Composite Solids → Composite Area, Surface Area, And Volume

1) A playground is a rectangle 12 m by 20 m with a semicircle attached on the side with diameter 12 m. What is the approximate total area?

296.5 m²

353.7 m²

413.7 m²

500.0 m²

Rectangle area = 12 * 20 = 240.

Semicircle area = 0.5 * π * 6^2 = 0.5 * 3.14 * 36 = 56.52.

Total ≈ 240 + 56.52 = 296.52 ≈ 296.5 m².

Explanation

Rectangle area = 12 * 20 = 240.

Semicircle area = 0.5 * π * 6^2 = 0.5 * 3.14 * 36 = 56.52.

Total ≈ 240 + 56.52 = 296.52 ≈ 296.5 m².

2) A square garden is 8 m by 8 m. A circular flower bed of radius 3 m is in the center. What is the approximate area not occupied by the circle?

46.3 m²

39.7 m²

35.7 m²

36.0 m²

Square area = 8 * 8 = 64.

Circle area = π * 3^2 = 3.14 * 9 = 28.26.

Remaining ≈ 64 − 28.26 = 35.74 ≈ 35.7 m².

Explanation

Square area = 8 * 8 = 64.

Circle area = π * 3^2 = 3.14 * 9 = 28.26.

Remaining ≈ 64 − 28.26 = 35.74 ≈ 35.7 m².

3) A basketball court is a 28 m by 15 m rectangle with a semicircle of diameter 15 m at each short end. What is the total approximate area?

420.0 m²

596.7 m²

660.0 m²

750.7 m²

Rectangle area = 28 * 15 = 420.

Two semicircles make one full circle (r = 7.5): circle area = 3.14 * 7.5^2 = 3.14 * 56.25 = 176.625.

Total ≈ 420 + 176.625 = 596.625 ≈ 596.7 m².

Explanation

Rectangle area = 28 * 15 = 420.

Two semicircles make one full circle (r = 7.5): circle area = 3.14 * 7.5^2 = 3.14 * 56.25 = 176.625.

Total ≈ 420 + 176.625 = 596.625 ≈ 596.7 m².

4) A swimming pool is a 20 m by 10 m rectangle with a semicircle attached along the 20 m side. What is the approximate total area?

200.0 m²

212.1 m²

357.1 m²

414.2 m²

Rectangle area = 20 * 10 = 200.

Semicircle radius = 10, area = 0.5 * 3.14 * 10^2 = 157.0.

Total ≈ 200 + 157.0 = 357.0 ≈ 357.1 m².

Explanation

Rectangle area = 20 * 10 = 200.

Semicircle radius = 10, area = 0.5 * 3.14 * 10^2 = 157.0.

Total ≈ 200 + 157.0 = 357.0 ≈ 357.1 m².

5) A rectangular field is 50 m by 30 m with a semicircle on a 30 m side. What is the approximate total area?

1500.0 m²

1570.5 m²

1853.4 m²

2000.0 m²

Rectangle = 50 * 30 = 1500.

Semicircle radius = 15: area = 0.5 * 3.14 * 15^2 = 0.5 * 3.14 * 225 = 353.25 m²

Total ≈ 1500 + 353.25 = 1853.25 ≈ 1853.3 – 1853.4 m².

Explanation

Rectangle = 50 * 30 = 1500.

Semicircle radius = 15: area = 0.5 * 3.14 * 15^2 = 0.5 * 3.14 * 225 = 353.25 m²

Total ≈ 1500 + 353.25 = 1853.25 ≈ 1853.3 – 1853.4 m².

6) A path is a 30 m by 2 m rectangle with a semicircle (diameter 2 m) on each end. What is the approximate total area?

60.0 m²

63.1 m²

70.3 m²

72.8 m²

Rectangle = 30 * 2 = 60.

Two semicircles (d = 2) form a circle (r = 1): circle area = 3.14 * 1^2 = 3.14.

Total ≈ 60 + 3.14 = 63.14 ≈ 63.1 m².

Explanation

Rectangle = 30 * 2 = 60.

Two semicircles (d = 2) form a circle (r = 1): circle area = 3.14 * 1^2 = 3.14.

Total ≈ 60 + 3.14 = 63.14 ≈ 63.1 m².

7) A running track is a 60 m by 20 m rectangle with a semicircle of radius 10 m at each short end. What is the approximate total area?

1514.0 m²

1828.5 m²

1920.0 m²

1205.4 m²

Rectangle = 60 * 20 = 1200.

Two semicircles form a circle area = 3.14 * 10^2 = 314.

