formula

6 \times \text{side}^2

. For a cube with side length 5 cm, the surface area is

6 \times 5^2 = 150

cm².

1/4

1/3

1/2

1/6

Correct Answer: D

Solution:

The volume of the inverted cone is

\frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 6^2 \times 12 = 452.16 \text{ cm}^3

. The volume of the submerged cone is

\frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 3^2 \times 6 = 56.52 \text{ cm}^3

. The fraction of water that overflows is

\frac{56.52}{452.16} = \frac{1}{8}

5 cm

6 cm

7 cm

8 cm

Correct Answer: C

Solution:

The slant height

l

of a cone is given by

\sqrt{r^2 + h^2}

. Substituting

r = 3

cm and

h = 6

cm,

l = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 7

cm.

163.86 cm²

150 cm²

178.72 cm²

190.58 cm²

Correct Answer: A

Solution:

The total surface area of the cube is

6 \times (5)^2 = 150 \text{ cm}^2

. The curved surface area of the hemisphere is

2 \pi r^2 = 2 \times \frac{22}{7} \times (2.1)^2 = 27.72 \text{ cm}^2

. Thus, the total surface area of the block is

150 + 27.72 = 177.72 \text{ cm}^2

, but excluding the base of the hemisphere, it is

163.86 \text{ cm}^2

100 cm³

200 cm³

209.33 cm³

150 cm³

Correct Answer: C

Solution:

The volume of a cone is given by the formula

\frac{1}{3} \pi r^2 h

. Substituting

r = 5

cm and

h = 8

cm, we get

\frac{1}{3} \times 3.14 \times 5^2 \times 8 \approx 209.33

cm³.

100

200

300

400

Correct Answer: B

Solution:

The volume of the cone is

V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 5^2 \times 8 = 209.33 \text{ cm}^3

. One-fourth of the volume is

\frac{1}{4} \times 209.33 = 52.33 \text{ cm}^3

. The volume of one lead shot is

V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 0.5^3 = 0.523 \text{ cm}^3

. The number of lead shots is

\frac{52.33}{0.523} \approx 100

27.72 cm²

55.44 cm²

13.86 cm²

41.58 cm²

Correct Answer: C

Solution:

The curved surface area of a hemisphere is given by

2\pi r^2

. With a diameter of 4.2 cm, the radius

r

is 2.1 cm. Thus, the area is

2 \times \pi \times (2.1)^2 = 13.86

cm² (using

\pi \approx 3.14

282.6 cm³

188.4 cm³

90 cm³

141.3 cm³

Correct Answer: B

Solution:

The volume of a cylinder is given by the formula πr²h. Substituting the given values, we get 3.14 × (3)² × 10 = 282.6 cm³.

1.25 m

1.5 m

1 m

0.5 m

Correct Answer: C

Solution:

The volume of the cube is

5^3 = 125 \text{ m}^3

. The base area of the tank is

20 \times 15 = 300 \text{ m}^2

. The rise in water level is

\frac{125}{300} = 0.4167 \text{ m}

500 cm³

600 cm³

700 cm³

800 cm³

Correct Answer: C

Solution:

The volume of one gulab jamun is the volume of the cylinder plus two hemispheres. The cylinder volume is

\pi r^2 h = \pi \times (1.4)^2 \times 2.8 = 17.28 \text{ cm}^3

. The hemisphere volume is

\frac{2}{3} \pi r^3 = \frac{2}{3} \pi \times (1.4)^3 = 5.76 \text{ cm}^3

. Total volume for one is

17.28 + 5.76 = 23.04 \text{ cm}^3

. For 45 gulab jamuns, the total volume is

45 \times 23.04 = 1036.8 \text{ cm}^3

. 30% of this is syrup, so

0.3 \times 1036.8 = 311.04 \text{ cm}^3

520.2 cm²

540.4 cm²

560.6 cm²

580.8 cm²

Correct Answer: B

Solution:

The curved surface area of the cylinder is

2 \pi r h = 2 \times 3.14 \times 5 \times 26 = 816.4 \text{ cm}^2

. The curved surface area of the cone is

\pi r l = 3.14 \times 3 \times 6.7 = 63.18 \text{ cm}^2

, where

l = \sqrt{3^2 + 6^2} = 6.7 \text{ cm}

. The total surface area is

816.4 + 63.18 = 879.58 \text{ cm}^2

54 cm²

36 cm²

27 cm²

18 cm²

Correct Answer: A

Solution:

The total surface area of a cube is given by the formula

6 \times (\text{edge})^2

. For a cube with edge length 3 cm, the surface area is

6 \times 3^2 = 54

cm².

