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.

#### 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.

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(lb) × h

= 2(20 + 12) × 10

= 2 × 32 × 10

= 640 cm2

Also,

Total surface area of the box

= 2(lbbhhl)

= 2(20 × 12 + 12 × 10 + 10 × 20)

= 2(240 + 120 + 200)

= 2 × 560

= 1120 cm2

Thus, the surface area of vertical faces and total surface area of the box is 640 cm2 and 1120 cm2, respectively.

#### 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 ?

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(lbbhhl) = 500 square units

Thus, the length of the box is 20 units.

#### 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.

Side of the cube, l = 4.5 cm

∴ Surface area of all vertical faces of the cube = 4l2 = 4 × (4.5 cm)2 = 4 × 20.25 cm2 = 81 cm2

Also,

Total surface area of the cube = 6l2 = 6 × (4.5 cm)2 = 6 × 20.25 cm2 = 121.50 cm2

Thus, the surface area of all the vertical faces and total surface area of the cube is 81 cm2 and 121.50 cm2, respectively.

#### Question 4:

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

Let the edge of the cube be l cm.

Total surface area of the cube = 5400 cm2

∴ Surface area of all vertical faces of the cube = 4l2 = 4 × (30 cm)2 = 4 × 900 cm2 = 3600 cm2

Thus, the surface area of all vertical faces of the cube is 3600 cm2.

#### 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.

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 m3

∴ l × b × h = 34.50 m3

⇒ l × 1.5 × 1.15 = 34.50

⇒ $\frac{34.50}{1.5×1.15}$ = 20 m

Thus, the length of the cuboid is 20 m.

#### Question 6:

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

Edge of the cube, l = 7.5 cm

∴ Volume of the cube = l3 = (7.5 cm)3 = 421.875 cm= 421.88 cm3 (Approx)

Thus, the volume of the cube is approximately 421.88 cm3.

#### 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)

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 cm2

Also,

Total surface area of the cylinder = 2$\mathrm{\pi }$r(rh) = 2 × 3.14 × 20 × (20 + 13) = 2 × 3.14 × 20 × 33 = 4144.80 cm2

Thus, the curved surface area and the total surface area of the cylinder is 1632.80 cm2 and 4144.80 cm2, respectively.

#### Question 8:

Curved surface area of a cylinder is 1980 cm2 and radius of its base is 15 cm. Find the height of the cylinder. ().

Let the height of the cylinder be h cm.

Radius of the cylinder, r = 15 cm

Curved surface area of the cylinder = 1980 cm2

∴ 2$\mathrm{\pi }$rh = 1980 cm2

Thus, the height of the cylinder is 21 cm.

#### 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.

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,

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

#### Question 2:

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

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 cm2

∴ $\mathrm{\pi }$r(rl) = 7128 cm2

Now,

∴ Volume of the cone = $\frac{1}{3}\mathrm{\pi }{r}^{2}h\mathit{=}\frac{1}{3}×\frac{22}{7}×{\left(28\right)}^{2}×45$ = 36960 cm3

Thus, the volume of the cone is 36960 cm3.

#### Question 3:

Curved surface area of a cone is 251.2 cm2 and radius of its base is 8 cm. Find its slant height and perpendicular height. ($\mathrm{\pi }$ = 3.14 )

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 cm2

∴ $\mathrm{\pi }$rl = 251.2 cm2

Now,

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

#### 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 ?

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(rl) = $\frac{22}{7}×6×\left(6+8\right)=\frac{22}{7}×6×14$ = 264 m2

Rate of making the cone of tin sheet = Rs 10/m2

∴ 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.

#### 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)

Let the perpendicular height of the cone be h cm.

Radius of the base of cone, r = 30 cm

Volume of the cone = 6280 cm3

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

#### 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)

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 cm2

∴ $\mathrm{\pi }$rl = 188.4 cm2

Now,

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

#### Question 7:

Volume of a cone is 1212 cm3 and its height is 24 cm. Find the surface area of the cone.($\mathrm{\pi }=\frac{22}{7}$)

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 cm3

Now,

∴ Surface area of the cone = $\mathrm{\pi }$rl$\frac{22}{7}×7×25$ = 550 cm2 (Approx)

Thus, the surface area of the cone is approximately 550 cm2.

#### 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.$\left(\mathrm{\pi }=\frac{22}{7}\right)$

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 cm2

∴ $\mathrm{\pi }$rl = 2200 cm2

∴ Total surface of the cone = $\mathrm{\pi }$r(rl) = $\frac{22}{7}×14×\left(14+50\right)=\frac{22}{7}×14×64$ = 2816 cm2

Thus, the total surface area of the cone is 2816 cm2.

#### 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.

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 m2

∴ Area occupied by 25 persons on the ground = 25 × 4 m2 = 100 m2

⇒ Area of the base of the cone = $\mathrm{\pi }$r2 = 100 m2         .....(1)

∴ Volume of the conical tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×100×18$ = 600 m3              [Using (1)]

Thus, the volume of the tent is 600 m3.

