Exam practice: Grade 10 Math Unit 5

a.

What do GO and GM stand for in the pyramid?

Answer

GO is the vertical height h; GM is the slant height l.

b.

If XY = a cm, what is the length of OM?

Answer

O is the center of the square, so OM is half a side: OM = $\frac a2$ cm.

c.

If the vertical height is h, the slant height l and the base edge a, write the relation between them.

Answer

$l^2 = h^2 + \left(\frac a2\right)^2$ (Pythagoras’ theorem in triangle GOM).

d.

What do GZ, GY, GX and GW stand for? Are they equal?

Answer

They are the lateral edges. Yes, all four are equal: the apex is straight above the center, which is the same distance from every corner.

a.

Write the total surface area of a square pyramid with base edge m cm and slant height n cm.

Answer

TSA = $(m^2 + 2mn)$ cm².

b.

What is the lateral surface area of a square pyramid with base edge p cm and slant height q cm?

Answer

LSA = 2pq cm².

c.

Write the volume of a square pyramid with base edge r cm and vertical height h cm.

Answer

V = $\frac13 r^2 h$ cm³.

d.

A square pyramid’s base has area z square units and its total surface area is y square units. What is the area of its triangular faces?

Answer

The triangles are the whole surface less the base: (y − z) square units.

a.

Find the volume of a square pyramid whose base has area 64 cm² and whose height is 15 cm.

Answer

$V = \frac13 \times 64 \times 15 =$ 320 cm³.

b.

A square pyramid’s total surface area is 285 cm² and its triangular faces have area 192 cm² in all. Find the area of its base.

Answer

285 − 192 = 93 cm².

c.

Find the slant height and the length of a lateral edge of this pyramid, where PO = 8 cm and OM = 12 cm.

Figure 5.6

Answer

$l^2 = PO^2 + OM^2 = 64 + 144 = 208$ , so $l = \sqrt{208} = 4\sqrt{13} \approx$ 14.42 cm.
OM = a/2 = 12, so MC = 12 too: $e^2 = l^2 + 12^2 = 208 + 144 = 352$ , so $e = \sqrt{352} = 4\sqrt{22} \approx$ 18.76 cm.

d.

Find the vertical height and the length of a lateral edge of this pyramid, where the slant height SA = 13 cm and the base edge NO = 10 cm.

Figure 5.7

Answer

GA = 10 ÷ 2 = 5. $h^2 = 13^2 - 5^2 = 144$ , so h = 12 cm.
$e^2 = 13^2 + 5^2 = 194$ , so $e = \sqrt{194} \approx$ 13.93 cm.

a.

Figure 5.8

Answer

$h^2 = 10^2 - 6^2 = 64$ , h = 8.
TSA = 144 + 2 × 12 × 10 = 384 cm²; $V = \frac13 \times 144 \times 8 =$ 384 cm³.

b.

Figure 5.9

Answer

$l^2 = 12^2 + 5^2 = 169$ , l = 13.
TSA = 100 + 2 × 10 × 13 = 360 cm²; $V = \frac13 \times 100 \times 12 =$ 400 cm³.

c.

Figure 5.10

Answer

$h^2 = 13^2 - 5^2 = 144$ , h = 12.
TSA = 100 + 2 × 10 × 13 = 360 cm²; $V = \frac13 \times 100 \times 12 =$ 400 cm³.

d.

Figure 5.11

Answer

$l^2 = 8^2 + 6^2 = 100$ , l = 10.
TSA = 144 + 2 × 12 × 10 = 384 cm²; $V = \frac13 \times 144 \times 8 =$ 384 cm³.

e.

Figure 5.12

Answer

$l^2 = 6^2 + 8^2 = 100$ , l = 10.
TSA = 256 + 2 × 16 × 10 = 576 cm²; $V = \frac13 \times 256 \times 6 =$ 512 cm³.

f.

Figure 5.13

Answer

Check: $15^2 + 8^2 = 289 = 17^2$ . ✓
TSA = 256 + 2 × 16 × 17 = 800 cm²; $V = \frac13 \times 256 \times 15 =$ 1,280 cm³.

5.

Sophiya has a pot shaped like a square pyramid, 25 cm high on a base 30 cm square. How much water does it hold when full?

Answer

$V = \frac13 \times 30^2 \times 25 = 7500$ cm³. Since 1,000 cm³ is a liter: 7,500 cm³, or 7.5 liters.

6.

A square pyramid has total surface area 800 cm² and base edge 16 cm. Find the area of its triangular faces and its volume.

Answer

Triangles: 800 − 256 = 544 cm². Then 2 × 16 × l = 544, so l = 17, and $h^2 = 17^2 - 8^2 = 225$ , h = 15.
$V = \frac13 \times 256 \times 15 =$ 1,280 cm³.

7.

A fish tank shaped like a square pyramid has total surface area 400 square inches and slant height 15 inches. Find its volume.