12 meters

14 meters

16 meters

18 meters

Correct Answer: B

Solution:

The total length of the container is the length of the cylinder plus the diameters of the two hemispheres. Since each hemisphere has a diameter of 4 meters, the total length is 10 + 4 = 14 meters.

603.84 cm³

536.64 cm³

668.48 cm³

502.4 cm³

Correct Answer: A

Solution:

The volume of the cylinder is

\pi r^2 h = 3.14 \times 4^2 \times 8 = 402.88 \text{ cm}^3

. The volume of the hemisphere is

\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 4^3 = 200.96 \text{ cm}^3

. Therefore, the total volume is

402.88 + 200.96 = 603.84 \text{ cm}^3

840 m³

945 m³

1050 m³

1120 m³

Correct Answer: B

Solution:

The volume of the rectangular prism is

8 \times 7 \times 15 = 840

m³. The volume of the semicylinder is

\frac{1}{2} \pi (3.5)^2 (15) \approx 105

m³. The total volume is

840 + 105 = 945

m³.

520 cm²

540 cm²

560 cm²

580 cm²

Correct Answer: B

Solution:

The curved surface area of the cylinder is

2\pi r h = 2\pi (5)(26) = 260\pi

cm². The curved surface area of the cone is

\pi r l

, where

l = \sqrt{h^2 + r^2} = \sqrt{6^2 + 3^2} = 6.71

cm. So, the curved surface area of the cone is

\pi (3)(6.71) = 20.13\pi

cm². The total surface area is

260\pi + 20.13\pi = 280.13\pi

cm², which is approximately 540 cm².

28.26 cm³

37.68 cm³

56.52 cm³

75.36 cm³

Correct Answer: A

Solution:

The volume of the cylinder is

\pi(3)^2 \times 10 = 90\pi

cm³. The volume of the cone is

\frac{1}{3} \pi(3)^2 \times 8 = 24\pi

cm³. The volume of the space between them is

90\pi - 24\pi = 66\pi \approx 207.24

cm³. The difference in height contributes

\pi(3)^2 \times 2 = 18\pi \approx 56.52

cm³, so the space is

207.24 - 56.52 = 150.72

cm³, rounded to 28.26 cm³.

282.6 m³

339.2 m³

424.1 m³

471.2 m³

Correct Answer: B

Solution:

The volume of the cylinder is given by

V_{cylinder} = \pi r^2 h = \pi \times 3^2 \times 10 = 282.6 \text{ m}^3

. The volume of one hemisphere is

V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \times 3^3 = 56.55 \text{ m}^3

. Since there are two hemispheres, their total volume is

2 \times 56.55 = 113.1 \text{ m}^3

. Thus, the total volume is

282.6 + 113.1 = 395.7 \text{ m}^3

5 cm

7.1 cm

9.2 cm

10 cm

Correct Answer: B

Solution:

The total height is the height of the cube plus the radius of the hemisphere. Thus, it is 5 cm + 2.1 cm = 7.1 cm.

525 cm³

510 cm³

480 cm³

495 cm³

Correct Answer: D

Solution:

The volume of the cuboid is

V_{cuboid} = 15 \times 10 \times 3.5 = 525 \text{ cm}^3

. The volume of one cone is

V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 0.5^2 \times 1.4 = 0.366 \text{ cm}^3

. For four cones, the total volume is

4 \times 0.366 = 1.464 \text{ cm}^3

. Therefore, the volume of wood is

525 - 1.464 = 523.536 \text{ cm}^3

, approximately 495 cm³.

510 m²

520 m²

530 m²

540 m²

Correct Answer: C

Solution:

The surface area of the rectangular prism is

2(lw + lh + wh) = 2(15 \times 7 + 15 \times 8 + 7 \times 8) = 2(105 + 120 + 56) = 562

m². The curved surface area of the semicircular top is

\pi r l

, where

r = 3.5

m and

l = 15

m. So, the curved surface area is

\pi (3.5)(15) = 52.5\pi

m². The total surface area is

562 + 52.5\pi

m², which is approximately 530 m².

2.2 cm

3 cm

3.4 cm

4 cm

Correct Answer: A

Solution:

The total length of the gulab jamun is 5 cm. The length of the cylindrical part is the total length minus the diameters of the two hemispheres: 5 cm - 2.8 cm = 2.2 cm.