Total ≈ 1200 + 314 = 1514 m² ≈ 1514.0 m².

Explanation

Rectangle = 60 * 20 = 1200.

Two semicircles form a circle area = 3.14 * 10^2 = 314.

Total ≈ 1200 + 314 = 1514 m² ≈ 1514.0 m².

8) A billboard is a 18 ft by 10 ft rectangle with a semicircle on the 18 ft side. What is the approximate total area?

280 ft²

300 ft²

307.2 ft²

320 ft²

Rectangle = 18 * 10 = 180.

Semicircle r = 9 ft: 0.5 * 3.14 * 9^2 = 127.17.

Total ≈ 180 + 127.17 = 307.17 ≈ 307.2 ft².

Explanation

Rectangle = 18 * 10 = 180.

Semicircle r = 9 ft: 0.5 * 3.14 * 9^2 = 127.17.

Total ≈ 180 + 127.17 = 307.17 ≈ 307.2 ft².

9) A cylinder (r = 5 cm, h = 10 cm) is attached to a cube (side 10 cm). Find the approximate total volume.

1285 cm³

1785 cm³

2000 cm³

2142 cm³

Cube volume = 10^3 = 1000.

Cylinder volume = 3.14 * 5^2 * 10 = 3.14 * 25 * 10 = 785. Total ≈ 1000 + 785 = 1785 cm³.

Explanation

Cube volume = 10^3 = 1000.

Cylinder volume = 3.14 * 5^2 * 10 = 3.14 * 25 * 10 = 785. Total ≈ 1000 + 785 = 1785 cm³.

10) A cone (r = 3 cm, h = 6 cm) sits on a cylinder (r = 3 cm, h = 8 cm). Approximate total volume?

282.7 cm³

226.2 cm³

339.3 cm³

452.4 cm³

Cylinder = 3.14 * 3^2 * 8 = 226.08.

Cone = (1/3) * 3.14 * 3^2 * 6 = 56.52.

Total ≈ 226.08 + 56.52 = 282.60 ≈ 282.7 cm³.

Explanation

Cylinder = 3.14 * 3^2 * 8 = 226.08.

Cone = (1/3) * 3.14 * 3^2 * 6 = 56.52.

Total ≈ 226.08 + 56.52 = 282.60 ≈ 282.7 cm³.

11) A hemisphere has radius 5 m. The volume of one half (the hemisphere) is _______.

Sphere volume = (4/3)πr^3 = (4/3) * 3.14 * 125 ≈ 523.33.

Hemisphere is half: ≈ 261.67 ≈ 261.8 m³.

Explanation

Sphere volume = (4/3)πr^3 = (4/3) * 3.14 * 125 ≈ 523.33.

Hemisphere is half: ≈ 261.67 ≈ 261.8 m³.

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12) A silo is a cylinder (r = 4 m, h = 10 m) topped by a hemisphere (r = 4 m). The approximate total volume is _______.

Cylinder = 3.14 * 4^2 * 10 = 3.14 * 16 * 10 = 502.4.

Hemisphere = (2/3) * 3.14 * 4^3 = (2/3) * 3.14 * 64 = 133.97.

Total ≈ 502.4 + 133.97 = 636.37 ≈ 636.4 m³.

Explanation

Cylinder = 3.14 * 4^2 * 10 = 3.14 * 16 * 10 = 502.4.

Hemisphere = (2/3) * 3.14 * 4^3 = (2/3) * 3.14 * 64 = 133.97.

Total ≈ 502.4 + 133.97 = 636.37 ≈ 636.4 m³.

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13) A rectangular prism 8 cm × 6 cm × 4 cm has a triangular prism roof (triangle base 8 cm × 4 cm, length 6 cm). The total volume is _______.

Rectangular prism = 8 * 6 * 4 = 192.

Triangular prism = (1/2 * 8 * 4) * 6 = 16 * 6 = 96.

Total = 192 + 96 = 288 cm³.

Explanation

Rectangular prism = 8 * 6 * 4 = 192.

Triangular prism = (1/2 * 8 * 4) * 6 = 16 * 6 = 96.

Total = 192 + 96 = 288 cm³.

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14) A 20 cm × 10 cm × 5 cm prism has a quarter-cylinder attached (r = 5 cm, height = 10 cm). The approximate total volume is _______.

Prism = 20 * 10 * 5 = 1000.

Quarter cylinder = (1/4) * π * 5^2 * 10 = 0.25 * 3.14 * 25 * 10 = 196.25.

Total ≈ 1000 + 196.25 = 1196.25 ≈ 1196.3 cm³.