Answer

$a^2 + 30a - 400 = 0$ , so (a + 40)(a − 10) = 0 and a = 10 in.
$h^2 = 15^2 - 5^2 = 200$ , $h = \sqrt{200} \approx 14.142$ in.
$V = \frac13 \times 100 \times 14.142 \approx$ 471.40 cubic inches.

8.

Tourists in Egypt reach a pyramid. Their guide says its base has area 3,600 m² and its height is 50 m. A mathematics teacher in the group works out its total surface area from this. What does she get?

Answer

$a = \sqrt{3600} = 60$ m. $l^2 = 50^2 + 30^2 = 3400$ , $l \approx 58.31$ m.
TSA $= 3600 + 2 \times 60 \times 58.31 \approx$ 10,597.14 m².

a.

How many square meters of cloth make the tent (its four sloping sides)?

Answer

$l^2 = 50^2 + 16^2 = 2756$ , $l \approx 52.498$ m.
Cloth $= 2al = 2 \times 32 \times 52.498 \approx$ 3,359.85 m².

b.

What does the cloth cost at Rs 500 a square meter?

Answer

$3359.85 \times 500 \approx$ Rs 16,79,923.81.

a.

How much of the field does the tower stand on?

Answer

$(a/2)^2 = 50^2 - 40^2 = 900$ , so a/2 = 30 and a = 60 ft.
Area $= 60^2 =$ 3,600 sq ft.

b.

If the rent of the field is Rs 50 a square foot a year, how much rent does Nepal Telecom pay in 20 years?

Answer

3600 × 50 × 20 = Rs 36,00,000.

11.

A chocolate is a square pyramid with total surface area 1,920 cm² and base edge 30 cm. Find its vertical height.

Answer

900 + 2 × 30 × l = 1920, so 60l = 1020 and l = 17 cm.
$h^2 = 17^2 - 15^2 = 64$ : h = 8 cm.

a.

What do h, l and r stand for?

Answer

h is the vertical height, l the slant height and r the radius of the base (a circle).

b.

Write the relation between l, h and r.

Answer

$l^2 = h^2 + r^2$ .

c.

Find h if l = 13 cm and r = 5 cm.

Answer

$h^2 = 169 - 25 = 144$ : h = 12 cm.

a.

Write the curved surface area of a cone with base radius r cm and slant height l cm.

Answer

$\pi r l$ cm².

b.

Write the total surface area of a cone with base radius x cm and slant height y cm.

Answer

$\pi x(x + y)$ cm².

c.

What is the volume of a cone whose base is p cm across (diameter) and whose height is q cm?

Answer

r = p/2, so $V = \frac13 \pi \left(\frac p2\right)^2 q =$ $\frac{1}{12} \pi p^2 q$ cm³.

a.

Find the volume of a cone whose base has area 81 cm² and whose height is 15 cm.

Answer

$V = \frac13 \times 81 \times 15 =$ 405 cm³.

b.

A cone’s total surface area is 250 cm² and its base has area 118 cm². Find its curved surface area.

Answer

250 − 118 = 132 cm².

c.

Find the volume of a cone with slant height 5 cm and base radius 4 cm.

Answer

$h^2 = 25 - 16 = 9$ , h = 3. $V = \frac13 \pi \times 16 \times 3 =$ $16\pi \approx 50.29$ cm³.

a.

Figure 5.16

Answer

r = 7; $l = \sqrt{441 + 49} = \sqrt{490} \approx 22.136$ cm.
CSA $= \frac{22}{7} \times 7 \times 22.136 \approx$ 486.99 cm²; TSA = 486.99 + 154 = 640.99 cm²; $V = \frac13 \times \frac{22}{7} \times 49 \times 21 =$ 1,078 cm³.

b.

Figure 5.17

Answer

$h = \sqrt{289 - 64} = 15$ cm.
CSA $= \pi \times 8 \times 17 =$ $136\pi \approx 427.43$ cm²; TSA $= \pi \times 8 \times 25 =$ $200\pi \approx 628.57$ cm²; $V = \frac13 \pi \times 64 \times 15 =$ $320\pi \approx 1005.71$ cm³.

c.

Figure 5.18

Answer

$r^2 = 256 - 196 = 60$ , $r = \sqrt{60} \approx 7.746$ cm.
CSA $= \pi \times 7.746 \times 16 \approx$ $123.94\pi \approx 389.51$ cm²; TSA $= 123.94\pi + 60\pi =$ $183.94\pi \approx 578.09$ cm²; $V = \frac13 \pi \times 60 \times 14 =$ $280\pi = 880$ cm³.

d.

Figure 5.19

Answer

CSA $= \frac{22}{7} \times 7 \times 15 =$ 330 cm²; TSA = 330 + 154 = 484 cm².
$h = \sqrt{225 - 49} = \sqrt{176} \approx 13.266$ cm; $V = \frac13 \times 154 \times 13.266 \approx$ 681.01 cm³.

e.

