A. 46 years

B. 52 years

C. 54 years

D. 56 years

### Solution(By Apex Team)

$\begin{array}{l}\text{Age of the instructor}\\ \text{= (34 × 11 – 32 × 10) years}\\ \text{= (374 – 320) years}\\ \text{= 54 years}\end{array}$

• NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes include the accurately designed wide range of solved exercise questions for an excellent understanding. These solutions in Maths for Class 9 is prepared considering the latest CBSE syllabus 2021-22 update of Term II examination. NCERT for Class 9 Maths Solutions is mainly created to be a guiding solution and advantageous reference to help students to clear doubts spontaneously and in an effective way. NCERT Solutions for Class 9 Maths is prepared by the teaching faculties having vast teaching experience along with subject matter experts to serve the purpose.

Class 9 NCERT Solutions are developed keeping in mind the concept-based approach along with the precise answering method for Term II examinations. Refer to NCERT Solutions for Class 9 for a better score. It is a detailed and well-structured solution for a good grasp of concept-based knowledge. NCERT for Class 9 Maths Solutions is made available in web and PDF format for ease of access.

 Box Dimensions (in cm) Total surface area (in cm2 ) Extra area required for overlapping (in cm^2) Total surface area for all overlaps (in cm 2) Area for 250 such boxes (in cm2) Bigger Box l = 25 b = 20 c = 5 1450 1450×5/100 = 72.5 (1450+72.5) = 1522.5 (1522.5×250) = 380625 Smaller Box l = 15 b = 12 h =5 630 630×5/100 = 31.5 (630+31.5) = 661.5 ( 250×661.5) = 165375

Now, Total cardboard sheet required = (380625+165375) cm2

= 546000 cm2

Given: Cost of 1000 cm2 cardboard sheet = Rs. 4

Therefore, Cost of 546000 cmcardboard sheet =Rs. (546000×4)/1000 = Rs. 2184

Therefore, the cost of cardboard required for supplying 250 boxes of each kind will be Rs. 2184.

8. Praveen wanted to make a temporary shelter for her car, by making a box – like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5m, with base dimensions 4m×3m?

Solution:

Let l, b and h be the length, breadth and height of the shelter.

Given:

l = 4m

b = 3m

h = 2.5m

Tarpaulin will be required for the top and four wall sides of the shelter.

Using formula, Area of tarpaulin required = 2(lh+bh)+lb

On putting the values of l, b and h, we get

= [2(4×2.5+3×2.5)+4×3] m2

= [2(10+7.5)+12]m2

= 47m2

Therefore, 47 m2 tarpaulin will be required.

## Exercise 13.2 Page No: 216

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 )

Solution:

Height of cylinder, h = 14cm

Let the diameter of the cylinder be d

Curved surface area of cylinder = 88 cm2

We know that, formula to find Curved surface area of cylinder is 2πrh.

So 2πrh =88 cm2 (r is the radius of the base of the cylinder)

2×(22/7)×r×14 = 88 cm2

2r = 2 cm

d =2 cm

Therefore, the diameter of the base of the cylinder is 2 cm.

2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? Assume π = 22/7

Solution:

Let h be the height and r be the radius of a cylindrical tank.

Height of cylindrical tank, h = 1m

Radius = half of diameter = (140/2) cm = 70cm = 0.7m

Area of sheet required = Total surface are of tank = 2πr(r+h) unit square

= [2×(22/7)×0.7(0.7+1)]

= 7.48

Therefore, 7.48 square meters of the sheet are required.

3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4cm. (see fig. 13.11). Find its (i) inner curved surface area,

(ii) outer curved surface area

(iii) total surface area

(Assume π=22/7)

Solution:

Let r1 and r2 Inner and outer radii of cylindrical pipe

r= 4/2 cm = 2 cm

r= 4.4/2 cm = 2.2 cm

Height of cylindrical pipe, h = length of cylindrical pipe = 77 cm

(i) curved surface area of outer surface of pipe = 2πr1h

= 2×(22/7)×2×77 cm2

= 968 cm2

(ii) curved surface area of outer surface of pipe = 2πr2h

= 2×(22/7)×2.2×77 cm2

= (22×22×2.2) cm2

= 1064.8 cm2

(iii) Total surface area of pipe = inner curved surface area+ outer curved surface area+ Area of both circular ends of pipe.

= 2r1h+2r2h+2π(r12-r22)

= 9668+1064.8+2×(22/7)×(2.22-22)

= 2031.8+5.28

= 2038.08 cm2

Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.

4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to

move once over to level a playground. Find the area of the playground in m2? (Assume π = 22/7)

Solution:

A roller is shaped like a cylinder.

Let h be the height of the roller and r be the radius.

h = Length of roller = 120 cm

Radius of the circular end of roller = r = (84/2) cm = 42 cm

Now, CSA of roller = 2πrh

= 2×(22/7)×42×120

= 31680 cm2

Area of field = 500×CSA of roller

= (500×31680) cm2

= 15840000 cm2

= 1584 m2.

Therefore, area of playground is 1584 m2. Answer!

5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2.

(Assume π = 22/7)

Solution:

Let h be the height of a cylindrical pillar and r be the radius.

Given:

Height cylindrical pillar = h = 3.5 m

Radius of the circular end of pillar = r = diameter/2 = 50/2 = 25cm = 0.25m

CSA of pillar = 2πrh

= 2×(22/7)×0.25×3.5

= 5.5 m2

Cost of painting 1 m2 area = Rs. 12.50

Cost of painting 5.5 m2 area = Rs (5.5×12.50)

= Rs.68.75

Therefore, the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m2 is Rs 68.75.

6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the base of the cylinder is 0.7 m, find its height. (Assume π = 22/7)

Solution:

Let h be the height of the circular cylinder and r be the radius.

Radius of the base of cylinder, r = 0.7m

CSA of cylinder = 2πrh

CSA of cylinder = 4.4m2

Equating both the equations, we have

2×(22/7)×0.7×h = 4.4

Or h = 1

Therefore, the height of the cylinder is 1 m.

7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find

(i) its inner curved surface area,

(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m2.

(Assume π = 22/7)

Solution:

Inner radius of circular well, r = 3.5/2m = 1.75m

Depth of circular well, say h = 10m

(i) Inner curved surface area = 2πrh

= (2×(22/7 )×1.75×10)

= 110

Therefore, the inner curved surface area of the circular well is 110 m2.

(ii)Cost of plastering 1 m2 area = Rs.40

Cost of plastering 110 m2 area = Rs (110×40)

= Rs.4400

Therefore, the cost of plastering the curved surface of the well is Rs. 4400.

8. In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find

the total radiating surface in the system. (Assume π = 22/7)

Solution:

Height of cylindrical pipe = Length of cylindrical pipe = 28m

Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m

Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of the cylinder

= 2×(22/7)×0.025×28 m2

= 4.4m2

The area of the radiating surface of the system is 4.4m2.

9. Find

(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in

diameter and 4.5m high.

(ii) How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank. (Assume π = 22/7)

Solution:

Height of cylindrical tank, h = 4.5m

Radius of the circular end , r = (4.2/2)m = 2.1m

(i) the lateral or curved surface area of cylindrical tank is 2πrh

= 2×(22/7)×2.1×4.5 m2

= (44×0.3×4.5) m2

= 59.4 m2

Therefore, CSA of tank is 59.4 m2.

(ii) Total surface area of tank = 2πr(r+h)

= 2×(22/7)×2.1×(2.1+4.5)

= 44×0.3×6.6

= 87.12 m2

Now, Let S m2 steel sheet be actually used in making the tank.