Explanation

Prism = 20 * 10 * 5 = 1000.

Quarter cylinder = (1/4) * π * 5^2 * 10 = 0.25 * 3.14 * 25 * 10 = 196.25.

Total ≈ 1000 + 196.25 = 1196.25 ≈ 1196.3 cm³.

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15) A cylinder (r = 4 cm, h = 10 cm) with a cone (r = 4 cm, h = 6 cm) on top has exposed outside surface area (exclude the shared circle) of _______.

Cylinder lateral = 2πrh = 2 * 3.14 * 4 * 10 = 251.2.

Cylinder bottom = πr^2 = 3.14 * 16 = 50.24.

Cone slant l = √(r^2 + h^2) = √(16 + 36) = √52 ≈ 7.21;

cone lateral = π r l ≈ 3.14 * 4 * 7.21 ≈ 90.5.

Total ≈ 251.2 + 50.24 + 90.5 ≈ 392.0–392.2 cm².

Explanation

Cylinder lateral = 2πrh = 2 * 3.14 * 4 * 10 = 251.2.

Cylinder bottom = πr^2 = 3.14 * 16 = 50.24.

Cone slant l = √(r^2 + h^2) = √(16 + 36) = √52 ≈ 7.21;

cone lateral = π r l ≈ 3.14 * 4 * 7.21 ≈ 90.5.

Total ≈ 251.2 + 50.24 + 90.5 ≈ 392.0–392.2 cm².

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16) Two semicircles attached to the short ends of a rectangle have the same combined area as one full circle whose diameter equals the rectangle’s width.

True

False

Two semicircles attached to the short ends of a rectangle have the same combined area as one full circle whose diameter equals the rectangle’s width. This works because each semicircle has radius w/2 and therefore area (1/2)π(w/2)², so adding two identical semicircles gives π(w/2)², which is exactly the area of one full circle with radius w/2, proving that the two semicircles together always form a complete circle of diameter w.

Explanation

Two semicircles attached to the short ends of a rectangle have the same combined area as one full circle whose diameter equals the rectangle’s width. This works because each semicircle has radius w/2 and therefore area (1/2)π(w/2)², so adding two identical semicircles gives π(w/2)², which is exactly the area of one full circle with radius w/2, proving that the two semicircles together always form a complete circle of diameter w.

17) When a cone is placed on top of a cylinder with the same radius, the circular interface between them should be included once in the total exposed surface area.

True

False

The circular face where the cone and cylinder touch is completely internal and never exposed to the outside, so it should not be counted in surface area at all, meaning only the external surfaces—the curved cylinder, the curved cone, and the cylinder’s uncovered base—contribute to the total external area.

Explanation

The circular face where the cone and cylinder touch is completely internal and never exposed to the outside, so it should not be counted in surface area at all, meaning only the external surfaces—the curved cylinder, the curved cone, and the cylinder’s uncovered base—contribute to the total external area.

18) The shared circle is internal and not exposed; it is not counted in the outside surface area. Only exposed surfaces are included.

True

False

The volume of a hemisphere is (2/3)πr³. Because a hemisphere is defined as exactly half of a sphere and the sphere’s volume is (4/3)πr³, dividing this by two gives (2/3)πr³, which correctly represents the total space contained in a half-spherical solid.

Explanation

The volume of a hemisphere is (2/3)πr³. Because a hemisphere is defined as exactly half of a sphere and the sphere’s volume is (4/3)πr³, dividing this by two gives (2/3)πr³, which correctly represents the total space contained in a half-spherical solid.

19) If a circular hole of radius r is cut out of a square of side s, the remaining area is s^2 − πr^2.

True

False

Removing a region of area πr² from the square’s area s² follows the standard composite-area method in which you subtract the area of the removed shape from the total, leaving the portion of the square that remains after the circular section is cut out.

Explanation

Removing a region of area πr² from the square’s area s² follows the standard composite-area method in which you subtract the area of the removed shape from the total, leaving the portion of the square that remains after the circular section is cut out.

20) The lateral surface area of a cylinder is πr^2h.

True

False

This expression is incorrect because πr²h belongs to the volume formula, while the lateral area comes from unwrapping the curved surface into a rectangle of height h and width equal to the circumference 2πr, giving the correct formula 2πrh, which depends on both radius and height.

Explanation

This expression is incorrect because πr²h belongs to the volume formula, while the lateral area comes from unwrapping the curved surface into a rectangle of height h and width equal to the circumference 2πr, giving the correct formula 2πrh, which depends on both radius and height.

Composite Solids → Composite Area, Surface Area, and Volume