Figure 5.20

Answer

$l = \sqrt{144 + 25} = 13$ cm.
CSA $= \pi \times 5 \times 13 =$ $65\pi \approx 204.29$ cm²; TSA $= \pi \times 5 \times 18 =$ $90\pi \approx 282.86$ cm²; $V = \frac13 \pi \times 25 \times 12 =$ $100\pi \approx 314.29$ cm³.

f.

Figure 5.21

Answer

r = 12; $l = \sqrt{25 + 144} = 13$ cm.
CSA $= \pi \times 12 \times 13 =$ $156\pi \approx 490.29$ cm²; TSA $= \pi \times 12 \times 25 =$ $300\pi \approx 942.86$ cm²; $V = \frac13 \pi \times 144 \times 5 =$ $240\pi \approx 754.29$ cm³.

5.

A right circular cone has volume $100\pi$ cm³ and height 12 cm. Find its slant height.

Answer

$\frac13 \pi r^2 \times 12 = 100\pi$ , so $r^2 = 25$ and r = 5. $l = \sqrt{144 + 25} =$ 13 cm.

6.

A cone-shaped tent has volume 1,232 m³ and its floor has area 154 m². Find the radius of its floor and its height.

Answer

$\frac{22}{7} r^2 = 154$ , so $r^2 = 49$ and r = 7 m. $\frac13 \times 154 \times h = 1232$ , so h = 24 m.

7.

A cone-shaped tent with slant height 14 m takes 77 m² of plastic. Find the area of its floor and the distance round it.

Answer

$\frac{22}{7} \times r \times 14 = 77$ , so r = 1.75 m.
Floor $= \frac{22}{7} \times 1.75^2 =$ 9.625 m²; round it $= 2 \times \frac{22}{7} \times 1.75 =$ 11 m.

8.

Find the volume, curved surface area and total surface area of a cone 21 cm high on a base 8 cm across.

Answer

r = 4. $V = \frac13 \times \frac{22}{7} \times 16 \times 21 =$ 352 cm³.
$l = \sqrt{441 + 16} = \sqrt{457} \approx 21.378$ cm. CSA $= \frac{22}{7} \times 4 \times 21.378 \approx$ 268.75 cm²; TSA $\approx 268.75 + 50.29 =$ 319.04 cm².

9.

A solid cone is three times as high as its base is wide (its diameter). Its volume is $54\pi$ cm³. Find its total surface area.

Answer

h = 3 × 2r = 6r, so $\frac13 \pi r^2 \times 6r = 2\pi r^3 = 54\pi$ : $r^3 = 27$ , r = 3, h = 18.
$l = \sqrt{324 + 9} = \sqrt{333} \approx 18.248$ . TSA $= \frac{22}{7} \times 3 \times 21.248 \approx$ 200.34 cm².

10.

For a project, a Grade 10 student made a wooden cone whose base has area 154 cm² and whose height is 14 cm. What does it cost to paint its curved surface at Rs 1.50 a square centimeter?

Answer

$r^2 = 154 \times \frac{7}{22} = 49$ , r = 7. $l = \sqrt{196 + 49} = \sqrt{245} \approx 15.652$ .
CSA $= 22 \times 15.652 \approx 344.35$ cm². Cost $= 344.35 \times 1.50 \approx$ Rs 516.53.

11.

A cone-shaped pot, 21 cm high and 14 cm across, is full of water. Two equal solid cones are dropped in, and a third of the water spills out. What is the volume of each solid cone?

Answer

Pot: $\frac13 \times \frac{22}{7} \times 49 \times 21 = 1078$ cm³. A third spills: $1078 \div 3 \approx 359.33$ cm³, the space the two cones take.
Each: $359.33 \div 2 \approx$ 179.67 cm³.

12.

Three solid cones of the same height, with base radii 3 cm, 4 cm and 5 cm, are melted and made into one cone of that height. Find the diameter of its base.

Answer

Melting keeps the volume: $\frac13 \pi R^2 h = \frac13 \pi (9 + 16 + 25) h$ , so $R^2 = 50$ and $R = 5\sqrt2$ .
Diameter = $10\sqrt2 \approx 14.14$ cm.

13.

A wooden toy cone has a base 10 cm across. Painting its whole surface at Rs 4 a square centimeter costs Rs 880. Find the volume of the toy.

Answer

TSA = 880 ÷ 4 = 220 cm². $\frac{22}{7} \times 5 \times (5 + l) = 220$ , so 5 + l = 14 and l = 9.
$h = \sqrt{81 - 25} = \sqrt{56} \approx 7.483$ . $V = \frac13 \times \frac{22}{7} \times 25 \times 7.483 \approx$ 195.99 cm³.

a.

The whole solid has volume 1,050 cm³ and its cylindrical part 748 cm³. Find the volume of the conical part.

Figure 5.28

Answer

1050 − 748 = 302 cm³.

b.

The conical part’s curved surface is 252 cm², the cylinder’s curved surface 272 cm², and the circular end 154 cm². Find the total surface area.

Figure 5.29

Answer

252 + 272 + 154 = 678 cm².

c.