S(1 -1/12) = 87.12 m2

This implies, S = 95.04 m2

Therefore, 95.04msteel was used in actual while making such a tank.

10. In fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth.

The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade. (Assume π = 22/7) Solution:

Say h = height of the frame of lampshade, looks like cylindrical shape

Total height is h = (2.5+30+2.5) cm = 35cm and

r = (20/2) cm = 10cm

Use curved surface area formula to find the cloth required for covering the lampshade which is 2πrh

= (2×(22/7)×10×35) cm2

= 2200 cm2

Hence, 2200 cm2 cloth is required for covering the lampshade.

11. The students of Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? (Assume π =22/7)

Solution:

Radius of the circular end of cylindrical penholder, r = 3cm

Height of penholder, h = 10.5cm

Surface area of a penholder = CSA of pen holder + Area of base of penholder

= 2πrh+πr2

= 2×(22/7)×3×10.5+(22/7)×32= 1584/7

Therefore, Area of cardboard sheet used by one competitor is 1584/7 cm2

So, Area of cardboard sheet used by 35 competitors = 35×1584/7 = 7920 cm2

Therefore, 7920 cm2 cardboard sheet will be needed for the competition.

## Exercise 13.3 Page No: 221

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area (Assume π=22/7)

Solution:

Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm

Slant height of cone, say l = 10 cm

CSA of cone is = πrl

= (22/7)×5.25×10 = 165

Therefore, the curved surface area of the cone is 165 cm2.

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (Assume π = 22/7)

Solution:

Radius of cone, r = 24/2 m = 12m

Slant height, l = 21 m

Formula: Total Surface area of the cone = πr(l+r)

Total Surface area of the cone = (22/7)×12×(21+12) m2

= 1244.57m2

3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find

(i) radius of the base and (ii) total surface area of the cone.

(Assume π = 22/7)

Solution:

Slant height of cone, l = 14 cm

Let the radius of the cone be r.

(i) We know, CSA of cone = πrl

Given: Curved surface area of a cone is 308 cm2

(308 ) = (22/7)×r×14

308 = 44 r

r = 308/44 = 7

Radius of a cone base is 7 cm.

(ii) Total surface area of cone = CSA of cone + Area of base (πr2)

Total surface area of cone = 308+(22/7)×72 = 308+154

Therefore, the total surface area of the cone is 462 cm2.

4. A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

(Assume π=22/7)

Solution: Let ABC be a conical tent

Height of conical tent, h = 10 m

Radius of conical tent, r = 24m

Let the slant height of the tent be l.

(i) In right triangle ABO, we have

AB= AO2+BO2(using Pythagoras theorem)

l2 = h2+r2

= (10)2+(24)2

= 676

l = 26

Therefore, the slant height of the tent is 26 m.

(ii) CSA of tent = πrl

= (22/7)×24×26 m2

Cost of 1 m2 canvas = Rs 70

Cost of (13728/7)m2 canvas is equal to Rs (13728/7)×70 = Rs 137280

Therefore, the cost of the canvas required to make such a tent is Rs 137280.

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. [Use π=3.14]

Solution:

Height of conical tent, h = 8m

Radius of base of tent, r = 6m

Slant height of tent, l2 = (r2+h2)

l= (62+82) = (36+64) = (100)

or l = 10

Again, CSA of conical tent = πrl

= (3.14×6×10) m2

= 188.4m2

Let the length of tarpaulin sheet required be L

As 20 cm will be wasted, therefore,

Effective length will be (L-0.2m).

Breadth of tarpaulin = 3m (given)

Area of sheet = CSA of tent

[(L–0.2)×3] = 188.4

L-0.2 = 62.8

L = 63

Therefore, the length of the required tarpaulin sheet will be 63 m.

6. The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2. (Assume π = 22/7)

Solution:

Slant height of conical tomb, l = 25m

Base radius, r = diameter/2 = 14/2 m = 7m

CSA of conical tomb = πrl

= (22/7)×7×25 = 550

CSA of conical tomb= 550m2

Cost of white-washing 550 m2 area, which is Rs (210×550)/100

= Rs. 1155

Therefore, cost will be Rs. 1155 while white-washing tomb.

7. A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. (Assume π =22/7)

Solution:

Radius of conical cap, r = 7 cm

Height of conical cap, h = 24cm

Slant height, l2 = (r2+h2)

= (72+242)

= (49+576)

= (625)

Or l = 25 cm

CSA of 1 conical cap = πrl

= (22/7)×7×25

= 550

CSA of 10 caps = (10×550) cm2 = 5500 cm2

Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) =1.02)

Solution:

Given:

Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m

Height of cone, h = 1m

Slant height of cone is l, and l= (r2+h2)

Using given values, l2 = (0.22+12)

= (1.04)

Or l = 1.02

Slant height of the cone is 1.02 m

Now,

CSA of each cone = πrl

= (3.14×0.2×1.02)

= 0.64056

CSA of 50 such cones = (50×0.64056) = 32.028

CSA of 50 such cones = 32.028 m2

Again,

Cost of painting 1 m2 area = Rs 12 (given)

Cost of painting 32.028 m2 area = Rs (32.028×12)

= Rs.384.336

= Rs.384.34 (approximately)

Therefore, the cost of painting all these cones is Rs. 384.34.

## Exercise 13.4 Page No: 225

1. Find the surface area of a sphere of radius:

(i) 10.5cm (ii) 5.6cm (iii) 14cm

(Assume π=22/7)

Solution:

Formula: Surface area of sphere (SA) = 4πr2

(i) Radius of sphere, r = 10.5 cm

SA = 4×(22/7)×10.5= 1386

Surface area of sphere is 1386 cm2

(ii) Radius of sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.6= 394.24

Surface area of sphere is 394.24 cm2

(iii) Radius of sphere, r = 14cm

SA = 4πr2

= 4×(22/7)×(14)2

= 2464

Surface area of sphere is 2464 cm2

2. Find the surface area of a sphere of diameter:

(i) 14cm (ii) 21cm (iii) 3.5cm

(Assume π = 22/7)

Solution:

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for Surface area of sphere = 4πr2

= 4×(22/7)×72 = 616

Surface area of a sphere is 616 cm2

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of sphere = 4πr2

= 4×(22/7)×10.5= 1386

Surface area of a sphere is 1386 cm2

Therefore, the surface area of a sphere having diameter 21cm is 1386 cm2

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr2

= 4×(22/7)×1.752 = 38.5

Surface area of a sphere is 38.5 cm2

3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]

Solution:

Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr2

= 3×3.14×102 = 942

The total surface area of given hemisphere is 942 cm2.

4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:

Let rand r2 be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So

r= 7cm

r= 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4r12/4r22

= (r1/r2)2

= (7/14)= (1/2)2 = ¼

Therefore, the ratio between the surface areas is 1:4.

5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7)

Solution:

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr2

= 2×(22/7)×(5.25)2 = 173.25

Surface area of hemispherical bowl is 173.25 cm2

Cost of tin-plating 100 cm2 area = Rs 16

Cost of tin-plating 1 cm2 area = Rs 16 /100

Cost of tin-plating 173.25 cmarea = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm2 is Rs 27.72.

6. Find the radius of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Solution:

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr= 154

r= (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

7. The diameter of the moon is approximately one fourth of the diameter of the earth.

Find the ratio of their surface areas.

