# Class 9 RD Sharma Solutions- Chapter 20 Surface Area And Volume of A Right Circular Cone – Exercise 20.1 | Set 2

**Question 11. A joker’s cap is in the form of a right circular cone of base radius 7cm and height 24 cm. Find the area of the sheet required to make 10 such caps.**

**Solution: **

As per question, r = 7 cm and h = 24 cm.

Therefore, slant height of cone,

l = √r= √(7^{2}+ h^{2}^{2}+24^{2}) = 25 cmNow, CSA of one cone =

πrl= (22 x 7 x 25)/7 = 550 cm

^{2}Therefore, area of the sheet required to make 10 such caps is 550 x 10 = 5500 cm

^{2}

**Question 12. Find the ratio of the curved surface area of two cones if their diameter of the bases are equal and slant height are in the ratio 4:3.**

**Solution: **

Given:

diameters of the cone are equal, so there radius will also be equal, thus r1 = r2 =r and their slant heights are in ratiol1:l2= 4:3Since, CSA of a cone =

πrl, therefore

π r1 l1 / π r2 l2⇒

π r l1 / π r l2⇒

l1 / l2= 4/3Hence, the ratio of the CSA of two cones is 4:3.

**Question 13. There are two cones the curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.**

**Solution: **

According to question,

CSA

_{1}/ CSA_{2}= 2/1, alsol1 / l2= 1/2⇒ π r1 l1 / π r2 l2 = 2/1

⇒ r1.1/r2.2 = 2/1

⇒ r1 / r2 = 4/1

Hence, their ratio of their radii is 4:1.

**Question 14. The diameters of two cones are equal. If there slant heights are in the ratio 5:4. Find the ratio of their curved surfaces.**

**Solution:**

Given that diameters of the cone are equal, so there radius will also be equal, thus r1 = r2 =r and their slant heights are in ratio

l1:l2= 5:4Since, CSA of a cone =

πrl, therefore

π r1 l1/π r2 l2⇒

π r l1 / π r l2⇒

l1 / l2= 5/4Hence, the ratio of the CSA of two cones is 5:4.

**Question 15. Curved surface area of a cone is 308 cm**^{2} and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

^{2}and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.

**Solution: **

Given:

l= 14 cm and CSA = 308 cm^{2}We know that, curved surface area of cone =

πrl⇒ 308 = 22/7 x 14 x

l⇒

l= 7 cmAlso, total surface area of a cone =

πr(l + r)= 22/7 x 7 x (14 + 7) = 462 cm

^{2}

**Question 16. The slant height and base diameter of conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface area at the rate of Rs. 210 per hundred meter square.**

**Solution:**

Given: slant height of the conical tomb = 25 m and base radius of tomb = 14/2 = 7 m.

CSA of the conical tomb =

πrl= 22 x 7 x 25 / 7 = 550 m

^{2}The cost of whitewashing the curves surface of the tomb = (210 x 550)/100 = Rs. 1155

**Question 17. A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent, if the cost of 1 m**^{2} canvas is Rs 70. Find the cost of canvas required for the tent.

^{2}canvas is Rs 70. Find the cost of canvas required for the tent.

**Solution:**

According to the question,

Height of conical tent, h = 10 m and radius of the conical tent, r = 24 m

Therefore, slant height of cone,

l= √r^{2}+ h^{2}= √(10^{2}+24^{2}) = 26 mThus, the slant height of the cone is 26 m.

Now, curved surface area of cone =

πrl

= 22 x 24 x 26 / 7 = 13728/7 m^{2}It is given cost of 1 m

^{2}canvas is Rs. 70. Therefore cost of 13728/7 m^{2}

^{ = }1378 * 70 / 7 = Rs.137280Hence, the cost of canvas required for the tent is Rs.137280.

**Question 18. A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.**

**Solution: **

Given: diameter of cylinder = 24 m, therefore radius= 12 m. The cone will also have the same base radius, since it is surmounted on the top of the cylinder, i.e, r = 12 m.

Now, height of the cylinder = 11 m and height of entire system =16 m.