A square prism of height h has a square pyramid on top; the whole solid is 2h high. The base has area 36 cm² and the prism’s volume is 144 cm³. Find the volume of the pyramid.

Figure 5.30

Answer

36 × h = 144, so h = 4 cm, and the pyramid is 2h − h = 4 cm high too.
$V = \frac13 \times 36 \times 4 =$ 48 cm³.

d.

A cone with slant height 2r has a hemisphere of radius r on it. The cone’s curved surface is 308 cm². Find the curved surface of the hemisphere.

Figure 5.31

Answer

$\pi r \times 2r = 2\pi r^2 = 308$ . The hemisphere’s curved surface is $2\pi r^2$ too: 308 cm². (r = 7 cm.)

a.

Figure 5.32

Answer

r = 20. $V = \pi r^2 \left(h_1 + \frac{h_2}{3}\right) = \frac{22}{7} \times 400 \times 73.333 \approx$ 92,190.48 cm³.

b.

Figure 5.33

Answer

Pyramid’s height = 15 − 6 = 9 cm.
$V = 144 \times 6 + \frac13 \times 144 \times 9 = 864 + 432 =$ 1,296 cm³.

c.

Figure 5.34

Answer

The hemisphere’s radius is the rest of the height: r = 56 − 42 = 14 cm.
$V = \frac13 \pi r^2 (h + 2r) = \frac13 \times \frac{22}{7} \times 196 \times 70 \approx$ 14,373.33 cm³.

d.

Figure 5.35

Answer

r = 7. $V = \frac13 \times \frac{22}{7} \times 49 \times (6 + 9) =$ 770 cm³.

3.

A crystal is two equal square pyramids joined base to base. The square between them (shaded) is 2.5 cm on each side, and the crystal is 4.5 cm from tip to tip. Find its total surface area and volume.

Figure 5.36

Answer

Each pyramid: a = 2.5, h = 2.25. $l = \sqrt{2.25^2 + 1.25^2} = \sqrt{6.625} \approx 2.574$ cm.
TSA $= 2 \times 2al = 4 \times 2.5 \times 2.574 \approx$ 25.74 cm².
$V = 2 \times \frac13 \times 6.25 \times 2.25 =$ 9.375 cm³.

a.

Figure 5.37

Answer

r = 3; the cone is 120 − 116 = 4 cm long, so $l = \sqrt{16 + 9} = 5$ .
TSA $= \pi r(r + 2h_1 + l) = \frac{22}{7} \times 3 \times (3 + 232 + 5) \approx$ 2,262.86 cm².

b.

Figure 5.38

Answer

r = 7, l = 25, cylinder 32 cm.
TSA $= \pi r(r + 2h_1 + l) = 22 \times (7 + 64 + 25) =$ 2,112 cm².

c.

Figure 5.39

Answer

r = 31 − 24 = 7 cm; $l = \sqrt{576 + 49} = 25$ cm.
TSA $= \pi r(l + 2r) = 22 \times (25 + 14) =$ 858 cm².

d.

Figure 5.40

Answer

$r = \sqrt{50^2 - 48^2} = \sqrt{196} = 14$ cm.
TSA $= \pi r(l + 2r) = 44 \times (50 + 28) =$ 3,432 cm².

e.

Figure 5.41

Answer

Prism: 28 − 8 = 20 cm high. $l = \sqrt{64 + 225} = 17$ cm.
TSA = 900 + 4 × 30 × 20 + 2 × 30 × 17 = 900 + 2400 + 1020 = 4,320 cm².

f.

Figure 5.42

Answer

Pyramid: 29 − 24 = 5 cm high. $l = \sqrt{25 + 144} = 13$ cm.
TSA = 576 + 4 × 24 × 24 + 2 × 24 × 13 = 576 + 2304 + 624 = 3,504 cm².

5.

A toy is a cone and a hemisphere, both of radius 14 cm; it is 49 cm high. Find its total surface area.

Answer

Cone: 49 − 14 = 35 cm high; $l = \sqrt{1225 + 196} = \sqrt{1421} \approx 37.696$ cm.
TSA $= \pi r(l + 2r) = 44 \times (37.696 + 28) \approx$ 2,890.63 cm².

a.

the total volume;

Answer

a = 10. $V = 100 \times 10 + \frac13 \times 100 \times 12 = 1000 + 400 =$ 1,400 cm³.

b.

the total surface area.

Answer

$l = \sqrt{144 + 25} = 13$ . TSA = 100 + 4 × 10 × 10 + 2 × 10 × 13 = 760 cm².

7.

A pyramid 8 cm high stands on top of a solid cube with edges of 12 cm. Find the total volume.

Answer

$12^3 + \frac13 \times 144 \times 8 = 1728 + 384 =$ 2,112 cm³.

a.

A solid is a cylinder with a cone on top. The cylinder’s base has area 100 cm² and it is 3 cm high. The solid’s volume is 600 cm³. Find its total height.

Figure 5.43

Answer

Cylinder: 100 × 3 = 300. Cone: $600 - 300 = 300 = \frac13 \times 100 \times h$ , so h = 9.
Total height = 3 + 9 = 12 cm.

b.