Solution:

If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)2

Surface area of earth = 4π(d/2)2

Ratio of their Surface area;

$\begin{array}{l}=\frac{4\pi\left(\frac{\mathrm{d}}{8}\right)^2}{4\pi\left(\frac{\mathrm{d}}{2}\right)^2}=\frac{4}{64}\\ =\frac{1}{16}\end{array}$

The ratio between their surface areas is 1:16.

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)

Solution:

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of hemispherical bowl = 2πr2, where r is radius of hemisphere

= 2×(22/7)×(5.25)2 = 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cm2.

9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in(i) and (ii). Solution:

(i) Surface area of sphere = 4πr2, where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r

CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr2

(iii) Ratio between areas = (Surface area of sphere)/CSA of Cylinder)

= 4r2/4r= 1/1

Ratio of the areas obtained in (i) and (ii) is 1:1.

## Exercise 13.5 Page No: 228

1. A matchbox measures 4 cm×2.5cm×1.5cm. What will be the volume of a packet containing 12 such boxes?

Solution:

Dimensions of a matchbox (a cuboid) are l×b×h = 4 cm×2.5 cm×1.5 cm respectively

Formula to find the volume of matchbox = l×b×h = (4×2.5×1.5) = 15

Volume of matchbox = 15 cm3

Now, volume of 12 such matchboxes = (15×12) cm3 = 180 cm3

Therefore, the volume of 12 matchboxes is 180cm3.

2. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? (1 m3= 1000 l)

Solution:

Dimensions of a cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m

Formula to find volume of tank, V = l×b×h

Put the values, we get

V = (6×5×4.5) = 135

Volume of water tank is 135 m3

Again,

We are given that, amount of water that 1mvolume can hold = 1000 l

Amount of water, 135 m3volume hold = (135×1000) litres = 135000 litres

Therefore, given cuboidal water tank can hold up to135000 litres of water.

3. A cuboidal vessel is 10m long and 8m wide. How high must it be made to hold 380 cubic metres of a liquid?

Solution:

Given:

Length of cuboidal vessel, l = 10 m

Width of cuboidal vessel, b = 8m

Volume of cuboidal vessel, V = 380 m3

Let the height of the given vessel be h.

Formula for Volume of a cuboid, V = l×b×h

Using formula, we have

l×b×h = 380

10×8×h= 380

Or h = 4.75

Therefore, the height of the vessels is 4.75 m.

4. Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3.

Solution:

The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m.

Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula)

Required Volume is 144 m3

Now,

Cost of digging per m3 volume = Rs 30

Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Solution:

Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.

To find: The value of breadth, say b.

Formula to find the volume of a tank = l×b×h = (2.5× b×10) m3= 25b m3

Capacity of tank= 25b m3, which is equal to 25000b litres

Also, capacity of a cuboidal tank is 50000 litres of water (Given)

Therefore, 25000 b = 50000

This implies, b = 2

Therefore, the breadth of the tank is 2 m.

6. A village, having a population of 4000, requires 150 litres of water per head per day.

It has a tank measuring 20 m×15 m×6 m. For how many days will the water of this tank last?

Solution:

Length of the tank = l = 20 m

Breadth of the tank = b = 15 m

Height of the tank = h = 6 m

Total population of a village = 4000

Consumption of the water per head per day = 150 litres

Water consumed by the people in 1 day = (4000×150) litres = 600000 litres …(1)

Formula to find the capacity of tank, C = l×b×h

Using given data, we have

C = (20×15×6) m3= 1800 m3

Or C = 1800000 litres

Let water in this tank last for d days.

Water consumed by all people in d days = Capacity of tank (using equation (1))

600000 d =1800000

d = 3

Therefore, the water of this tank will last for 3 days. Answer

7. A godown measures 40 m×25m×15 m. Find the maximum number of wooden crates each

measuring 1.5m×1.25 m×0.5 m that can be stored in the godown.

Solution:

From statement, we have

Length of the godown = 40 m

Height = 15 m

Whereas,

Length of the wooden crate = 1.5 m

Height = 0.5 m

Since godown and wooden crate are in cuboidal shape. Find the volume of each using formula, V = lbh.

Now,

Volume of godown = (40×25×15) m3 = 15000 m3

Volume of a wooden crate = (1.5×1.25×0.5) m3 = 0.9375 m3

Let us consider that, n wooden crates can be stored in the godown, then

Volume of n wooden crates = Volume of godown

0.9375×n =15000

Or n= 15000/0.9375 = 16000

Hence, the number of wooden crates that can be stored in the godown is 16,000.

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Solution:

Side of a cube = 12cm (Given)

Find the volume of cube:

Volume of cube = (Side)= (12)3cm3= 1728cm3

Surface area of a cube with side 12 cm = 6a= 6(12) 2 cm2 …(1)

Cube is cut into eight small cubes of equal volume, say side of each cube is p.

Volume of a small cube = p3

Surface area = 6p2 …(2)

Volume of each small cube = (1728/8) cm3 = 216 cm3

Or (p)3 = 216 cm3

Or p = 6 cm

Now, Surface areas of the cubes ratios = (Surface area of bigger cube)/(Surface area of smaller cubes)

From equation (1) and (2), we get

Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4

Therefore, the required ratio is 4 : 1.

9. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will fall into the sea in a minute?

Solution:

Given:

Depth of river, h = 3 m

Width of river, b = 40 m

Rate of water flow = 2km per hour = 2000m/60min = 100/3 m/min

Now, Volume of water flowed in 1 min = (100/3) × 40 × 3 = 4000m3

Therefore, 4000 m3water will fall into the sea in a minute.

## Exercise 13.6 Page No: 230

1. The circumference of the base of cylindrical vessel is 132cm and its height is 25cm.

How many litres of water can it hold? (1000 cm3= 1L) (Assume π = 22/7)

Solution:

Circumference of the base of cylindrical vessel = 132 cm

Height of vessel, h = 25 cm

Let r be the radius of the cylindrical vessel.

Step 1: Find the radius of vessel

We know that, circumference of base = 2πr, so

2πr = 132 (given)

r = (132/(2 π))

r = 66×7/22 = 21

Step 2: Find the volume of vessel

Formula: Volume of cylindrical vessel = πr2h

= (22/7)×212×25

= 34650

Therefore, volume is 34650 cm3

Since, 1000 cm= 1L

So, Volume = 34650/1000 L= 34.65L

Therefore, vessel can hold 34.65 litres of water.

2. The inner diameter of a cylindrical wooden pipe is 24cm and its outer diameter is 28 cm. The length of the pipe is 35cm.Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6g. (Assume π = 22/7)

Solution:

Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm

Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm

Height of pipe, h = Length of pipe = 35cm

Now, the Volume of pipe = π(r22-r12)h cm3

Substitute the values.

Volume of pipe = 110×52 cm= 5720 cm3

Since, Mass of 1 cm3 wood = 0.6 g

Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg. Answer!

3. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5cm and width 4cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7cm and height 10cm. Which container has greater capacity and by how much? (Assume π=22/7)

Solution:

1. tin can will be cuboidal in shape Dimensions of tin can are

Length, l = 5 cm

Height, h = 15 cm

Capacity of tin can = l×b×h= (5×4×15) cm= 300 cm3

1. Plastic cylinder will be cylindrical in shape. Dimensions of plastic can are:

Radius of circular end of plastic cylinder, r = 3.5cm

Height , H = 10 cm

Capacity of plastic cylinder = πr2H

Capacity of plastic cylinder = (22/7)×(3.5)2×10 = 385

Capacity of plastic cylinder is 385 cm3

From results of (i) and (ii), plastic cylinder has more capacity.