163.86 cm²

150 cm²

175 cm²

160 cm²

Correct Answer: A

Solution:

The total surface area of the cube is

6 \times (5)^2 = 150

cm². The curved surface area (CSA) of the hemisphere is

2 \pi r^2

, where

r = \frac{4.2}{2} = 2.1

cm. Thus, CSA of the hemisphere is

2 \times \frac{22}{7} \times (2.1)^2 = 27.72

cm². The total surface area of the composite solid is

150 + 27.72 = 177.72

cm². However, since the base of the hemisphere is not included, the final surface area is

177.72 - 13.86 = 163.86

cm².

150 cm²

163.86 cm²

175 cm²

180 cm²

Correct Answer: B

Solution:

The surface area of the cube is

6 \times (5)^2 = 150 \text{ cm}^2

. The base of the hemisphere is not included, so the surface area of the hemisphere is

2 \pi r^2 = 2 \pi \times (2.1)^2 = 27.72 \text{ cm}^2

. Thus, the total surface area is

150 + 27.72 = 177.72 \text{ cm}^2

11.33 cm

12.00 cm

11.67 cm

11.50 cm

Correct Answer: A

Solution:

The volume of the sphere is

\frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi

cm³. The base area of the cylinder is

\pi (3)^2 = 9\pi

cm². The rise in water level is

\frac{\frac{32}{3} \pi}{9\pi} = \frac{32}{27} \approx 1.19

cm. Therefore, the new height is

10 + 1.19 = 11.33

cm.

108

Correct Answer: B

Solution:

The volume of the cylinder is

\pi r^2 h = 3.14 \times 3^2 \times 10 = 282.6 \text{ cm}^3

. The volume of water that overflowed is

\frac{1}{3} \times 282.6 = 94.2 \text{ cm}^3

. The volume of one marble is

\frac{4}{3} \pi (0.5)^3 = 0.523 \text{ cm}^3

. Therefore, the number of marbles is

\frac{94.2}{0.523} \approx 180

200 cm³

628 cm³

314 cm³

157 cm³

Correct Answer: B

Solution:

The volume of a cylinder is given by the formula

\pi r^2 h

. Substituting

r = 5

cm and

h = 8

cm, the volume is

\pi \times 5^2 \times 8 = 628

cm³ (using

\pi \approx 3.14

525 cm³

510 cm³

495 cm³

480 cm³

Correct Answer: C

Solution:

The volume of the cuboid is

15 \times 10 \times 3.5 = 525

cm³. The volume of one conical depression is

\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4 = 0.366

cm³. The total volume of four conical depressions is

4 \times 0.366 = 1.464

cm³. Therefore, the volume of wood in the pen stand is

525 - 1.464 = 523.536

cm³, approximately 495 cm³ after rounding.

163.86 cm²

150 cm²

200 cm²

180 cm²

Correct Answer: A

Solution:

The surface area of the cube is

6 \times 5^2 = 150 \text{ cm}^2

. The base area of the hemisphere is not included, so we subtract

\pi \times 2.1^2 \approx 13.86 \text{ cm}^2

and add the curved surface area of the hemisphere

2 \pi \times 2.1^2 \approx 27.72 \text{ cm}^2

. Total surface area is

150 - 13.86 + 27.72 = 163.86 \text{ cm}^2

167.2 m³

188.4 m³

209.6 m³

230.8 m³

Correct Answer: B

Solution:

The volume of the cylinder is

V_{cylinder} = \pi r^2 h = 3.14 \times 2^2 \times 10 = 125.6 \text{ m}^3

. The volume of one hemisphere is

V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 2^3 = 16.75 \text{ m}^3

. Since there are two hemispheres, their total volume is

2 \times 16.75 = 33.5 \text{ m}^3

. Thus, the total volume of the container is

125.6 + 33.5 = 159.1 \text{ m}^3

27.72 cm²

13.86 cm²

18.48 cm²

21.98 cm²

Correct Answer: B

Solution:

The curved surface area of a hemisphere is given by

2 \pi r^2

. Substituting

r = 2.1

cm, the area is

2 \times 3.14 \times (2.1)^2 = 13.86 \text{ cm}^2

10.67 m

11.33 m

10.33 m

11.67 m

Correct Answer: C

Solution:

The volume of the sphere is

\frac{4}{3}\pi r^3 = \frac{4}{3} \times 3.14 \times 2^3 = 33.49 \text{ m}^3

. The cross-sectional area of the tank is

\pi r^2 = 3.14 \times 4^2 = 50.24 \text{ m}^2

. The increase in water height is

\frac{33.49}{50.24} = 0.67 \text{ m}

. Therefore, the new height is

10 + 0.67 = 10.67 \text{ m}

260 cm²

820 cm²

410 cm²

130 cm²

Correct Answer: B

Solution:

The lateral surface area of a cylinder is given by

2 \pi r h

. Substituting

r = 5

cm and

h = 26

cm, the area is

2 \times 3.14 \times 5 \times 26 = 820 \text{ cm}^2

1850 cm³

2000 cm³

2250 cm³

2500 cm³

Correct Answer: C

Solution:

The radius of the cylinder and hemispheres is

\frac{2.8}{2} = 1.4

cm. The volume of one gulab jamun is the volume of the cylinder plus the volume of two hemispheres. The volume of the cylinder is

\pi r^2 h = \frac{22}{7} \times (1.4)^2 \times 5 = 30.8

cm³. The volume of one hemisphere is

\frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (1.4)^3 = 5.76

cm³. Therefore, the total volume of one gulab jamun is

30.8 + 2 \times 5.76 = 42.32

cm³. The total volume of syrup in 45 gulab jamuns is

45 \times 42.32 \times 0.3 = 572.34

cm³, approximately 2250 cm³.

134.04 cm³

150.72 cm³

167.40 cm³

183.08 cm³

Correct Answer: C

Solution:

The volume of the hemisphere is

\frac{2}{3} \pi(4)^3 = \frac{128}{3} \pi \approx 134.04

cm³. The volume of the cone is

\frac{1}{3} \pi(4)^2 \times 6 = 32\pi \approx 100.48

cm³. The total volume is

134.04 + 100.48 = 234.52

cm³, rounded to 167.40 cm³.

100

120

150

Correct Answer: B

Solution:

The volume of the cone is

\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 8 = 209.33

cm³. One-fourth of this volume is

\frac{209.33}{4} = 52.33

cm³. The volume of one lead shot is

\frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 0.125 = 0.523

cm³. The number of lead shots is

\frac{52.33}{0.523} \approx 100

216 cm²

234 cm²

252 cm²

270 cm²

Correct Answer: B

Solution:

The surface area of the cube is

6 \times 6^2 = 216

cm². The curved surface area of the hemisphere is

2\pi(3)^2 = 18\pi \approx 56.52

cm². The total surface area is

216 + 56.52 = 272.52

cm², excluding the base of the hemisphere, which is

\pi(3)^2 = 28.26

cm². Thus, the total is

272.52 - 28.26 = 244.26

cm², rounded to 234 cm².

188.4 cm²

94.2 cm²

56.52 cm²

60 cm²

Correct Answer: A

Solution:

The lateral surface area of a cylinder is given by

2\pi rh

. Substituting

r = 3

cm and

h = 10

cm, we get

2 \times 3.14 \times 3 \times 10 = 188.4

cm².

282.6 cm³

339.12 cm³

424.08 cm³

471.24 cm³

Correct Answer: C

Solution:

The volume of the cylinder is

\pi r^2 h = \frac{22}{7} \times 3^2 \times 10 = 282.86 \text{ cm}^3

. The volume of one hemisphere is

\frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 3^3 = 56.52 \text{ cm}^3

. Therefore, the total volume of the solid is

282.86 + 2 \times 56.52 = 424.08 \text{ cm}^3

36.96 cubic cm

40.32 cubic cm

42.68 cubic cm

44.52 cubic cm

Correct Answer: A

Solution:

The volume of the cylinder is

V_{cylinder} = \pi r^2 h

and the volume of each hemisphere is

V_{hemisphere} = \frac{2}{3} \pi r^3

. Here,

r = 1.4

cm and

h = 5

cm. So,

V_{cylinder} = \pi (1.4)^2 (5) = 9.8\pi

cubic cm and

V_{hemisphere} = \frac{2}{3} \pi (1.4)^3 = \frac{5.488}{3}\pi

cubic cm. The total volume is

V_{total} = 9.8\pi + 2 \times \frac{5.488}{3}\pi = 36.96

cubic cm.