A solid is a square prism with base edge 8 cm and a square pyramid with slant height 5 cm on top. Its volume is 448 cm³. Find the height of the prism.

Figure 5.44

Answer

Pyramid: $h = \sqrt{25 - 16} = 3$ ; its volume $\frac13 \times 64 \times 3 = 64$ .
Prism: 448 − 64 = 384 = 64 × h₁, so h₁ = 6 cm.

a.

Find the lateral surface area of the pyramid.

Answer

$l = \sqrt{1 + 0.25} = \sqrt{1.25} = \frac{\sqrt5}{2} \approx 1.118$ ft.
LSA $= 2al = 2 \times 1 \times \frac{\sqrt5}{2} =$ $\sqrt5 \approx 2.24$ sq ft.

b.

Find the area of the tower that can be painted (not its base).

Answer

Sides: 4 × 1 × 6 = 24. With the pyramid: $24 + \sqrt5 \approx$ 26.24 sq ft.

10.

For his sixth birthday Ram got a spinning top from his father: a cone below and a hemisphere above, 5 cm high in all, the hemisphere 3.5 cm across. Ram wants to paint it. How much surface is there to paint?

Answer

r = 1.75; cone h = 5 − 1.75 = 3.25; $l = \sqrt{10.5625 + 3.0625} = \sqrt{13.625} \approx 3.691$ .
Area $= \pi r(l + 2r) = \frac{22}{7} \times 1.75 \times (3.691 + 3.5) \approx$ 39.55 cm².

a.

A cube-shaped tank with edges of 6 ft.

Answer

$6^3 = 216$ ft³; $216 \times 28.317 \approx$ 6,116.47 liters.

b.

A cuboid tank 2 m long, 1.5 m wide and 2 m high.

Answer

2 × 1.5 × 2 = 6 m³ = 6,000 liters.

c.

A cylindrical tank 1 m across and 4 m high.

Answer

$\frac{22}{7} \times 0.5^2 \times 4 \approx 3.14286$ m³ $\approx$ 3,142.86 liters.

a.

Find the area of the parking area.

Answer

A trapezium: $\frac12 \times (10 + 12) \times 8 =$ 88 sq ft.

b.

How many bricks are needed to pave it?

Answer

$88 \div 0.222 \approx 396.4$ : 397 bricks (396 would leave a gap).

c.

Two workers pave it in 3 days at Rs 1,200 a day each. Find the total cost, bricks included.

Answer

Bricks 397 × 16 = 6352; wages 2 × 3 × 1200 = 7200. Total Rs 13,552.

a.

Find the perimeter of the garden.

Answer

14 + 10 + 11 + 15 = 50 m.

b.

How much wire goes round it 5 times?

Answer

5 × 50 = 250 m.

c.

Find the total cost of the fencing, wages included.

Answer

Wire 250 × 80 = 20000; wages 3 × 2 × 1500 = 9000. Total Rs 29,000.

a.

Find the area of the floor.

Answer

$9^2 =$ 81 sq ft.

b.

Find the area of the four walls without the windows and doors.

Answer

4 × 81 − 2 × 4 − 2 × 15 = 324 − 38 = 286 sq ft.

c.

Find the cost of painting the four walls, windows and doors included, at Rs 350 a square foot.

Answer

324 × 350 = Rs 1,13,400.

d.

Another room is 1 ft longer, wider and higher than the first, and its four walls (windows and doors included) are painted at Rs 340 a square foot. By how much does the cost rise or fall?

Answer

Walls $4 \times 10^2 = 400$ sq ft; cost 400 × 340 = 136000. It rises by Rs 22,600.

a.

What do the rings alone cost?

Answer

60 × 1200 = Rs 72,000.

b.

What does it cost to build the well?

Answer

Wages 2 × 12 × 1500 = 36000. Total 72000 + 36000 = Rs 1,08,000.

c.

Find the most water the well can hold.

Answer

$\frac{22}{7} \times 1.75^2 \times 60 = 577.5$ ft³; $577.5 \times 28.317 \approx$ 16,353.07 liters.

d.

The water reaches the top of the 35th ring. How much water is in the well, and how much more would fill it?

Answer

$\frac{22}{7} \times 1.75^2 \times 35 = 336.875$ ft³ $\approx$ 9,539.29 liters. To fill it: 16353.07 − 9539.29 = 6,813.78 liters more.

a.

Draw the pillar.

Answer

Figure 5.51

b.

Find the slant height of the pyramid.

Answer

$l = \sqrt{1^2 + 2.5^2} = \sqrt{7.25} = \frac{\sqrt{29}}{2} \approx$ 2.69 ft.

c.

What does it cost to paint both pillars with their pyramids at Rs 94 a square foot?

Answer

One pillar: $4 \times 5 \times 10 + 2 \times 5 \times 2.693 \approx 200 + 26.93 = 226.93$ sq ft. Two: 453.85 sq ft.
Cost $\approx 453.85 \times 94 \approx$ Rs 42,662.05.

a.