Difference in capacity = (385-300) cm3 = 85cm3

4. If the lateral surface of a cylinder is 94.2cm2 and its height is 5cm, then find

(i) radius of its base (ii) its volume.[Use π= 3.14]

Solution:

CSA of cylinder = 94.2 cm2

Height of cylinder, h = 5cm

(i) Let radius of cylinder be r.

Using CSA of cylinder, we get

2πrh = 94.2

2×3.14×r×5 = 94.2

r = 3

(ii) Volume of cylinder

Formula for volume of cylinder = πr2h

Now, πr2h = (3.14×(3)2×5) (using value of r from (i))

= 141.3

Volume is 141.3 cm3

5. It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10m deep. If the cost of painting is at the rate of Rs 20 per m2, find

(i) inner curved surface area of the vessel

(iii) capacity of the vessel

(Assume π = 22/7)

Solution:

(i) Rs 20 is the cost of painting 1 marea.

Rs 1 is the cost to paint 1/20 marea

So, Rs 2200 is the cost of painting = (1/20×2200) m2

= 110 m2 area

The inner surface area of the vessel is 110m2.

(ii) Radius of the base of the vessel, let us say r.

Height (h) = 10 m and

Surface area formula = 2πrh

Using result of (i)

2πrh = 110 m2

2×22/7×r×10 = 110

r = 1.75

(iii) Volume of vessel formula = πr2h

Here r = 1.75 and h = 10

Volume = (22/7)×(1.75)2×10 = 96.25

Volume of vessel is 96.25 m3

Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

6. The capacity of a closed cylindrical vessel of height 1m is15.4 liters. How many square meters of metal sheet would be needed to make it? (Assume π = 22/7)

Solution:

Height of cylindrical vessel, h = 1 m

Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3

Let r be the radius of the circular end.

Now,

Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154

After simplifying, we get, r = 0.07 m

Again, total surface area of vessel = 2πr(r+h)

= 2×22/7×0.07(0.07+1)

= 0.44×1.07

= 0.4708

Total surface area of vessel is 0.4708 m2

Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.

7. A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. (Assume π = 22/7)

Solution: Radius of pencil, r= 7/2 mm = 0.7/2 cm = 0.35 cm

Radius of graphite, r= 1/2 mm = 0.1/2 cm = 0.05 cm

Height of pencil, h = 14 cm

Formula to find, volume of wood in pencil = (r12-r22)h cubic units

Substitute the values, we have

= [(22/7)×(0.352-0.052)×14]

= 44×0.12

= 5.28

This implies, volume of wood in pencil = 5.28 cm3

Again,

Volume of graphite = r22h cubic units

Substitute the values, we have

= (22/7)×0.052×14

= 44×0.0025

= 0.11

So, the volume of graphite is 0.11 cm3.

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7cm. If the bowl is filled with soup to a height of 4cm, how much soup the hospital has to prepare daily to serve 250 patients? (Assume π = 22/7)

Solution:

Diameter of cylindrical bowl = 7 cm

Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm

Bowl is filled with soup to a height of4cm, so h = 4 cm Volume of soup in one bowl= πr2h

(22/7)×3.52×4 = 154

Volume of soup in one bowl is 154 cm3

Therefore,

Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3

## Exercise 13.7 Page No: 233

1. Find the volume of the right circular cone with

(i) radius 6cm, height 7 cm (ii) radius 3.5 cm, height 12 cm (Assume π = 22/7)

Solution:

Volume of cone = (1/3) πr2h cube units

Where r be radius and h be the height of the cone

(i) Radius of cone, r = 6 cm

Height of cone, h = 7cm

Say, V be the volume of the cone, we have

V = (1/3)×(22/7)×36×7

= (12×22)

= 264

The volume of the cone is 264 cm3.

(ii) Radius of cone, r = 3.5cm

Height of cone, h = 12cm

Volume of cone = (1/3)×(22/7)×3.52×7 = 154

Hence,

The volume of the cone is 154 cm3.

2. Find the capacity in litres of a conical vessel with

(i) radius 7cm, slant height 25 cm (ii) height 12 cm, slant height 12 cm

(Assume π = 22/7)

Solution:

(i) Radius of cone, r =7 cm

Slant height of cone, l = 25 cm or h = 24

Height of the cone is 24 cm

Now,

Volume of cone, V = (1/3) πr2h (formula)

V = (1/3)×(22/7) ×72×24

= (154×8)

= 1232

So, the volume of the vessel is 1232 cm3

Therefore, capacity of the conical vessel = (1232/1000) liters (because 1L = 1000 cm3)

= 1.232 Liters.

(ii) Height of cone, h = 12 cm

Slant height of cone, l = 13 cm r = 5

Hence, the radius of cone is 5 cm.

Now, Volume of cone, V = (1/3)πr2h

V = (1/3)×(22/7)×52×12 cm3

= 2200/7

Volume of cone is 2200/7 cm3

Now, Capacity of the conical vessel= 2200/7000 litres (1L = 1000 cm3)

= 11/35 litres

3. The height of a cone is 15cm. If its volume is 1570cm3, find the diameter of its base. (Use π = 3.14)

Solution:

Height of the cone, h = 15 cm

Volume of cone =1570 cm3

Let r be the radius of the cone

As we know: Volume of cone, V = (1/3) πr2h

So, (1/3) πr2h = 1570

(1/3)×3.14×r×15 = 1570

r2 = 100

r = 10

Radius of the base of cone 10 cm.

4. If the volume of a right circular cone of height 9cm is 48πcm3, find the diameter of its base.

Solution:

Height of cone, h = 9cm

Volume of cone =48π cm3

Let r be the radius of the cone.

As we know: Volume of cone, V = (1/3) πr2h

So, 1/3 π r2(9) = 48 π

r2 = 16

r = 4

Radius of cone is 4 cm.

So, diameter = 2×Radius = 8

Thus, diameter of base is 8cm.

5. A conical pit of top diameter 3.5m is 12m deep. What is its capacity in kiloliters?

(Assume π = 22/7)

Solution:

Diameter of conical pit = 3.5 m

Radius of conical pit, r = diameter/ 2 = (3.5/2)m = 1.75m

Height of pit, h = Depth of pit = 12m

Volume of cone, V = (1/3) πr2h

V = (1/3)×(22/7) ×(1.75)2×12 = 38.5

Volume of cone is 38.5 m3

Hence, capacity of the pit = (38.5×1) kiloliters = 38.5 kiloliters.

6. The volume of a right circular cone is 9856cm3. If the diameter of the base is 28cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

(Assume π = 22/7)

Solution:

Volume of a right circular cone = 9856 cm3

Diameter of the base = 28 cm

(i) Radius of cone, r = (28/2) cm = 14 cm

Let the height of the cone be h

Volume of cone, V = (1/3) πr2h

(1/3) πr2h = 9856

(1/3)×(22/7) ×14×14×h = 9856

h = 48

The height of the cone is 48 cm. Slant height of the cone is 50 cm.

(iii) curved surface area of cone = πrl

= (22/7)×14×50

= 2200

curved surface area of the cone is 2200 cm2.

7. A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Solution:

Height (h)= 12 cm

Radius (r) = 5 cm, and

Slant height (l) = 13 cm Volume of cone, V = (1/3) πr2h

V = (1/3)×π×52×12

= 100π

Volume of the cone so formed is 100π cm3.

