__NCERT Solutions For Class 9 Math Chapter – 13 Exercise – 13.3__

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**Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm Find its curved surface area.**

**Solution:-**

** **D = 10.5 cm

R = 10.5/2

L = 10 cm

Curved surface area = πrl

= 22/7 x 10.5/2 x 10

= 165 cm^{2}.

**Q2. Find the total surface area of a cone If its slant height is 21 m and diameter of its base is 24 m.**

** ****Solution:-**

** **

D = 24 m

R = 24/2 m = 12 m

L = 21 m

TSA of the cone = πr(l+r)

= 22/7x12mx(12m + 21m)

= 8712/7 m^{2}

= 1244.57 m^{2}

**Q3.
**

**Curved surface area of the cone is 308 m**

^{2}and its slant height is 14 cm Find**(i) Radius of the base**

**(ii) TSA of the cone.**

** ****Solution:-**

R = r cm

L = 14 cm

Curved surface area = 308 cm^{2}.

Πrl = 308

r = 308 cm^{2}/πl

r = 308 cm^{2}/14 cm x 7/22

r = 7 cm

Total Surface Area = π(l+r)

= 22/7 x7 x (7 + 14)

= 22 cm x 21 cm

= 462 cm^{2}.

**Q4. 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 costs of 1 m ^{2} canvas is Rs 70.**

** **

** **

** **

** **

**Solution:-**

R = 24 m

H = 10 m

L = √r^{2} + h^{2}

L = √(24 m)^{2} + (10 m)^{2}

L = √576 m^{2} + 100 m^{2}

L = √676 m^{2}

L = 26 m

CSA of the cone = πrl

22/7 x 24 x 26

= 13728/7 m^{2}

The cost of canvas required to make the tent at Rs 70 per m^{2} = 70 x 13728/7

= Rs 137280

**Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.**

**Solution:-**

R = 6 m

H = 8 m

L = √(6)^{2} + (8)^{2}

= 10 m

Curved Surface Area = πrl

= 3.14 x 6 x 10

= 188.4 m^{2}

Width of the tarpaulin is 3 m

Area of the tarpaulin = 188.4 m^{2}

Length of the tarpaulin = 188.4 / 3 m^{2} = 62.8 m

Extra length of the material = 20 cm = 20/100 = 0.2 m

Actual length required = 62.8 m + 0.2 m

= 63 m

**Q6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface at the rate of Rs 210 per 100 m ^{2}.**

**Solution:-**

D = 14 m

R = 14/2 = 7 m

L = 25 m

CSA = πrl

= 22/7 x 7 x 25

= 550 m^{2}

Cost of the white washing at Rs 210 per 100 m^{2} = 210/100 x 550 = 1155

**Q7. A joker cap is in the form of a right circular cone base radius 7 cm and height 24 cm Find the area of the sheet required to make 10 such caps.**

**Solution:-**

** **

R = 7 cm

H = 24 cm

L = √(7)^{2} + (24)^{2}

L = √625 cm^{2}

L = 25 cm

Area of the sheet required to make each cap = πrl

= 22/7 x 7 x 25

= 550 cm^{2}.

Area of the sheet required to make 10 such caps = 10 x 550 cm^{2} = 5500 cm^{2}

**Q8. 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 m ^{2} What wil be the cost of painting all these cones**

**Solution:-**

** **

D = 40 cm = 40/100 m = 0.04 m

R = 0.4/2 = 0.2 m

H = 1 m

L = √(0.2m)^{2} + (1m)^{2}

= √0.04 m^{2} + 1 m^{‑2}

= 1.02 m^{2}

CSA = πrl

= 3.14 x 0.2 m x 1.02 m

= 0.64056 m^{2}

CSA of 50 cones = 50 x 0.64056

= 32.028 m^{2}

Cost of painting of 50 cones at Rs 12 per m^{2} = 32.028 x 12

= Rs 384.34