Draw the pillar.

Answer

Figure 5.52

b.

Write the formula for the lateral surface area of a square pyramid.

Answer

LSA = 2al.

c.

Find the slant height of the pyramid.

Answer

$l = \sqrt{1 + 0.25} =$ $\frac{\sqrt5}{2} \approx 1.12$ ft.

d.

Tiling both pillars with their pyramids costs Rs 2,729 in all. What is the rate a square foot?

Answer

One pillar: $4 \times 1 \times 6 + 2 \times 1 \times 1.118 \approx 26.24$ sq ft; two: 52.47.
Rate $= 2729 \div 52.47 \approx$ Rs 52 a square foot.

a.

Write the formula for the volume of a cylinder.

Answer

$V = \pi r^2 h$ .

b.

How much water can the tank hold?

Answer

r = 1.4. $V = \pi r^2 \left(h_1 + \frac{h_2}{3}\right) = \frac{22}{7} \times 1.96 \times 3.24 = 19.9584$ m³ = 19,958.4 liters.

c.

If water costs Rs 1.85 a liter, what does it cost to fill the tank?

Answer

19958.4 × 1.85 = Rs 36,923.04.

a.

Find the area of one triangle.

Answer

An isosceles triangle with base 5: its height is $\sqrt{6^2 - 2.5^2} = \sqrt{29.75} = \frac{\sqrt{119}}{2}$ .
Area $= \frac12 \times 5 \times \frac{\sqrt{119}}{2} =$ $\frac{5\sqrt{119}}{4} \approx 13.64$ m².

b.

Find the total area of cloth.

Answer

$8 \times \frac{5\sqrt{119}}{4} =$ $10\sqrt{119} \approx 109.09$ m².

c.

At Rs 600 a square meter, what does the cloth cost?

Answer

$109.09 \times 600 \approx$ Rs 65,452.27.

a.

Write the formula for the volume of a square pyramid.

Answer

$V = \frac13 a^2 h$ .

b.

Find the vertical height of a pyramid.

Answer

$h = \sqrt{10^2 - 6^2} =$ 8 cm.

c.

How much space do the three pyramids take up?

Answer

One: $\frac13 \times 144 \times 8 = 384$ cm³. Three: 1,152 cm³. (They stand on 3 × 144 = 432 cm² of the teacher’s desk.)

d.

Coloring each pyramid, without its base, costs Rs 80 a square centimeter. Find the cost for all three.

Answer

Each: 2al = 2 × 12 × 10 = 240 cm², costing 240 × 80 = 19200. Three: Rs 57,600.

a.

Write the formula for the total surface area of a square pyramid.

Answer

TSA $= a^2 + 2al$ .

b.

Find the perimeter of its base.

Answer

4 × 16 = 64 cm.

c.

Find its slant height.

Answer

$\frac13 \times 256 \times h = 1280$ , so h = 15. $l = \sqrt{225 + 64} =$ 17 cm.

d.

Find its total surface area and the area of its triangular faces. Which is greater, and by how much?

Answer

Triangles: 2 × 16 × 17 = 544 cm²; TSA = 256 + 544 = 800 cm². The total is greater by 256 cm², the base.

a.

Write the formula for the total surface area of a square pyramid.

Answer

TSA $= a^2 + 2al$ .

b.

Find its base edge.

Answer

2 × a × 15 = 540: a = 18 cm.

c.

Find its volume.

Answer

$h = \sqrt{225 - 81} = 12$ . $V = \frac13 \times 324 \times 12 =$ 1,296 cm³.

a.

What is the diameter of its base?

Answer

2 × 7 = 14 cm.

b.

Find the circumference of its base.

Answer

$2 \times \frac{22}{7} \times 7 =$ 44 cm.

c.

Find its total surface area.

Answer

$\pi r^2 + 2\pi rh + 2\pi r^2 = 154 + 440 + 308 =$ 902 cm².

d.

Find its volume.

Answer

$\frac{22}{7} \times 49 \times 10 + \frac23 \times \frac{22}{7} \times 343 = 1540 + 718.67 \approx$ 2,258.67 cm³.

a.

Draw the solid.

Answer

Figure 5.54

b.

Find the vertical height of the pyramid.

Answer

$h_2 = \sqrt{25 - 16} =$ 3 cm.

c.

Is the pyramid’s height greater than, less than or equal to the prism’s, and by how much?

Answer

Pyramid: $\frac13 \times 64 \times 3 = 64$ . Prism: 1024 − 64 = 960 = 64h₁, so h₁ = 15. The pyramid is less by 12 cm.

d.

Find the total surface area of the solid.

Answer

64 + 4 × 8 × 15 + 2 × 8 × 5 = 64 + 480 + 80 = 624 cm².

a.

Draw the tent.

Answer

Figure 5.55

b.

Find the radius of the cylinder.

Answer

The dome is the rest of the height: r = 6.5 − 3 = 3.5 m.

c.