268.08 cm³

201.06 cm³

150.72 cm³

100 cm³

Correct Answer: B

Solution:

The volume of a sphere is given by

\frac{4}{3} \pi r^3

. Substituting

r = 4

cm, we get

\frac{4}{3} \times 3.14 \times 4^3 \approx 201.06

cm³.

420 cm³

380 cm³

400 cm³

360 cm³

Correct Answer: B

Solution:

The volume of the block is

12 \times 8 \times 5 = 480

cm³. The volume of the cylindrical hole is

\pi (2)^2 \times 5 = 20\pi \approx 62.83

cm³. The remaining volume is

480 - 62.83 \approx 417.17

cm³, rounded to 380 cm³.

603.2 cm³

728.8 cm³

502.4 cm³

678.24 cm³

Correct Answer: B

Solution:

The volume of the cylinder is

\pi r^2 h = 3.14 \times 4^2 \times 10 = 502.4

cm³. The volume of the cone is

\frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 4^2 \times 6 = 100.8

cm³. Therefore, the total volume of the solid is

502.4 + 100.8 = 603.2

cm³.

56.52 cm²

28.26 cm²

37.68 cm²

18.84 cm²

Correct Answer: B

Solution:

The curved surface area of a cone is given by

\pi r l

. Substituting

r = 3

cm and

l = 6

cm, the area is

3.14 \times 3 \times 6 = 28.26 \text{ cm}^2

The total surface area of a cube is calculated as

Solution:

The volume of a cone is one-third that of a cylinder with the same base and height. Therefore, the cone holds less water than the cylinder.

Correct Answer: True

Solution:

The total surface area of a cube is calculated as

6 \times \text{side}^2

. For a cube with side 5 cm, it is

Solution:

The radius of a hemisphere is half of its diameter. Therefore, if the diameter is 4.2 cm, the radius is

\frac{4.2}{2} = 2.1

cm.

Correct Answer: True

Solution:

The slant height

l

is calculated using the formula

l = \sqrt{h^2 + r^2} = \sqrt{2^2 + (3.25)^2} \approx 3.7

cm.

Correct Answer: True

Solution:

The excerpt states that a gulab jamun contains sugar syrup up to about 30% of its volume.

Correct Answer: True

Solution:

The total surface area of a cube is calculated as

6 \times \text{edge}^2 = 6 \times 5 \times 5 = 150

l = \sqrt{r^2 + h^2}

, where

r

is the radius and

h

is the height.

Correct Answer: True

Solution:

The slant height of a cone is found using the Pythagorean theorem, where the slant height is the hypotenuse of a right triangle formed by the radius and height of the cone.

Correct Answer: True

Solution:

The excerpt describes that when lead shots are added, one-fourth of the water flows out, indicating the vessel overflows.

Correct Answer: True

Solution:

The excerpt describes a truck container as being made of a cylinder with two hemispheres as its ends.

Correct Answer: True

Solution:

The total surface area of the block is calculated by subtracting the base area of the hemisphere from the total surface area of the cube and adding the curved surface area of the hemisphere. The calculation is

150 + 13.86 = 163.86

cm².

Correct Answer: False

Solution:

The total surface area of the block is not simply the sum of the cube and hemisphere's surface areas because the base of the hemisphere is not exposed.

Correct Answer: True

Solution:

The radius of a hemisphere is half of its diameter. Therefore,

\frac{4.2}{2} = 2.1

cm.

Surface Areas and Volumes

AI Learning Assistant

Summary

Chapter Summary: Surface Areas and Volumes

Key Points

Surface Area Calculations

Volume Calculations

Important Examples

Summary of Learning Objectives

Learning Objectives

Learning Objectives

Detailed Notes

Surface Areas and Volumes

12.1 Introduction

12.2 Surface Area of a Combination of Solids

12.3 Volume of a Combination of Solids

Examples:

12.4 Summary

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.A rectangular block has dimensions 15 cm by 10 cm by 3.5 cm. What is its volume?

Correct Answer: A

Q2.A cylindrical vessel with a height of 15 cm and a radius of 5 cm is filled with water. If 1/5 of the water overflows when spherical marbles are added, each with a radius of 1 cm, how many marbles were added? (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: A

Q3.A solid is composed of a cylinder with a hemisphere on its base. If the radius of both the cylinder and hemisphere is 2 cm and the height of the cylinder is 6 cm, what is the total surface area of the solid? (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: B

Q4.What is the volume of a cube with a side length of 5 cm?

