Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 9 Surface Area And Volume are provided here with simple step-by-step explanations. These solutions for Surface Area And Volume are extremely popular among Class 9 students for Maths Surface Area And Volume Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 115:

#### Question 1:

Length, breadth and height of a cuboid shape box of medicine is 20 cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.

#### Answer:

Lenght of the box, *l* = 20 cm

Breadth of the box, *b* = 12 cm

Height of the box, *h* = 10 cm

∴ Surface area of the vertical faces of the box

= 2(*l* + *b*) × *h*

= 2(20 + 12) × 10

= 2 × 32 × 10

= 640 cm^{2}

Also,

Total surface area of the box

= 2(*lb* + *bh* + *h**l*)

= 2(20 × 12 + 12 × 10 + 10 × 20)

= 2(240 + 120 + 200)

= 2 × 560

= 1120 cm^{2}

Thus, the surface area of vertical faces and total surface area of the box is 640 cm^{2} and 1120 cm^{2}, respectively.

#### Page No 115:

#### Question 2:

Total surface area of a box of cuboid shape is 500 sq. unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box ?

#### Answer:

Let the length of the box be *l* units.

Breadth of the box, *b* = 6 units

Height of the box, *h* = 5 units

Total surface area of the box = 500 square units

∴ 2(*lb*+ *bh* + *h**l*) = 500 square units

$\Rightarrow l\times 6+6\times 5+5\times l=\frac{500}{2}=250\phantom{\rule{0ex}{0ex}}\Rightarrow 6l+30+5l=250\phantom{\rule{0ex}{0ex}}\Rightarrow 11l=250-30=220\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{220}{11}=20\mathrm{units}$

Thus, the length of the box is 20 units.

#### Page No 115:

#### Question 3:

Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.

#### Answer:

Side of the cube, *l* = 4.5 cm

∴ Surface area of all vertical faces of the cube = 4*l*^{2} = 4 × (4.5 cm)^{2} = 4 × 20.25 cm^{2} = 81 cm^{2}

Also,

Total surface area of the cube = 6*l*^{2} = 6 × (4.5 cm)^{2} = 6 × 20.25 cm^{2} = 121.50 cm^{2}

Thus, the surface area of all the vertical faces and total surface area of the cube is 81 cm^{2} and 121.50 cm^{2}, respectively.

#### Page No 115:

#### Question 4:

Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.

#### Answer:

Let the edge of the cube be *l* cm.

Total surface area of the cube = 5400 cm^{2}

$\therefore 6{l}^{2}=5400\phantom{\rule{0ex}{0ex}}\Rightarrow {l}^{2}=\frac{5400}{6}=900\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{900}=30\mathrm{cm}$

∴ Surface area of all vertical faces of the cube = 4*l*^{2} = 4 × (30 cm)^{2} = 4 × 900 cm^{2} = 3600 cm^{2}

Thus, the surface area of all vertical faces of the cube is 3600 cm^{2}.

#### Page No 115:

#### Question 5:

Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5m and 1.15m respectively. Find its length.

#### Answer:

Let the length of the cuboid be *l* m.

Breadth of the cuboid, *b* = 1.5 m

Height of the cuboid, *h* = 1.15 m

Volume of the cuboid = 34.50 m^{3}

∴ *l* × *b* × *h* = 34.50 m^{3}

⇒ *l* × 1.5 × 1.15 = 34.50

⇒ *l *= $\frac{34.50}{1.5\times 1.15}$ = 20 m

Thus, the length of the cuboid is 20 m.

#### Page No 115:

#### Question 6:

What will be the volume of a cube having length of edge 7.5 cm ?

#### Answer:

Edge of the cube, *l* = 7.5 cm

∴ Volume of the cube = *l*^{3} = (7.5 cm)^{3} = 421.875 cm^{3 }= 421.88 cm^{3} (Approx)

Thus, the volume of the cube is approximately 421.88 cm^{3}.