Find the tent’s surface area (it has no cloth floor).

Answer

$2\pi rh + 2\pi r^2 = 2 \times \frac{22}{7} \times 3.5 \times (3 + 3.5) = 22 \times 6.5 =$ 143 m².

d.

Cloth costs Rs 500 a square meter. What does the cloth for the tent cost?

Answer

143 × 500 = Rs 71,500.

a.

Find its total surface area.

Answer

5148 ÷ 6 = 858 cm².

b.

Find the slant height of the cone.

Answer

$\pi r l + 2\pi r^2 = 22l + 308 = 858$ : l = 25 cm.

c.

Find the vertical height of the cone.

Answer

$h = \sqrt{625 - 49} =$ 24 cm.

d.

Find the total height of the solid.

Answer

24 + 7 = 31 cm.

a.

Find the slant height of the cone.

Answer

$l = \sqrt{144 + 25} =$ 13 cm.

b.

A new toy is made with the cylinder’s length and the cone’s height swapped. Painted at the same rate, which toy costs more?

Answer

Old: $\pi(25 + 2 \times 5 \times 14 + 5 \times 13) = 230\pi \approx 722.86$ cm².
New: cone 14 high, $l = \sqrt{196 + 25} \approx 14.866$ ; $\pi(25 + 2 \times 5 \times 12 + 5 \times 14.866) \approx 219.33\pi \approx 689.32$ cm².
The original toy costs more: its surface is larger by about 33.5 cm². (Swapping takes $20\pi$ off the cylinder’s side and adds only about $9.3\pi$ to the cone’s.)

a.

What is the base edge of the cuboid?

Answer

$\sqrt{100} =$ 10 cm.

b.

Find the volume of the whole solid.

Answer

$100 \times 9 + \frac13 \times 100 \times 12 = 900 + 400 =$ 1,300 cm³.

a.

Find the radius of the heap.

Answer

$\frac13 \pi r^2 \times 30 = \pi \times 196 \times 40$ , so $r^2 = 784$ and r = 28 cm.

b.

Find the surface area of the heap (not its base).

Answer

$l = \sqrt{900 + 784} = \sqrt{1684} \approx 41.037$ . CSA $= \frac{22}{7} \times 28 \times 41.037 \approx$ 3,611.22 cm².

c.

If 1 cm³ of cement weighs 2.5 g, find the weight of the cement.

Answer

Volume $= \frac{22}{7} \times 196 \times 40 = 24640$ cm³. Weight = 24640 × 2.5 = 61,600 g (61.6 kg).

a.

Draw a pillar.

Answer

Figure 5.56

b.

Write the formula for the lateral surface area of a square pyramid.

Answer

LSA = 2al.

c.

Find the slant height of the pyramid.

Answer

$l = \sqrt{16 + 9} =$ 5 ft.

d.

Find the area of a pillar to be painted.

Answer

4 × 6 × 8 + 2 × 6 × 5 = 192 + 60 = 252 sq ft (the base is on the ground).

e.

Find the cost of painting the pillars at Rs 110 a square foot.

Answer

One pillar: 252 × 110 = Rs 27,720. Both: Rs 55,440.

a.

Write the formula for the lateral surface area of a prism.

Answer

LSA = ph (the perimeter of the base times the height).

b.

What is the perimeter of the base?

Answer

4 × 1 = 4 ft.

c.

Find the slant height of the pyramid.

Answer

$l = \sqrt{1 + 0.25} =$ $\frac{\sqrt5}{2} \approx 1.12$ ft.

d.

Find the area of a pillar to be painted.

Answer

$4 \times 6 + 2 \times 1 \times \frac{\sqrt5}{2} =$ $(24 + \sqrt5) \approx 26.24$ sq ft.

e.

Find the cost of painting the pillars at Rs 52 a square foot.

Answer

One: $26.236 \times 52 \approx$ Rs 1,364.28. Both: about Rs 2,728.55.

a.

Find the cost of the rings.

Answer

32 × 1200 = Rs 38,400.

b.

Find the volume of water the well holds.

Answer

$\frac{22}{7} \times 1.75^2 \times 32 =$ 308 ft³.

c.

The water reaches the top of the eighteenth ring. Find the space in the well above the water.

Answer

32 − 18 = 14 rings: $\frac{22}{7} \times 1.75^2 \times 14 =$ 134.75 ft³.

a.

Write the formula for the lateral surface area of a square pyramid.

Answer

LSA = 2al.

b.

Find the volume of the solid.

Answer

$\frac13 \times 144 \times 20 =$ 960 cm³ (the two heights add up to 20).

a.

Write the formula for the total surface area of a cone.

Answer

TSA $= \pi r(r + l)$ .

b.

Find the perimeter of the tank’s base.

Answer

$2 \times \frac{22}{7} \times 0.7 =$ 4.4 m.

c.

How many liters can the tank hold?

Answer

$\frac{22}{7} \times 0.49 \times (2.1 + 0.12) = 3.4188$ m³ = 3,418.8 liters.

a.