8. If the triangle ABC in the Question 7 is revolved about the side 5cm, then find the volume of the solids so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution: A right-angled ΔABC is revolved about its side 5cm, a cone will be formed of radius as 12 cm, height as 5 cm, and slant height as 13 cm.

Volume of cone = (1/3) πr2h; where r is the radius and h be the height of cone

= (1/3)×π×12×12×5

= 240 π

The volume of the cones of formed is 240π cm3.

So, required ratio = (result of question 7) / (result of question 8) = (100π)/(240π) = 5/12 = 5:12.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas.

(Assume π = 22/7)

Solution:

Radius (r) of heap = (10.5/2) m = 5.25

Height (h) of heap = 3m

Volume of heap = (1/3)πr2h

= (1/3)×(22/7)×5.25×5.25×3

= 86.625

The volume of the heap of wheat is 86.625 m3.

Again, = (22/7)×5.25×6.05

= 99.825

Therefore, the area of the canvas is 99.825 m2.

## Exercise 13.8 Page No: 236

1. Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m

(Assume π =22/7)

Solution:

(i) Radius of sphere, r = 7 cm

Using, Volume of sphere = (4/3) πr3

= (4/3)×(22/7)×73

= 4312/3

Hence, volume of the sphere is 4312/3 cm3

(ii) Radius of sphere, r = 0.63 m

Using, volume of sphere = (4/3) πr3

= (4/3)×(22/7)×0.633

= 1.0478

Hence, volume of the sphere is 1.05 m(approx).

2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

(Assume π =22/7)

Solution:

(i) Diameter = 28 cm

Radius, r = 28/2 cm = 14cm

Volume of the solid spherical ball = (4/3) πr3

Volume of the ball = (4/3)×(22/7)×14= 34496/3

Hence, volume of the ball is 34496/3 cm3

(ii) Diameter = 0.21 m

Radius of the ball =0.21/2 m= 0.105 m

Volume of the ball = (4/3 )πr3

Volume of the ball = (4/3)× (22/7)×0.1053 m3

Hence, volume of the ball = 0.004851 m3

3.The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3? (Assume π=22/7)

Solution:

Given,

Diameter of a metallic ball = 4.2 cm

Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm

Volume formula = 4/3 πr3

Volume of the metallic ball = (4/3)×(22/7)×2.1 cm3

Volume of the metallic ball = 38.808 cm3

Now, using relationship between, density, mass and volume,

Density = Mass/Volume

Mass = Density × volume

= (8.9×38.808) g

= 345.3912 g

Mass of the ball is 345.39 g (approx).

4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution:

Let the diameter of earth be “d”. Therefore, the radius of earth will be will be d/2

Diameter of moon will be d/4 and the radius of moon will be d/8

Find the volume of the moon :

Volume of the moon = (4/3) πr= (4/3) π (d/8)3 = 4/3π(d3/512)

Find the volume of the earth :

Volume of the earth = (4/3) πr3= (4/3) π (d/2)3 = 4/3π(d3/8)

Fraction of the volume of the earth is the volume of the moon Answer: Volume of moon is of the 1/64 volume of earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5cm hold? (Assume π = 22/7)

Solution:

Diameter of hemispherical bowl = 10.5 cm

Radius of hemispherical bowl, r = 10.5/2 cm = 5.25 cm

Formula for volume of the hemispherical bowl = (2/3) πr3

Volume of the hemispherical bowl = (2/3)×(22/7)×5.253 = 303.1875

Volume of the hemispherical bowl is 303.1875 cm3

Capacity of the bowl = (303.1875)/1000 L = 0.303 litres(approx.)

Therefore, hemispherical bowl can hold 0.303 litres of milk.

6. A hemi spherical tank is made up of an iron sheet 1cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. (Assume π = 22/7)

Solution:

Inner Radius of the tank, (r ) = 1m

Outer Radius (R ) = 1.01m

Volume of the iron used in the tank = (2/3) π(R3– r3)

Put values,

Volume of the iron used in the hemispherical tank = (2/3)×(22/7)×(1.013– 13) = 0.06348

So, volume of the iron used in the hemispherical tank is 0.06348 m3.

7. Find the volume of a sphere whose surface area is 154 cm2. (Assume π = 22/7)

Solution:

Let r be the radius of a sphere.

Surface area of sphere = 4πr2

4πr= 154 cm2 (given)

r2 = (154×7)/(4 ×22)

r = 7/2

Now,

Volume of the sphere = (4/3) πr3 8. A dome of a building is in the form of a hemi sphere. From inside, it was white-washed at the cost of Rs. 4989.60. If the cost of white-washing isRs20 per square meter, find the

(i) inside surface area of the dome (ii) volume of the air inside the dome

(Assume π = 22/7)

Solution:

(i) Cost of white-washing the dome from inside = Rs 4989.60

Cost of white-washing 1m2 area = Rs 20

CSA of the inner side of dome = 498.96/2 m2  = 249.48 m2

(ii) Let the inner radius of the hemispherical dome be r.

CSA of inner side of dome = 249.48 m2 (from (i))

Formula to find CSA of a hemi sphere = 2πr2

2πr2 = 249.48

2×(22/7)×r= 249.48

r= (249.48×7)/(2×22)

r= 39.69

r = 6.3

Volume of air inside the dome = Volume of hemispherical dome

Using formula, volume of the hemisphere = 2/3 πr3

= (2/3)×(22/7)×6.3×6.3×6.3

= 523.908

= 523.9(approx.)

Answer: Volume of air inside the dome is 523.9 m3.

9. Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the

(i) radius r’ of the new sphere,

(ii) ratio of Sand S’.

Solution:

Volume of the solid sphere = (4/3)πr3

Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3

(i) New solid iron sphere radius = r’

Volume of this new sphere = (4/3)π(r’)3

(4/3)π(r’)= 36 π r3

(r’)= 27r3

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr2

Surface area of iron sphere of radius r’= 4π (r’)2

Now

S/S’ = (4πr2)/( 4π (r’)2)

S/S’ = r2/(3r’)2 = 1/9

The ratio of S and S’ is 1: 9.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5mm. How much medicine (in mm3) is needed to fill this capsule? (Assume π = 22/7)

Solution:

Diameter of capsule = 3.5 mm

Radius of capsule, say r = diameter/ 2 = (3.5/2) mm = 1.75mm

Volume of spherical capsule = 4/3 πr3

Volume of spherical capsule = (4/3)×(22/7)×(1.75)3 = 22.458

Answer: The volume of the spherical capsule is 22.46 mm3.

## Exercise 13.9 Page No: 236

1. A wooden bookshelf has external dimensions as follows: Height = 110cm, Depth = 25cm,

Breadth = 85cm (see fig. 13.31). The thickness of the plank is 5cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf. Solution:

External dimensions of book self,

Length, l = 85cm

Height, h = 110 cm

External surface area of shelf while leaving out the front face of the shelf

= lh+2(lb+bh)

= [85×110+2(85×25+25×110)] = (9350+9750) = 19100

External surface area of shelf is 19100 cm2

Area of front face = [85×110-75×100+2(75×5)] = 1850+750

So, area is 2600 cm2

Area to be polished = (19100+2600) cm= 21700 cm2.

Cost of polishing 1 cm2 area = Rs 0.20

Cost of polishing 21700 cm2 area Rs. (21700×0.20) = Rs 4340

Dimensions of row of the book shelf

Length(l) = 75 cm

Breadth (b) = 20 cm and

Height(h) = 30 cm

Area to be painted in one row= 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450

So, area is 6450 cm2.

