# Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone – Exercise 20.2

**Question 1. Find the volume of the right circular cone with:**

**(i) Radius 6cm, height 7cm**

**(ii) Radius 3.5cm, height 12cm**

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**(iii) Height is 21cm and slant height 28cm**

**Solution:**

(i)Radius of Cone(r)=6cmHeight of Cone(h)=7cm

As we know that the Volume of a Right Circular Cone = 1/3 πr

^{2}hBy putting the values in formula we get,

= 1/3 x 3.14 x 6

^{2}x 7 =264

Hence, Volume of a Right Circular Cone is 264 cm^{3}

(ii)Radius of Cone(r)=3.5 cmHeight of Cone(h)=12cm

Volume of a Right Circular Cone = 1/3 πr

^{2}hBy putting the values in formula we get,

= 1/3 x 3.14 x 3.5

^{2}x 12 =154

Hence, Volume of a Right Circular Cone is 154 cm^{3}

(iii)Height of Cone(h)=21 cmSlant height of Cone(l) = 28 cm

As we know that, l

^{2 }= r^{2}+ h^{2}282 = r

^{2}+ 212r = 7√7

As we know that Volume of a Right Circular Cone = 1/3 πr

^{2}hBy putting the values in formula we get,

= 1/3 x 3.14 x (7√7)

^{2}x 21 =7546

Hence, Volume of a Right Circular Cone is 7546 cm^{3}

**Question 2. Find the capacity in liters of a Conical Vessel with :**

**(i) radius 7 cm, slant height 25 cm**

**(ii) height 12 cm, slant height 13 cm.**

**Solution:**

**(i) **Radius of the Cone(r) =7 cm

Slant height of the Cone (l) =25 cm

As we know that l^{2} = r^{2} + h^{2}

252 = 72 + h^{2}

h = 24

We know that Volume of a right circular cone = = 1/3 πr^{2}h

By putting the values in formula we get,

= 1/3 x 3.14 x (7)^{2} x 24 = 1232

Hence Volume of a right circular cone is 1232 cm^{3} or 1.232 liters

**(ii) **Height of Cone(h)=12 cm

Slant height of Cone(l)=13 cm

As we know that l^{2} = r^{2 }+ h^{2}

132 = r^{2} + 122

r = 5

We know that Volume of a Right Circular Cone = 1/3 πr^{2}h

By putting the values in formula we get,

= 1/3 x 3.14 x (5)^{2} x 12 = **314.28**

**Hence, Volume of a right circular cone is 314.28 cm ^{3} or 0.314 liters.**

**Question 3. Two cones have their heights in the ratio 1:3 and the radii of their bases in the ratio 3:1. Find the ratio of their volumes.**

**Solution: **

Let us assume that the heights of the cones is h and 3h and radii of their bases is 3r and r. Then, their volumes are

Volume of 1st Cone (V1) = 1/3 π(3r)

^{2}hVolume of 2nd Cone (V2) = 1/3 πr

^{2}(3h)V1/V2 =

3/1

Hence, Ratio of two volumes is 3:1.

**Question 4. The radius and the height of a right circular cone are in the ratio 5:12. If its volume is 314 cubic meter, find the slant height and the radius. (Use **π **= 3.14).**

**Solution: **

Let us assume that the ratio of radius and the height of a Right Circular Cone to be x.

Then the radius be 5x and height be 12x

As we know that l

^{2 }= r^{2}+ h^{2}= (5x)

^{ 2}+ (12x)^{2}= 25 x

^{2 }+ 144 x^{2}l = 13x

Hence, Slant Height is 13 m.

Now it is given that the Volume of Cone = 314 m

^{3}= 1/3πr

^{2}h = 314= 1/3 x 3.14 x (25x

^{2 }) x (12x) = 314=x

^{3}=1x = 1

Hence,

Radius = 5x 1 = 5 m

Therefore, Slant height = 13m

**Question 5. The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use **π** = 3.14).**

**Solution: **

Let us assume that the ratio of radius and height of a right circular cone be y.

Radius of Cone(r) = 5y

Height of Cone (h) =12y

As we know that, l

^{2 }= r^{2}+ h^{2}= (5y)

^{2}+ (12y)^{2}= 25 y

^{2}+ 144 y^{2}= l = 13y

Volume of Cone, given 2512cm

^{3}= 1/3πr

^{2}h = 2512= 1/3 x 3.14 x (5y)

^{2}x 12y = 2512= y

^{3}= (2512 x 3)/(3.14 x 25 x 12) = 8= y = 2

Therefore,

Radius of Cone = 5y = 5×2 = 10cm

Slant Height (l) =13y = 13×2 = 26cm

**Question 6. The ratio of volumes of two cones is 4:5 and the ratio of the radii of their bases is 2:3. Find the ratio of their vertical heights.**

**Solution:**

Let us assume that the ratio of the Radius is x and Ratio of the Volume is y.

Radius of 1st Cone (r1) =2x

Radius of 2nd Cone (r2) =3x

Volume of 1st Cone (V1)= 4y

Volume of 2nd Cone (V2)= 5y

As we know that the formula for Volume of a Cone = 1/3πr

^{2}hLet’s assume that h1 and h2 be the heights of respective cones.