Therefore, height of the cone = 16-11 = 5m.

Slant height of cone,

l = √r= √(12^{2}+ h^{2}^{2}+5^{2}) = 13 mNow, curved surface area of cone =

πrl= 22 x 12 x 13 / 7 = 3432/7 m

^{2}Similarly, CSA of the cylindrical portion = 2πrh = 2 x 22 x 12 x 11 / 7 = 5808/7 m

^{2}Therefore, area of the canvas required for the tent = (3432+5808)/7

= 9240/7 = 1320 m

^{2}

**Question 19. A circus tent is cylindrical to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m. Calculate the length of the canvas 5 m wide to make the required tent.**

**Solution:**

Given: diameter = 105 m, so radius = 52.5 m and slant height of the conical portion = 53 m

^{2}Now, curved surface area of cone =

πrl= 22 x 52.5 x 53 / 7 = 8745 m

^{2}Similarly, CSA of the cylindrical portion = 2πrh = 2 x 22 x 52.5 x 3 / 7 = 990 m

^{2}So, area of the canvas required for the tent = 8745 + 990

= 9735 m

^{2}Therefore, length of canvas required = 9735/5 = 1947 m

**Question 20. The circumference of the base of a 10 m height conical tent is 44 m. Calculate the length of the canvas used in making the tent if width of canvas is 2 m.**

**Solution: **

Given: circumference of the base of conical tent = 2πr

⇒ 44 = 2 x 22 x r / 7

⇒ r = 7 m

Now, slant height of the cone,

l= √r^{2}+ h^{2}= √(7^{2}+10^{2}) = √149 mSo, curved surface area of cone =

πrl= 22 x 7 x √149 / 7 = 22√149 m

^{2}Thus, the length of the canvas required to make the tent = 22√149/2 = 11√149 = 134.2 m

**Question 21. What length of tarpaulin 4 m wide will be required to make a conical tent of height 8 m and base radius 6 m. Assume that the extra length of material will be required for stitching margins and wastage is cutting is approximately 20 cm.**

**Solution:**

Given: height of conical tent, h = 8 m and radius of conical tent, r = 6 m

So, slant height of the cone,

l = √r= √(6^{2}+ h^{2}^{2}+8^{2}) = 10 mSo, curved surface area of cone =

πrl= 3.14 x 6 x 10 = 188.4 m

^{2}Now, let the length of tarpaulin sheet required be x m. Also, 20 cm will get wasted, so effective length of tarpaulin required = (

x-0.2) m and breadth of tarpaulin = 3m.Area of the sheet = CSA of sheet

⇒ (

x – 0.2) * 3 = 188.4^{ }⇒ x = 62.8 + 0.2 = 63 m

^{2}Hence, the length of tarpaulin sheet required will be 63 m.

^{2}

**Question 22. A bus stop is dedicated 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 meter square. What will be the cost of painting all these cones.**

**Solution:**

Slant height of the cone,

l = √r= √(0.2^{2}+ h^{2}^{2}+1^{2}) = √1.04 m = 1.02 mCSA of a single cone =

πrl

= 3.14 x 0.2 x1.02 = 0.64056 m^{2}Therefore, total area for 50 such cones = 50 x 0.64056 = 32.028 m

^{2}Now, the total cost of painting all the cones = 12 x 32.028 = 384.336

Therefore, the total cost of painting all the cones is Rs. 384.336

**Question 23. A cylinder and cone have equal radii of their base and equal heights. If their curved surface area are in the ratio 8:5, show that the radius of each is to the height of each as 3:4.**

**Solution: **

Given that both cylinder have equal radii and equal height.

So, lets assume base radius as r, height as h and slant height of the cone be

l.We know that, CSA of cone =

πrland CSA of the cylinder = 2πrh⇒

2πrh/πrl= 8/5⇒

h / l= 4/5⇒ h / √r

^{2}+ h^{2}= 4/5⇒ h

^{2}/ r^{2}+ h^{2}= 16/25⇒ r

^{2}/ h^{2 }= 9 / 16⇒ r / h = 3/4