Area to be painted in 3 rows = (3×6450)cm= 19350 cm2.

Cost of painting 1 cm2 area = Rs. 0.10

Cost of painting 19350 cm2 area = Rs (19350 x 0.1) = Rs 1935

Total expense required for polishing and painting = Rs. (4340+1935) = Rs. 6275

Answer: The cost for polishing and painting the surface of the book shelf is Rs. 6275.

2. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in fig. 13.32. Eight such spheres are used forth is purpose, and are to be painted silver. Each support is a cylinder of radius 1.5cm and height 7cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2. Solution:

Diameter of wooden sphere = 21 cm

Radius of wooden sphere, r = diameter/ 2 = (21/2) cm = 10.5 cm

Formula: Surface area of wooden sphere = 4πr2

= 4×(22/7)×(10.5)= 1386

So, surface area is 1386 cm3

Radius of the circular end of cylindrical support = 1.5 cm

Height of cylindrical support = 7 cm

Curved surface area = 2πrh

= 2×(22/7)×1.5×7 = 66

So, CSA is 66 cm2

Now,

Area of the circular end of cylindrical support = πr2

= (22/7)×1.52

= 7.07

Area of the circular end is 7.07 cm2

Again,

Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44

Area to be painted is 11031.44 cm2

Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86

Area to be painted black = (8×66) cm2 = 528 cm2

Cost for painting with black colour =Rs (528×0.05) = Rs26.40

Therefore, the total painting cost is:

= Rs(2757.86 +26.40)

= Rs 2784.26

3. The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?

Solution:

Let the diameter of the sphere be “d”.

Radius of sphere, r1 = d/2

New radius of sphere, say r2 = (d/2)×(1-25/100) = 3d/8

Curved surface area of sphere, (CSA)1 = 4πr12 = 4π×(d/2)2 = πd2 …(1)

Curved surface area of sphere when radius is decreased (CSA)= 4πr22 = 4π×(3d/8)2 = (9/16)πd2 …(2)

From equation (1) and (2), we have

Decrease in surface area of sphere = (CSA)1 – (CSA)2

= πd– (9/16)πd2

= (7/16)πd2

Percentage decrease in Surface area of sphere;

$=\frac{(C S A)_{1}-(\mathrm{CSA})_{2}}{(\mathrm{CSA})_{1}} \times 100$

= (7d2/16d2)×100 = 700/16 = 43.75% .

Therefore, the percentage decrease in the surface area of the sphere is 43.75% .

### NCERT Solutions for Class 9 Maths Chapter 13

NCERT Solutions Class 9 Maths Chapter 13 helps you find the surface areas and volumes of cuboids and cylinders, cones, and spheres. This chapter explains how the area is found by multiplying the length and breadth of various objects. NCERT Solutions Class 9 Maths Chapter 13 includes all types of exercise problems from basic to advance level questions to make students prepare for the Term II and competitive examinations. NCERT Solutions for Class 9 Maths Chapter 13 also includes the activities explained in a better way for the understanding of the concepts before solving the questions. NCERT Solutions for Class 9 Maths Chapter 13 includes the following topics:

• Surface area of a cuboid and a cube
• Surface area of a right circular cylinder
• Surface area of a right circular cone
• Surface area of a sphere
• Volume of a cuboid
• Volume of a cylinder
• Volume of a right circular cone
• Volume of a sphere
• NCERT Solutions for Class 9 Chapter 13 explains the formation of various geometrical objects.
• It signifies the importance of different formulas in finding the surface area and volume for various objects.
• Details of the cuboid, cube, right circular cone, cylinder, hemisphere, and sphere are given.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes:

• Lists the important formula to find surface areas and volumes of the cube, cuboid, cylinder, cone, and sphere.
• It is designed for the students to remember the formulas and apply them relevantly.
• These solutions will be useful for CBSE Term II exams, competitive exams, and even for Maths Olympiads too.
• Answers are aimed at providing an effortless solution in finding the surface area and volumes.
• Provides completely solved solutions to all the questions present in the respective NCERT textbooks.

Apart from the questions from the NCERT Textbook, students are also advised to solve questions from other study materials. It will not only provide a strong conceptual knowledge but also improves their abilities to solve difficult questions effortlessly.

• RD Sharma Solutions for Class 9 Maths Chapter 18 Surface Area and Volume of a Cuboid and Cube

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 13

### Why should we follow NCERT Solutions for Class 9 Maths Chapter 13?

NCERT Solutions for Class 9 Maths Chapter 13 is the best reference material that offers complete and quality information about different Math concepts. The questions that have been given in the solutions have been solved in an easy-to-remember format, which further helps students to clearly understand and remember the answers. So, it’s clear that NCERT Textbooks for Class 9 are essential reference books to score high in term-wise examinations. To score good marks, practising these solutions for Class 9 Maths can help to a great extent.

### How Our Team NCERT Solutions for Class 9 Maths Chapter 13 help the students in preparing for CBSE Term II exams?

Our solution module uses various examples and diagrams to explain the questions, wherever necessary. For CBSE term II examination students aiming at securing an excellent score, solving NCERT Solutions for Class 9 is a must. These NCERT Solutions for Class 9 Maths Chapter 13 helps the students in gaining a better knowledge of the topics covered. Solving the questions from each exercise will ensure that the students score high marks in the CBSE Term II exams.

### What are the topics covered under NCERT Solutions for Class 9 Maths Chapter 13?

NCERT Solutions for Class 9 Maths Chapter 13 includes the following topics:
1. Surface area of a cuboid and a cube
2. Surface area of a right circular cylinder
3. Surface area of a right circular cone
4. Surface area of a sphere
5. Volume of a cuboid
6. Volume of a cylinder
7. Volume of a right circular cone
8. Volume of a sphere
• NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2

The best way to understand Heron’s Formula with ease is through NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula Exercise 12.2 and 12.1. These exercises are designed to test your conceptual grasp of the subject.

From an educational point of view, this elementary concept is significant in various fields of study. Therefore, it is essential to have a good grasp of the concept. Another key point is that these solutions are designed by highly competent teachers with many years of relevant experience.

Therefore, NCERT Solutions is one of the best guides for studies. Important topics are presented in an easy-to-understand format. Moreover, the use of complex jargons is barred throughout the content. Furthermore, content is updated as per the prescribed CBSE syllabus.