#### Page No 115:

#### Question 7:

Radius of base of a cylinder is 20 cm and its height is 13 cm, find its curved surface area and total surface area. ( $\mathrm{\pi}$ = 3.14)

#### Answer:

Radius of the cylinder, *r* = 20 cm

Height of the cylinder, *h* = 13 cm

∴ Curved surface area of the cylinder = 2$\mathrm{\pi}$*rh* = 2 × 3.14 × 20 × 13 = 1632.80 cm^{2}

Also,

Total surface area of the cylinder = 2$\mathrm{\pi}$*r*(*r* + *h*) = 2 × 3.14 × 20 × (20 + 13) = 2 × 3.14 × 20 × 33 = 4144.80 cm^{2}

Thus, the curved surface area and the total surface area of the cylinder is 1632.80 cm^{2} and 4144.80 cm^{2}, respectively.

#### Page No 115:

#### Question 8:

Curved surface area of a cylinder is 1980 cm^{2 }and radius of its base is 15 cm. Find the height of the cylinder. ($\mathrm{\pi}=\frac{22}{7}$).

#### Answer:

Let the height of the cylinder be *h* cm.

Radius of the cylinder, *r* = 15 cm

Curved surface area of the cylinder = 1980 cm^{2}

∴ 2$\mathrm{\pi}$*rh* = 1980 cm^{2}

$\Rightarrow 2\times \frac{22}{7}\times 15\times h=1980\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{1980\times 7}{660}=21\mathrm{cm}$

Thus, the height of the cylinder is 21 cm.

#### Page No 119:

#### Question 1:

Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.

#### Answer:

Let the radius of the base of the cone be *r* cm.

Height of the cone, *h* = 12 cm

Slant height of the cone, *l* = 13 cm

Now,

${r}^{2}+{h}^{2}={l}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}+{\left(12\right)}^{2}={\left(13\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}+144=169\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=169-144=25\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{25}=5\mathrm{cm}$

Thus, the radius of the base of the cone is 5 cm.

#### Page No 119:

#### Question 2:

Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm.$\left(\mathrm{\pi}=\frac{22}{7}\right)$

#### Answer:

Let the perpendicular height and the slant height of the cone be *h* cm and *l* cm, respectively.

Radius of the base of cone, *r* = 28 cm

Total surface area of the cone = 7128 cm^{2}

∴ $\mathrm{\pi}$*r*(*r* + *l*) = 7128 cm^{2}

$\Rightarrow \frac{22}{7}\times 28\times \left(28+l\right)=7128\phantom{\rule{0ex}{0ex}}\Rightarrow 28+l=\frac{7128}{22\times 4}=81\phantom{\rule{0ex}{0ex}}\Rightarrow l=81-28=53\mathrm{cm}$

Now,

${r}^{2}+{h}^{2}={l}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(28\right)}^{2}+{h}^{2}={\left(53\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}^{2}=2809-784=2025\phantom{\rule{0ex}{0ex}}\Rightarrow h=\sqrt{2025}=45\mathrm{cm}$

∴ Volume of the cone = $\frac{1}{3}\mathrm{\pi}{r}^{2}h\mathit{=}\frac{1}{3}\times \frac{22}{7}\times {\left(28\right)}^{2}\times 45$ = 36960 cm^{3}

Thus, the volume of the cone is 36960 cm^{3}.

#### Page No 119:

#### Question 3:

Curved surface area of a cone is 251.2 cm^{2} and radius of its base is 8 cm. Find its slant height and perpendicular height. ($\mathrm{\pi}$ = 3.14 )

#### Answer:

Let the perpendicular height and slant height of the cone be *h* cm and *l* cm, respectively.

Radius of the base of cone, *r* = 8 cm

Curved surface area of the cone = 251.2 cm^{2}

∴ $\mathrm{\pi}$*rl* = 251.2 cm^{2}

$\Rightarrow 3.14\times 8\times l=251.2\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{251.2}{25.12}=10\mathrm{cm}$

Now,

${r}^{2}+{h}^{2}={l}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(8\right)}^{2}+{h}^{2}={\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 64+{h}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow {h}^{2}=100-64=36\phantom{\rule{0ex}{0ex}}\Rightarrow h=\sqrt{36}=6\mathrm{cm}$

Thus, the slant height and perpendicular height of the cone is 10 cm and 6 cm, respectively.

#### Page No 119:

#### Question 4:

What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is

Rs.10 per sq.m ?