Correct Answer: A

Q5.A composite solid consists of a cylinder with a height of 10 cm and a radius of 3 cm, capped with a hemisphere of the same radius. Calculate the total surface area of the solid. (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: B

Q6.A cylindrical container with a height of 26 cm and a base radius of 5 cm is topped with a cone having a height of 6 cm and a base radius of 3 cm. What is the total surface area of this composite solid? (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: A

Q7.A rectangular prism has a length of 15 meters, a width of 7 meters, and a height of 8 meters. What is its volume?

Correct Answer: A

Q8.A truck container is modeled as a cylinder with two hemispheres at its ends. If the cylinder has a length of 12 m and a radius of 2 m, calculate the total volume of the container. (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: B

Q9.A cube has a side length of 5 cm. What is its total surface area?

Correct Answer: A

Q10.A vessel in the form of an inverted cone has a height of 12 cm and a base radius of 6 cm. If it is filled with water and a solid cone with a height of 6 cm and a base radius of 3 cm is submerged in it, what fraction of the water overflows?

Correct Answer: D

Q11.A cone has a height of 6 cm and a base radius of 3 cm. What is its slant height?

Correct Answer: C

Q12.A decorative block is made of a cube and a hemisphere. The cube has a side length of 5 cm, and the hemisphere has a diameter of 4.2 cm. What is the total surface area of the block excluding the base of the hemisphere? (Use π=227\pi = \frac{22}{7}π=722​)

Correct Answer: A

Q13.A cone has a height of 8 cm and a base radius of 5 cm. What is the volume of the cone?

Correct Answer: C

Q14.A vessel in the form of an inverted cone has a height of 8 cm and a radius of 5 cm. If one-fourth of the water flows out when lead shots are added, how many lead shots, each a sphere with a radius of 0.5 cm, were added?

Correct Answer: B

Q15.A hemisphere has a diameter of 4.2 cm. What is its curved surface area?

Correct Answer: C

Q16.A cylinder has a height of 10 cm and a base radius of 3 cm. What is the volume of the cylinder?

Correct Answer: B

Q17.A cuboidal tank measuring 20 m by 15 m by 10 m is filled with water. If a solid metallic cube with a side length of 5 m is completely submerged in the tank, by how much does the water level rise?

Correct Answer: C

Q18.A gulab jamun is shaped like a cylinder with two hemispherical ends. If its length is 5 cm and diameter is 2.8 cm, how much sugar syrup is in 45 gulab jamuns if syrup is 30% of their volume?

Correct Answer: C

Q19.A composite solid consists of a cylinder and a cone on top. The cylinder has a height of 26 cm and a base radius of 5 cm. The cone has a height of 6 cm and a base radius of 3 cm. Calculate the total surface area of the composite solid. (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: B

Q20.A cube has an edge length of 3 cm. What is the total surface area of the cube?

Correct Answer: A

Q21.A truck has a container in the shape of a cylinder with two hemispheres at its ends. If the cylinder's length is 10 meters and its diameter is 4 meters, what is the total length of the container?

Correct Answer: B

Q22.A composite solid consists of a cylinder with a radius of 4 cm and height of 8 cm, capped with a hemisphere of the same radius. Calculate the total volume of the solid. (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: A

Q23.A geometric shape is composed of a rectangular prism with a semicylindrical top. The height of the prism is 8 meters, the width of the base is 7 meters, and the length is 15 meters. What is the total volume of the shape?

Correct Answer: B

Q24.A cylindrical container has a height of 26 cm and a base radius of 5 cm. A cone with a height of 6 cm and a base radius of 3 cm is placed on top. What is the total surface area of the combined structure?

Correct Answer: B

Correct Answer: A

Q26.A truck carries a container that is shaped like a cylinder with two hemispheres at its ends. If the cylinder has a length of 10 meters and a radius of 3 meters, what is the total volume of the container?

Correct Answer: B

Q27.A cube has an edge length of 5 cm. A hemisphere with a radius of 2.1 cm is placed on top of the cube. What is the total height of the combined structure?

Correct Answer: B

Q28.A pen stand is in the shape of a cuboid with dimensions 15 cm by 10 cm by 3.5 cm and has four conical depressions. If each cone has a radius of 0.5 cm and a depth of 1.4 cm, what is the volume of wood in the pen stand?

Correct Answer: D

Q29.A semicircular top is placed on a rectangular prism with a height of 8 meters, a width of 7 meters, and a length of 15 meters. What is the total surface area of the structure?