### Access other exercise solutions of Class 9 Maths Chapter 12 – Heron’s Formula

Exercise 12.1 Solutions – 6 Questions

### Access Answers of Maths NCERT Class 9 Chapter 12 – Heron’s Formula Exercise 12.2

1. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? Solution: First, construct a quadrilateral ABCD and join BD. We know that C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m The diagram is: Now, apply Pythagoras theorem in ΔBCD BD2 = BC+CD2 ⇒ BD2 = 122+52 ⇒ BD2 = 169 ⇒ BD = 13 m Now, the area of ΔBCD = (½ ×12×5) = 30 m2 The semi perimeter of ΔABD (s) = (perimeter/2) = (8+9+13)/2 m = 30/2 m = 15 m Using Heron’s formula, Area of ΔABD = 6√35 m= 35.5 m2 (approximately) ∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD = 30 m2+35.5m2 = 65.5 m2 2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. Solution: First, construct a diagram with the given parameter. Now, apply Pythagorean theorem in ΔABC, AC2 = AB2+BC2 ⇒ 52 = 32+42 ⇒ 25 = 25 Thus, it can be concluded that ΔABC is a right angled at B. So, area of ΔBCD = (½ ×3×4) = 6 cm2 The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m Now, using Heron’s formula, Area of ΔABD = 2√21 cm= 9.17 cm2 (approximately) Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm+9.17 cm2 = 15.17 cm2 3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used. Solution: For the triangle I section: It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm Perimeter = 5+5+1 = 11 cm So, semi perimeter = 11/2 cm = 5.5 cm Using Heron’s formula, Area = √[s(s-a)(s-b)(s-c)] = √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm2 = √[5.5×0.5×0.5×4.5] cm2 = 0.75√11 cm2 = 0.75 × 3.317cm2 = 2.488cm2 (approx) For the quadrilateral II section: This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively. ∴ Area = 6.5×1 cm2=6.5 cm2 For the quadrilateral III section: It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm. Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle The perpendicular height of the parallelogram will be = 0.86 cm And, the area of the equilateral triangle will be (√3/4×a2) = 0.43 ∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm(approximately). For triangle IV and V: These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm Area triangles IV and V = 2×(½×6×1.5) cm= 9 cm2 So, the total area of the paper used = (2.488+6.5+1.3+9) cm= 19.3 cm2 4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram. Solution: Given, It is given that the parallelogram and triangle have equal areas. The sides of the triangle are given as 26 cm, 28 cm and 30 cm. So, the perimeter = 26+28+30 = 84 cm And its semi perimeter = 84/2 cm = 42 cm Now, by using Heron’s formula, area of the triangle = = √[42(42-26)(42-28)(42-30)] cm2 = √[42×16×14×12] cm2 = 336 cm2 Now, let the height of parallelogram be h. As the area of parallelogram = area of the triangle, 28 cm× h = 336 cm2 ∴ h = 336/28 cm So, the height of the parallelogram is 12 cm. 5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? Solution: Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows. Consider the triangle BCD, Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m Using Heron’s formula, Area of the ΔBCD = = 432 m2 ∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m= 864 m2 Thus, the area of the grass field that each cow will be getting = (864/18) m= 48 m2 6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella? Solution: For each triangular piece, The semi perimeter will be s = (50+50+20)/2 cm = 120/2 cm = 60cm Using Heron’s formula, Area of the triangular piece = = √[60(60-50)(60-50)(60-20)] cm2 = √[60×10×10×40] cm2 = 200√6 cm2 ∴ The area of all the triangular pieces = 5 × 200√6 cm= 1000√6 cm2 7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it? Solution: As the kite is in the shape of a square, its area will be A = (½)×(diagonal)2 Area of the kite = (½)×32×32 = 512 cm2. The area of shade I = Area of shade II 512/2 cm= 256 cm2 So, the total area of the paper that is required in each shade = 256 cm2 For the triangle section (III), The sides are given as 6 cm, 6 cm and 8 cm Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm By using Heron’s formula, the area of the III triangular piece will be = = √[10(10-6)(10-6)(10-8)] cm2 = √(10×4 ×4×2) cm2 = 8√5 cm2 = 17.92 cm2 (approx.) 8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2 . Solution: The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm By using Heron’s formula, The area of each triangular shape will be = 36√6 cm= 88.2 cm2 Now, the total area of 16 tiles = 16×88.2 cm= 1411.2 cm2 It is given that the polishing cost of tiles = 50 paise/cm2 ∴ The total polishing cost of the tiles = Rs. (1411.2×0.5) = Rs. 705.6 9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. Solution: First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD. Now, it can be seen that the quadrilateral ABED is a parallelogram. So, AB = ED = 10 m AD = BE = 13 m EC = 25-ED = 25-10 = 15 m Now, consider the triangle BEC, Its semi perimeter (s) = (13+14+15)/2 = 21 m By using Heron’s formula, Area of ΔBEC = = 84 m2 We also know that the area of ΔBEC = (½)×CE×BF 84 cm= (½)×15×BF BF = (168/15) cm = 11.2 cm So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m2 ∴ Area of the field = 84+112 = 196 m2

This formula is used to find the area of a triangle when the length of the 3 sides is the only data available.

Heron also contributed in several other ways, the most notable one being the invention of the Aeolipile. The Aeolipile was the very first steam engine.

Though it was technically a steam engine, it didn’t resemble an engine. And no practical applications could be found for this invention. It was projected as a toy and a novelty item for the ancient Greeks.

Explore other important concepts related to Heron’s Formula, learn how to solve numerical problems and more, only on the best resources online to help you study better.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2

1. Well – structured content
2. All relevant formulas highlighted
3. Jargon-free and simple language used
4. Solutions drafted by well-qualified teachers
5. Questions are updated regularly
6. Breakdown of tough questions for easy comprehension
• NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula Exercise 12.1
NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula Exercise 12.1 helps you understand Heron’s Formula with ease. From an academic perspective, this fundamental concept finds significance in a myriad of areas. Therefore, it is necessary to have a clear grasp of the concept. These solutions are designed by highly knowledgeable teachers with countless years of experience. In other words, NCERT Solutions is one of the best guides for your study needs. Relevant topics are presented in an easy-to-understand format. Moreover, we have barred the use of complex jargons. Furthermore, its content is updated regularly as per prescribed CBSE syllabus.

### Access Answers of Maths NCERT Class 9 Chapter 12 – Heron’s Formula Exercise 12.1

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board? Solution: Given, Side of the signal board = a Perimeter of the signal board = 3a = 180 cm ∴ a = 60 cm Semi perimeter of the signal board (s) = 3a/2 By using Heron’s formula, Area of the triangular signal board will be = 2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay? Solution: The sides of the triangle ABC are 122 m, 22 m and 120 m respectively. Now, the perimeter will be (122+22+120) = 264 m Also, the semi perimeter (s) = 264/2 = 132 m Using Heron’s formula, Area of the triangle = =1320 m2

We know that the rent of advertising per year = ₹ 5000 per m2

∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000 3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. Solution: It is given that the sides of the wall as 15 m, 11 m and 6 m. So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m Using Heron’s formula, Area of the message = = √[16(16-15)(16-11) (16-6)] m2 = √[16×1×5×10] m= √800 m2 = 20√2 m2 4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm. Solution: Assume the third side of the triangle to be “x”. Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm It is given that the perimeter of the triangle = 42cm So, x = 42-(18+10) cm = 14 cm ∴ The semi perimeter of triangle = 42/2 = 21 cm Using Heron’s formula, Area of the triangle, = = √[21(21-18)(21-10)(21-14)] cm2 = √[21×3×11×7] m2 = 21√11 cm2 5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area. Solution: The ratio of the sides of the triangle are given as 12 : 17 : 25 Now, let the common ratio between the sides of the triangle be “x” ∴ The sides are 12x, 17x and 25x It is also given that the perimeter of the triangle = 540 cm 12x+17x+25x = 540 cm 54x = 540cm So, x = 10 Now, the sides of triangle are 120 cm, 170 cm, 250 cm. So, the semi perimeter of the triangle (s) = 540/2 = 270 cm Using Heron’s formula, Area of the triangle = 9000 cm2 6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle. Solution: First, let the third side be x. It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm. So, 30 = 12+12+x ∴ The length of the third side = 6 cm Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm Using Heron’s formula, Area of the triangle = = √[15(15-12)(15-12)(15-6)] cm2 = √[15×3×3×9] cm2 = 9√15 cm2

Heron’s formula is used to find the area of a triangle, provided if the length of the 3 sides are given. Apart from the formula, Heron contributed in many other ways, and the most notable was the invention of the very first steam engine. It was called the Aeolipile and technically, it didn’t resemble an engine. However, no practical applications could be identified and instead, it was used as a toy and a novelty for the ancient Greeks. Explore more about Heron’s Formula, learn how to solve related problems and more, only on NCERT Solutions For Class 9 Maths.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1

1. Structured content
2. Important formulas are highlighted
3. Simple language with jargon-free explanations
4. Solutions provided by qualified teachers
5. Updated questions with solutions
6. A thorough breakdown of tough questions

• NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula is provided here. Heron’s formula is a fundamental concept that finds significance in countless areas and is included in the first term CBSE Syllabus of Class 9 Maths. Therefore, it is imperative to have a clear grasp of the concept. And one of the best ways to do just that is by referring to the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula. These solutions are designed by knowledgeable teachers with years of experience in accordance with the latest update on CBSE term-wise syllabus 2021-22. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.