#### Answer:

Radius of the base of cone, *r* = 6 m

Slant height of the cone, *l* = 8 m

∴ Total surface area of the cone = $\mathrm{\pi}$*r*(*r* + *l*) = $\frac{22}{7}\times 6\times \left(6+8\right)=\frac{22}{7}\times 6\times 14$ = 264 m^{2}

Rate of making the cone of tin sheet = Rs 10/m^{2}

∴ Total cost of making the cone of tin sheet

= Total surface area of the cone × Rate of making the cone of tin sheet

= 264 × 10

= Rs 2640

Thus, the cost of making the cone of tin sheet is Rs 2640.

#### Page No 119:

#### Question 5:

Volume of a cone is 6280 cubic cm and base radius of the cone is 30 cm. Find its perpendicular height. ($\mathrm{\pi}$= 3.14)

#### Answer:

Let the perpendicular height of the cone be *h* cm.

Radius of the base of cone, *r* = 30 cm

Volume of the cone = 6280 cm^{3}

$\therefore \frac{1}{3}\mathrm{\pi}{r}^{2}h=6280{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\times 3.14\times {\left(30\right)}^{2}\times h=6280\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{6280\times 3}{3.14\times 900}=6.67\mathrm{cm}$

Thus, the perpendicular height of the cone is 6.67 cm approximately.

#### Page No 119:

#### Question 6:

Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height ( $\mathrm{\pi}$= 3.14)

#### Answer:

Let the radius of the base and perpendicular height of the cone be *r* cm and *h* cm, respectively.

Slant height of the cone, *l* = 10 cm

Surface area of the cone = 188.4 cm^{2}

∴ $\mathrm{\pi}$*r**l* = 188.4 cm^{2}

$\Rightarrow 3.14\times r\times 10=188.4\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{188.4}{3.14\times 10}=6\mathrm{cm}$

Now,

${r}^{2}+{h}^{2}={l}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(6\right)}^{2}+{h}^{2}={\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 36+{h}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow {h}^{2}=100-36=64\phantom{\rule{0ex}{0ex}}\Rightarrow h=\sqrt{64}=8\mathrm{cm}$

Thus, the perpendicular height of the cone is 8 cm.

#### Page No 119:

#### Question 7:

Volume of a cone is 1212 cm^{3} and its height is 24 cm. Find the surface area of the cone.($\mathrm{\pi}=\frac{22}{7}$)

#### Answer:

Let the radius of the base and slant height of the cone be *r* cm and *l* cm, respectively.

Height of the cone, *h* = 24 cm

Volume of the cone = 1212 cm^{3}

$\therefore \frac{1}{3}\mathrm{\pi}{r}^{2}h=1212{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\times \frac{22}{7}\times {r}^{2}\times 24=1212\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{1212\times 7}{22\times 8}=48.2\phantom{\rule{0ex}{0ex}}\Rightarrow r=7\mathrm{cm}\left(\mathrm{Approx}\right)$

Now,

${l}^{2}={r}^{2}+{h}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}^{2}={\left(7\right)}^{2}+{\left(24\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}^{2}=49+576=625\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{625}=25\mathrm{cm}\left(\mathrm{Approx}\right)$

∴ Surface area of the cone = $\mathrm{\pi}$*rl* = $\frac{22}{7}\times 7\times 25$ = 550 cm^{2} (Approx)

Thus, the surface area of the cone is approximately 550 cm^{2}.

#### Page No 119:

#### Question 8:

The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone.$\left(\mathrm{\pi}=\frac{22}{7}\right)$

#### Answer:

Let the radius of the base of the cone be *r* cm.

Slant height of the cone, *l* = 50 cm

Curved surface area of the cone = 2200 cm^{2}

∴ $\mathrm{\pi}$*rl* = 2200 cm^{2}

$\Rightarrow \frac{22}{7}\times r\times 50=2200\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{2200\times 7}{22\times 50}=14\mathrm{cm}$

∴ Total surface of the cone = $\mathrm{\pi}$*r*(*r* + *l*) = $\frac{22}{7}\times 14\times \left(14+50\right)=\frac{22}{7}\times 14\times 64$ = 2816 cm^{2}

Thus, the total surface area of the cone is 2816 cm^{2}.

#### Page No 119:

#### Question 9:

There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m. of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.

#### Answer:

Let the radius of the base of the cone be *r* m.