V1/V2 = 4/5 = (1/3 x π x (r1)

^{2 }x h1) / (1/3 x π x (r2)^{2 }x h2)4/5 = 4h1/9h2 =

9/5

Hence, Heights are in the ratio of 9 : 5.

**Question 7. A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3:1.**

**Solution: **

Given that,

Cylinder and Cone are having equal radii of their bases and heights.

Let’s assume that Radius of the Cone = Radius of the Cylinder = r &

Height of the Cone = Height of the Cylinder = h

Volume of Cylinder / Volume of Cone = (πr

^{2}h) / (1/3 x π x r^{2 }x h) = 3:1

Hence, The Ratio of their Volumes is 3:1.

**Question 8. If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?**

**Solution:**

Let us assume that r be the radius and h be the height of the cone,

As we know that Volume of Cone = 1/3 πr

^{2}hPut the values of radius and same height we get,

Volume = 1/3 π(r/2)

^{2}h = 1/3 x π x r^{2}/2 x h= 1/4 x (1/3 x π x r

^{2 }x h)Ratio of two cone’s = 1/3 πr

^{2}h : 1/4 x (1/3 πr^{2}h) = 1 : 1/4 =4:1

Hence, the ratio between Cones are 1:4

**Question 9. A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the** **heap? **

**Solution:**

Given that,

Diameter of conical heap of wheat = 9 m

Radius = 9/2m and height = 3.5m

As we know that volume of cone = 1/3 πr

^{2}h = 1/3 x 22/7 x (4.5)^{2}x 3.5 = 74.18 m^{3}As we know that Curved Surface area = πrl

l = √r

^{2 }+ h^{2}= √(4.5)^{2 }+ (3.5)^{2}= √130/2 mCurved Surface area = π x 4.5 x √130/2 = 22/7 x 4.5 x √130/2 =

80.54 m^{2}

**Question 10. Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.**

**Solution: **

Given that,

Diameter of the base of solid cone = 14 cm and vertical height (h) = 51 cm

Radius(r) = 7cm

As we know that volume of cone = 1/3 πr

^{2}h = 1/3 x 22/7 x 7^{2}x 51 = 2618 cm^{3}Weight of 1 cm

^{3 }= 10 gramsThen total weight will = 2618 x 10gm =

26180 gm=26.180 kg

**Question 11. A right-angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find** **the volume of the solid, thus generated. Also, find its curved surface area.**

**Solution:**

Given that,

Length of sides of a right-angled triangle are 6.3 cm and 10 cm

By turning around the longer side a Cone is formed in which radius (r) is = 6.3 cm and height (h) = 10 cm

As we know that l = √r

^{2}+ h^{2}= √(6.3)^{2}+ (10)^{2 }= 11.82 cmAs we know that volume of cone = 1/3 πr

^{2}h = 1/3 x 22/7 x (6.3^{)2}x10 =415.8 cm^{3}We know that Curved Surface Area of Cone = πrl = 22/7 x 6.3 x 11.82 =

234.03 cm^{2}

**Question 12. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.**

**Solution: **

Given that,

Side of Cube = 14 cm,

Radius(r) of the largest cone that can be fitted in Cube = Side / 2 = 14/2 = 7cm,

Height(h) = 14cm

As we know that Volume of Cone = 1/3 πr

^{2}h = 1/3 x 22/7 x 7^{2}x 14 =718.67 cm^{3}

**Question 13. The volume of a right circular cone is 9856 cm**^{3}. If the diameter of the base is 28 cm, find:

^{3}. If the diameter of the base is 28 cm, find:

**(i) Height of the cone**

**(ii) Slant height of the cone**

**(iii) Curved surface area of the cone.**

**Solution:**

Given that,

Volume of a Right Circular Cone = 9856 cm3,

Diameter of the base = 28 cm,

Radius(r) = 28/2 = 14cm

1. Height of cone(h) = Volume x 3/πr

^{2 }= (9856x3x7)/(22x14x14) =48cm2. Slant Height(l) = √r

^{2 }+ h^{2}= √(14)^{2}+ (48)^{2}=50cm3. Curved Surface area = πrl = 22/7 x 14 x 50 =

2200 cm^{2}

**Question 14. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? [NCERT]**

**Solution: **

Given that,

Diameter of the top of the conical pit = 3.5 m,

Radius(r) = 3.5/2 = 1.75 m and depth(h) = 12 m

As we know that volume of pit = 1/3 πr

^{2}h= 1/3 x 22/7 x (1.75)

^{2}x 12 = 38.5 m^{3}Volume in Kiloliters = (38.5×1000) / 1000 =

38.5 Kiloliters

**Question 15. Monica has a piece of Canvas whose area is 551 m**^{2}. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m^{2}. Find the volume of the tent that can be made with it.

^{2}. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m

^{2}. Find the volume of the tent that can be made with it.

**Solution: **

Given that,

Area of Canvas = 551 m

^{2},Area of wastage = 1 m

^{2},Actual area = 551 – 1 = 550 m

^{2},Base radius of the conical tent = 7 m

Let us assume l is the slant height and h is the vertical height then the Slant Height(l) = Area / πr

= (550 x 7) / 22×7 = 25m

As we know that Vertical height (h) = √l

^{2}– r^{2 }= √25

^{2}– 7^{2}= √576 =24 mAs we know that volume of Tent = 1/3 πr

^{2}h= 1/3 x 22/7 x 7 x 7 x 24 =

1232 m^{3}