Hence, one of the best guides you could adapt for your study needs is NCERT Solutions. Relevant topics are presented in an easy to understand format, barring the use of any complicated jargons. Furthermore, its content is updated as per the last CBSE term-wise syllabus and its guidelines.

List of Exercises in NCERTclass 9 Maths Chapter 12:
• Area of a Triangle – by Heron’s Formula
• Application of Heron’s Formula in finding Areas of Quadrilaterals

Heron’s formula helps us to find the area of a triangle with 3 side lengths. Besides the formula, Heron also contributed in other ways – the most notable one being the inventor of the very first steam engine called the Aeolipile. However, Heron couldn’t find any practical applications for it, instead, it ended up being used as a toy and an object of curiosity for the ancient Greeks.

Explore more about Heron’s Formula and learn how to solve various kinds of problems only on NCERT Solutions For Class 9 Maths. It is also one of the best academic resources to revise for your first term exams.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

1. Well-structured content
2. Relevant formulas are highlighted
3. Easy-to-understand language and jargon-free explanations
4. Designed by qualified teachers
5. Latest questions with solutions from the updated term-wise syllabus
6. A thorough analysis of previous year question papers
7. Access to other learning resources such as sample papers and more
The expert faculty team of members have formulated the solutions in a lucid manner to improve the problem-solving abilities among the students. For a more clear idea about Heron’s Formula, students can refer to the study materials available at SSC CGL APEX COACHING.
• RD Sharma Solutions for Class 9 Maths Chapter 12 – Heron’s Formula

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 12

### What is the main aim of NCERT Solutions for Class 9 Maths Chapter 12?

The students are encouraged to go through NCERT Solutions for Class 9 Maths Chapter 12 thoroughly before the first term exams to score well and intensify their problem-solving abilities. Understanding the concepts and clearing the doubts immediately during self-study is the main aim of these solutions.

### What are the key benefits of learning NCERT Solutions for Class 9 Maths Chapter 12?

Key benefits of NCERT Solutions for Class 9 Maths Chapter 12 are provided here:
1. Students can easily access the solutions for each exercise from the chapters available.
2. The solutions also provide graphs and illustrations that help to understand the concepts clearly.
3. These solutions are prepared by our expert team focussing on accuracy.

### Why should we refer to NCERT Solutions for Class 9 Maths Chapter 12?

Students can refer to these NCERT Solutions for Class 9 Maths Chapter 12 which helps in gaining knowledge and for better guidance its best to solve these solutions. Solving these exercises in this chapter will assure that students can score good marks in the first term exams.
• NCERT Solutions for Class 9 Maths Exercise 11.2 Chapter 11 Construction
Construction of various shapes, especially in Geometry, is important as we learn various concepts through this topic. It is helpful as it has significance in other fields of study too. Moreover, this particular concept is important from an academic perspective as it carries a significant number of marks (28 marks from Unit 4, to be precise). Therefore, it is important to learn this concept for exams, and one of the best guides to use for the same is the NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2 and 11.1. Also, NCERT Solutions is held in high regards by many students due to its ease-of-use and versatile content. Consequently, NCERT Solutions is one of the best guides for studies. We have ensured that the conceptual aspects are well-articulated in the chapter. Problems are explained with easy examples and a step-by-step guide for tougher questions. Besides our detailed guide, content is regularly refreshed with immediate changes with respect to CBSE guidelines (if any).

### Access Answers to NCERT Class 9 Maths Chapter 11 – Constructions Exercise 11.2

1. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB+AC = 13 cm. Construction Procedure: The steps to draw the triangle of given measurement is as follows: 1. Draw a line segment of base BC = 7 cm 2. Measure and draw ∠B = 75° and draw the ray BX 3. Take a compass and measure AB+AC = 13 cm. 4. With B as the centre, draw an arc at the point be D 5. Join DC 6. Now draw the perpendicular bisector of the line DC and the intersection point is taken as A. 7. Now join AC 8. Therefore, ABC is the required triangle. 2. Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB–AC = 3.5 cm. Construction Procedure: The steps to draw the triangle of given measurement is as follows: 1. Draw a line segment of base BC = 8 cm 2. Measure and draw ∠B = 45° and draw the ray BX 3. Take a compass and measure AB-AC = 3.5 cm. 4. With B as centre and draw an arc at the point be D on the ray BX 5. Join DC 6. Now draw the perpendicular bisector of the line CD and the intersection point is taken as A. 7. Now join AC 8. Therefore, ABC is the required triangle. 3. Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR–PQ = 2cm. Construction Procedure: The steps to draw the triangle of given measurement is as follows: 1. Draw a line segment of base QR = 6 cm 2. Measure and draw ∠Q = 60° and let the ray be QX 3. Take a compass and measure PR–PQ = 2cm. 4. Since PR–PQ is negative, QD will below the line QR. 5. With Q as centre and draw an arc at the point be D on the ray QX 6. Join DR 7. Now draw the perpendicular bisector of the line DR and the intersection point is taken as P. 8. Now join PR 9. Therefore, PQR is the required triangle. 4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY+YZ+ZX = 11 cm. Construction Procedure: The steps to draw the triangle of given measurement is as follows: 1. Draw a line segment AB which is equal to XY+YZ+ZX = 11 cm. 2. Make an angle ∠Y = 30° from the point A and the angle be∠LAB 3. Make an angle ∠Z = 90° from the point B and the angle be ∠MAB 4. Bisect ∠LAB and ∠MAB at the point X. 5. Now take the perpendicular bisector of the line XA and XB and the intersection point be Y and Z, respectively. 6. Join XY and XZ 7. Therefore, XYZ is the required triangle 5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm. Construction Procedure: The steps to draw the triangle of given measurement is as follows: 1. Draw a line segment of base BC = 12 cm 2. Measure and draw ∠B = 90° and draw the ray BX 3. Take a compass and measure AB+AC = 18 cm. 4. With B as centre and draw an arc at the point be D on the ray BX 5. Join DC 6. Now draw the perpendicular bisector of the line CD and the intersection point is taken as A. 7. Now join AC 8. Therefore, ABC is the required triangle. Constructions Exercise 11.2 includes 5 practical questions that are based on the concept of geometrical shapes and their process of construction. These exercise questions are crucial from an examination perspective as it helps the students to understand and solve the various questions that might arise from the topic. It also helps students to realize a particular concept’s real-world applications and its significance. Discover various concepts such as a bisector, perpendicular bisector and more, only on NCERT Solutions For Class 9 Maths.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 11 – Constructions Exercise 11.2

1. Well-articulated content
2. Topics presented in a structured format
3. Important formulas are explained
4. Comprehensive content
5. Content created by experienced teachers