Height of the conical tent, *h* = 18 m

Area occupied by each person on the ground = 4 m^{2}

∴ Area occupied by 25 persons on the ground = 25 × 4 m^{2} = 100 m^{2}

⇒ Area of the base of the cone = $\mathrm{\pi}$*r*^{2} = 100 m^{2} .....(1)

∴ Volume of the conical tent = $\frac{1}{3}\mathrm{\pi}{r}^{2}h=\frac{1}{3}\times 100\times 18$ = 600 m^{3} [Using (1)]

Thus, the volume of the tent is 600 m^{3}.

#### Page No 120:

#### Question 10:

In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1m. and diameter of base is 7.2 m. Find the volume of the fodder. if it is to be covered by polythin in rainy season then how much minimum polythin sheet is needed ?

($\mathrm{\pi}=\frac{22}{7}$ and $\sqrt{17.37}=4.17$).

#### Answer:

Height of the conical heap of fodder, *h* = 2.1 m

Radius of the conical heap of fodder, *r* = $\frac{7.2}{2}$ = 3.6 m

∴ Volume of the fodder = $\frac{1}{3}\mathrm{\pi}{r}^{2}h=\frac{1}{3}\times \frac{22}{7}\times {\left(3.6\right)}^{2}\times 2.1$ = 28.51 m^{3} (Approx)

Let the slant height of the conical heap of fodder be *l* m.

$\therefore {l}^{2}={r}^{2}+{h}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}^{2}={\left(2.1\right)}^{2}+{\left(3.6\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {l}^{2}=4.41+12.96=17.37\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{17.37}=4.17\mathrm{m}$

∴ Minimum area of the polythin needed to cover the fodder = $\mathrm{\pi}$*rl* = $\frac{22}{7}\times 3.6\times 4.17$ = 47.18 m^{2} (Approx)

Thus, the volume of the fodder is 28.51 m^{3} and the minimum area of the polythin needed to cover the fodder in rainy season is 47.18 m^{2}.

#### Page No 123:

#### Question 1:

#### Answer:

(i)

Radius of the sphere, *r* = 4 cm

Surface area of the sphere = 4$\mathrm{\pi}$*r*^{2} = 4 × 3.14 × (4 cm)^{2} = 200.96 cm^{2}

Volume of the sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\times 3.14\times {\left(4\mathrm{cm}\right)}^{3}$ = 267.95 cm^{3}

(ii)

Radius of the sphere, *r* = 9 cm

Surface area of the sphere = 4$\mathrm{\pi}$*r*^{2} = 4 × 3.14 × (9 cm)^{2} = 1017.36 cm^{2}

Volume of the sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\times 3.14\times {\left(9\mathrm{cm}\right)}^{3}$ = 3052.08 cm^{3}

(iii)

Radius of the sphere, *r* = 3.5 cm

Surface area of the sphere = 4$\mathrm{\pi}$*r*^{2} = 4 × 3.14 × (3.5 cm)^{2} = 153.86 cm^{2}

Volume of the sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\times 3.14\times {\left(3.5\mathrm{cm}\right)}^{3}$ = 179.50 cm^{3}

#### Page No 123:

#### Question 2:

If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area. ( $\mathrm{\pi}=3.14$ )

#### Answer:

Radius of the hemisphere, *r* = 5 cm

∴ Curved surface area of the hemisphere = 2$\mathrm{\pi}$*r*^{2 }=^{ }2 × 3.14 × (5 cm)^{2} = 157 cm^{2}

Total surface area of the hemisphere = 3$\mathrm{\pi}$*r*^{2 }=^{ }3 × 3.14 × (5 cm)^{2} = 235.5 cm^{2}

Thus, the curved surface area and total surface area of the solid hemisphere is 157 cm^{2} and 235.5 cm^{2}, respectively.

#### Page No 123:

#### Question 3:

If the surface area of a sphere is 2826 cm^{2 }then find its volume. ( $\mathrm{\pi}$= 3.14)

#### Answer:

Let the radius of the sphere be *r* cm.

Surface area of the sphere = 2826 cm^{2}

∴ 4$\mathrm{\pi}$*r*^{2} = 2826 cm^{2}

$\Rightarrow {r}^{2}=\frac{2826}{4\times 3.14}=225\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{225}=15\mathrm{cm}$

∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\times 3.14\times {\left(15\mathrm{cm}\right)}^{3}$ = 14130 cm^{3}

Thus, the volume of the sphere is 14130 cm^{3}.