#### 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 ).

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 = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(3.6\right)}^{2}×2.1$ = 28.51 m3  (Approx)

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

∴ Minimum area of the polythin needed to cover the fodder = $\mathrm{\pi }$rl$\frac{22}{7}×3.6×4.17$ = 47.18 m2 (Approx)

Thus, the volume of the fodder is 28.51 m3 and the minimum area of the polythin needed to cover the fodder in rainy season is 47.18 m2.

#### Question 1:

Find the surface areas and volumes of spheres of the following radii.
(i) 4 cm (ii) 9 cm (iii) 3.5 cm. ( )

(i)
Radius of the sphere, r = 4 cm

Surface area of the sphere = 4$\mathrm{\pi }$r2 = 4 × 3.14 × (4 cm)2 = 200.96 cm2

Volume of the sphere =  = 267.95 cm3

(ii)
Radius of the sphere, r = 9 cm

Surface area of the sphere = 4$\mathrm{\pi }$r2 = 4 × 3.14 × (9 cm)2 = 1017.36 cm2

Volume of the sphere =  = 3052.08 cm3

(iii)
Radius of the sphere, r = 3.5 cm

Surface area of the sphere = 4$\mathrm{\pi }$r2 = 4 × 3.14 × (3.5 cm)2 = 153.86 cm2

Volume of the sphere =  = 179.50 cm3

#### Question 2:

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

Radius of the hemisphere, r = 5 cm

∴ Curved surface area of the hemisphere = 2$\mathrm{\pi }$r2 = 2 × 3.14 × (5 cm)2 = 157 cm2

Total surface area of the hemisphere = 3$\mathrm{\pi }$r= 3 × 3.14 × (5 cm)2 = 235.5 cm2

Thus, the curved surface area and total surface area of the solid hemisphere is 157 cm2 and 235.5 cm2, respectively.

#### Question 3:

If the surface area of a sphere is 2826 cm2 then find its volume. ( $\mathrm{\pi }$= 3.14)

Let the radius of the sphere be r cm.

Surface area of the sphere = 2826 cm2

∴ 4$\mathrm{\pi }$r2 = 2826 cm2

∴ Volume of the sphere =  = 14130 cm3

Thus, the volume of the sphere is 14130 cm3.

#### Question 4:

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

Let the radius of the sphere be r cm.

Volume of the sphere = 38808 cm3

∴ Surface area of the sphere = 4$\mathrm{\pi }$r2 = 5544 cm2

Thus, the surface area of the sphere is 5544 cm2.

#### Question 5:

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

Let the radius of the hemisphere be r cm.

Volume of the hemisphere = 18000$\mathrm{\pi }$ cm3

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

Thus, the diameter of the hemisphere is 60 cm.

#### 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 ?

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

$2×\frac{22}{7}×0.45×1.4$

= 3.96 m2

∴ Area of the field pressed in 500 rotations = 500 × Area of the field pressed in one rotation = 500 × 3.96 = 1980 m2

Thus, the area of the field pressed in 500 rotations is 1980 m2.

#### 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 ?

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 cm3

Thus, the maximum volume of water contained in the tank is 96000 cm3.

#### 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
()

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 5x and 12x, respectively.

Volume of the cone = 314 m3

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

Radius of the cone = 5x = 5 × 1 = 5 m

Now,

(Slant height)2 = (Perpendicular height)2 + (Radius)2

⇒ (Slant height)2 = (12)+ (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.

#### Question 4:

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

Let the radius of the sphere be r cm.

Volume of the sphere = 904.32 cm3

Thus, the radius of the sphere is 6 cm.

#### Question 5:

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

Let the edge of the cube be l cm.

Total surface area of the cube = 864 cm2

∴ 6l2 = 864 cm2

∴ Volume of the cube = l3 = (12)3 = 1728 cm3

Thus, the volume of the cube is 1728 cm3.

#### Question 6:

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

Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm2

∴ 4$\mathrm{\pi }$r2 = 154 cm2

∴ Volume of the sphere = $\frac{4}{3}\mathrm{\pi }{r}^{3}=\frac{4}{3}×\frac{22}{7}×{\left(3.5\right)}^{3}$ = 179.67 cm3

Thus, the volume of the sphere is 179.67 cm3.

#### 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.

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 = 3r

Total surface area of the cone = 616 cm2

∴ $\mathrm{\pi }$r(rl) = 616 cm2

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

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

#### 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.

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$2×\frac{22}{7}×2.1×10$ = 132 m2

Rate of plastering = Rs 52/m2

∴ 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 m2 and the cost of plastering the well from inside is Rs 6,864.

#### 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.

Radius of the road roller, = $\frac{1.4}{2}$ = 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

$2×\frac{22}{7}×0.7×2.1$

= 9.24 m2

∴ 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 m2

Rate of levelling the ground = Rs 7/m2

∴ 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 m2 and the cost of levelling the ground is Rs 32,340.

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