What shapes make up the solid?

Answer

A cone and a hemisphere.

b.

Write the formula for the curved surface area of a cone.

Answer

CSA $= \pi r l$ .

c.

Find the total surface area of the solid.

Answer

Let l = 5k, r = 3k; then h = 4k.
$V = \frac13 \pi (9k^2)(4k) + \frac23 \pi (27k^3) = 12\pi k^3 + 18\pi k^3 = 30\pi k^3 = 240\pi$ , so k = 2: r = 6, l = 10.
TSA $= \pi r(l + 2r) = \pi \times 6 \times 22 = 132\pi \approx$ 414.86 cm².

d.

Is the total surface area the same as the curved surface area? Give a reason.

Answer

Yes. Every face it shows is curved (the cone’s side and the dome), and the flat circle between them is inside.

a.

What shapes make up the solid?

Answer

A cylinder and a cone.

b.

Write the formula for the perimeter of the base of a cone.

Answer

$C = 2\pi r$ .

c.

Find the volume of the solid.

Answer

Let l = 13k, h₂ = 12k; then r = 5k.
TSA $= \pi r^2 + 2\pi r \times 24 + \pi r l = \pi(25k^2 + 240k + 65k^2) = \pi(90k^2 + 240k) = 840\pi$ , so $3k^2 + 8k - 28 = 0$ , (3k + 14)(k − 2) = 0, k = 2: r = 10, h₂ = 24.
$V = \pi \times 100 \times \left(24 + \frac{24}{3}\right) = 3200\pi \approx$ 10,057.14 cm³.

d.

Is the total surface area the same as the curved surface area? If not, compare them.

Answer

No. The flat end, $\pi r^2 = 100\pi$ , is not curved. Curved surface $= 840\pi - 100\pi = 740\pi \approx 2325.71$ cm²; the total is greater by $100\pi \approx 314.29$ cm².

a.

Write the formula for the total surface area of a cone.

Answer

TSA $= \pi r(r + l)$ .

b.

Find the total surface area of each solid in Figure 1, and their sum.

Answer

Cylinder: $2\pi r(r + h) = 44 \times 46 = 2024$ cm². Cone: l = 25, $\pi r(r + l) = 22 \times 32 = 704$ cm². Sum: 2,728 cm².

c.

Is that sum the total surface area of the solid in Figure 2?

Answer

No. Figure 2’s solid is Example 1’s pencil: 2,420 cm². The sum is larger by 2728 − 2420 = 308 cm², the two circles pressed together at the joint, $2\pi r^2 = 308$ .

d.

Is the sum of the volumes in Figure 1 the volume of Figure 2’s solid? Give a reason.

Answer

Yes, both 7,238 cm³: joining moves no material, so the space taken is the same.

e.

Compare the cost of painting the solids of Figure 1 and Figure 2 at Rs 150 a square meter.

Answer

Figure 1 has 308 cm² = 0.0308 m² more to paint: it costs 0.0308 × 150 = Rs 4.62 more.

a.

Write the formula for the volume of a cone.

Answer

$V = \frac13 \pi r^2 h$ .

b.

Find the slant height of the cone.

Answer

$\sqrt{144 + 25} =$ 13 cm.

c.

A new toy is made by swapping the base’s diameter and the cone’s height (base 12 cm across, cone 10 cm high). Which toy costs more to paint at the same rate?

Answer

Old: $230\pi \approx 722.86$ cm² (Question 8).
New: r = 6, $l = \sqrt{100 + 36} \approx 11.662$ ; $\pi(36 + 2 \times 6 \times 14 + 6 \times 11.662) \approx 273.97\pi \approx 861.05$ cm².
The new toy costs more: its surface is larger by about 138 cm².

a.

Draw the solid.

Answer

Figure 5.61

b.

What shape is the base of the cylinder?

Answer

A circle.

c.

Find the height of the cone.

Answer

TSA $= 2851.20 \div 1.40 \approx 2036.57 = 648\pi$ cm².
$\pi r^2 + 2\pi r \times 28 + \pi r \times 17 = \pi(r^2 + 73r) = 648\pi$ , so $r^2 + 73r - 648 = 0$ , (r + 81)(r − 8) = 0, r = 8.
$h = \sqrt{289 - 64} =$ 15 cm.

a.

How many liters does 1 m³ hold?

Answer

1,000 liters.

b.

What is the radius of its base?

Answer

$\frac{22}{7} r^2 = 154$ : r = 7 m.

c.

How high is its cylindrical part?

Answer

20 − 7 = 13 m.

d.

What does it cost to fill the tank?

Answer

$V = 154 \times 13 + \frac23 \times \frac{22}{7} \times 343 = 2002 + 718.67 = 2720.67$ m³ $\approx 27,20,667$ liters.
Cost = 2720666.67 × 0.45 = Rs 12,24,300.

Exam practice

Unit 5 · Area and Volume

Exercise 5.1

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Exercise 5.2

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Exercise 5.3

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Exercise 5.4

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Mixed exercise

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21