#### Page No 123:

#### Question 4:

Find the surface area of a sphere, if its volume is 38808 cubic cm.($\mathrm{\pi}=\frac{22}{7}$)

#### Answer:

Let the radius of the sphere be *r* cm.

Volume of the sphere = 38808 cm^{3}

$\therefore \frac{4}{3}\mathrm{\pi}{r}^{3}=38808{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{3}\times \frac{22}{7}\times {r}^{3}=38808\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}=\frac{38808\times 21}{88}=9261\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}={\left(21\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow r=21\mathrm{cm}$

∴ Surface area of the sphere = 4$\mathrm{\pi}$*r*^{2} = $4\times \frac{22}{7}\times {\left(21\mathrm{cm}\right)}^{2}$ = 5544 cm^{2}

Thus, the surface area of the sphere is 5544 cm^{2}.

#### Page No 123:

#### Question 5:

Volume of a hemisphere is 18000 $\mathrm{\pi}$ cubic cm. Find its diameter.

#### Answer:

Let the radius of the hemisphere be *r* cm.

Volume of the hemisphere = 18000$\mathrm{\pi}$ cm^{3}

$\therefore \frac{2}{3}\mathrm{\pi}{r}^{3}=18000\mathrm{\pi}{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}=\frac{18000\times 3}{2}=27000\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}={\left(30\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow r=30\mathrm{cm}$

∴ Diameter of the hemisphere = 2*r* = 2 × 30 = 60 cm

Thus, the diameter of the hemisphere is 60 cm.

#### Page No 123:

#### Question 1:

If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations ?

#### Answer:

Length of the road roller, *h* = 1.4 m

Radius of the road roller, *r* = $\frac{0.9}{2}$ = 0.45 m

Now,

Area of the field pressed in one rotation = Curved surface area of the road roller

= 2$\mathrm{\pi}$*rh*

= $2\times \frac{22}{7}\times 0.45\times 1.4$

= 3.96 m^{2}

∴ Area of the field pressed in 500 rotations = 500 × Area of the field pressed in one rotation = 500 × 3.96 = 1980 m^{2}

Thus, the area of the field pressed in 500 rotations is 1980 m^{2}.

#### Page No 123:

#### Question 2:

To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?

#### Answer:

Thickness of the glass = 2 mm = $\frac{2}{10}$ = 0.2 cm (1 cm = 10 mm)

Inner length of the tank, *l* = 60.4 − 0.2 − 0.2 = 60.4 − 0.4 = 60 cm

Inner breadth of the tank, *b* = 40.4 − 0.2 − 0.2 = 40.4 − 0.4 = 40 cm

Inner height of the tank, *h* = 40.2 − 0.2 = 40 cm (Tank is open at the top)

∴ Maximum volume of water contained in the tank

= Inner volume of the tank

= *l* × *b* × *h*

= 60 × 40 × 40

= 96000 cm^{3}

Thus, the maximum volume of water contained in the tank is 96000 cm^{3}.

#### Page No 123:

#### Question 3:

If the ratio of radius of base and height of a cone is 5:12 and its volume is 314 cubic metre. Find its perpendicular height and slant height

($\mathrm{\pi}=3.14$)

#### Answer:

The ratio of radius of base and perpendicular height of a cone is 5 : 12.

Let the radius of base and perpendicular height of the cone be 5*x* and 12*x*, respectively.

Volume of the cone = 314 m^{3}

$\therefore \frac{1}{3}\mathrm{\pi}{r}^{2}h=314{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\times 3.14\times {\left(5x\right)}^{2}\times 12x=314\phantom{\rule{0ex}{0ex}}\Rightarrow 314{x}^{3}=314\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\mathrm{m}$

∴ Perependicular height of the cone = 12*x* = 12 × 1 = 12 m

Radius of the cone = 5*x* = 5 × 1 = 5 m

Now,

(Slant height)^{2} = (Perpendicular height)^{2 }+ (Radius)^{2}

⇒ (Slant height)^{2} = (12)^{2 }+ (5)^{2}

⇒ (Slant height)^{2} = 144^{ }+ 25 = 169

⇒ (Slant height)^{2} = (13)^{2}

⇒ Slant height = 13 m

Thus, the perpendicular height and slant height of the cone is 12 m and 13 m, respectively.

#### Page No 123:

#### Question 4:

Find the radius of a sphere if its volume is 904.32 cubic cm. ($\mathrm{\pi}$= 3.14)

#### Answer:

Let the radius of the sphere be *r* cm.

Volume of the sphere = 904.32 cm^{3}

$\therefore \frac{4}{3}\mathrm{\pi}{r}^{3}=904.32{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}=\frac{904.32\times 3}{4\times 3.14}=216\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}={\left(6\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow r=6\mathrm{cm}$

Thus, the radius of the sphere is 6 cm.

#### Page No 123:

#### Question 5:

Total surface area of a cube is 864 sq.cm. Find its volume.

#### Answer:

Let the edge of the cube be *l* cm.

Total surface area of the cube = 864 cm^{2}

∴ 6*l*^{2} = 864 cm^{2}

$\Rightarrow {l}^{2}=\frac{864}{6}=144\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{144}=12\mathrm{cm}$

∴ Volume of the cube = *l*^{3} = (12)^{3} = 1728 cm^{3}

Thus, the volume of the cube is 1728 cm^{3}.

#### Page No 123:

#### Question 6:

Find the volume of a sphere, if its surface area is 154 sq.cm.

#### Answer:

Let the radius of the sphere be *r *cm.

Surface area of the sphere = 154 cm^{2}

∴ 4$\mathrm{\pi}$*r*^{2} = 154 cm^{2}

$\Rightarrow 4\times \frac{22}{7}\times {r}^{2}=154\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{154\times 7}{88}=12.25\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}={\left(3.5\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow r=3.5\mathrm{cm}$

∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{3}$ = 179.67 cm^{3}

Thus, the volume of the sphere is 179.67 cm^{3}.

#### Page No 123:

#### Question 7:

Total surface area of a cone is 616 sq.cm. If the slant height of the cone is three times the radius of its base, find its slant height.

#### Answer:

Let the radius of base and slant height of the cone be *r* cm and *l* cm, respectively.

Slant height of the cone = 3 × Radius of the cone (Given)

∴ *l* = 3*r*

Total surface area of the cone = 616 cm^{2}

∴ $\mathrm{\pi}$*r*(*r* + *l*) = 616 cm^{2}

$\Rightarrow \frac{22}{7}\times r\times \left(r+3r\right)=616\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times r\times 4r=616\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{88}{7}{r}^{2}=616\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{616\times 7}{88}=49\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{49}=7\mathrm{cm}$

∴ Slant height of the cone, *l* = 3*r* = 3 × 7 = 21 cm

Thus, the slant height of the cone is 21 cm.

#### Page No 123:

#### Question 8:

The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs.52 per sq.m.

#### Answer:

Inner radius of the well, *r* = $\frac{4.2}{2}$ = 2.1 m

Depth of the well, *h* = 10 m

∴ Inner surface area of the well = 2$\mathrm{\pi}$*rh* = $2\times \frac{22}{7}\times 2.1\times 10$ = 132 m^{2}

Rate of plastering = Rs 52/m^{2}

∴ Cost of plastering the well from inside

= Inner surface area of the well × Rate of plastering

= 132 × 52

= Rs 6,864

Thus, the inner surface area of the well is 132 m^{2} and the cost of plastering the well from inside is Rs 6,864.

#### Page No 123:

#### Question 9:

The length of a road roller is 2.1m and its diameter is 1.4m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m.

#### Answer:

Radius of the road roller, *r *= $\frac{1.4}{2}$ = 0.7 m

Length of the road roller, *h *= 2.1 m

∴ Area levelled by the road roller in one rotation = Curved surface area of the road roller

= 2$\mathrm{\pi}$*rh*

= $2\times \frac{22}{7}\times 0.7\times 2.1$

= 9.24 m^{2}

∴ Area of the ground levelled by the road roller = Area levelled by the road roller in 500 rotations

= 500 × Area levelled by the road roller in one rotation

= 500 × 9.24

= 4620 m^{2}

Rate of levelling the ground = Rs 7/m^{2}

∴ Cost of levelling the ground

= Area of the ground levelled by the road roller × Rate of levelling the ground

= 4620 × 7

= Rs 32,340

Thus, the area of ground levelled by the road roller is 4620 m^{2} and the cost of levelling the ground is Rs 32,340.

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