NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 13 |

Chapter Name |
Surface Areas and Volumes |

Exercise |
Ex 13.3 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

**Question 1.**

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

**Solution:**

We have, diameter = 10.5 cm

Radius (r) = \(\frac { 10.5 }{ 2 }\) = 5.25 cm

and slant height l= 10 cm

Curved surface area = πrl= \(\frac { 22 }{ 7 }\) x 5.25 x 10 = 165 cm^{2}

**Question 2.**

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

**Solution:**

We have, slant height l = 21 m

and diameter = 24 cm

**Question 3.**

**Curved surface area of a cone is 308 cm ^{2} and its slant height is 14 cm. Find**

**(i) radius of the base and**

**(ii) total surface area of the cone.**

**Solution:**

We have, slant height, l= 14cm

Curved surface area of a cone = 308 cm

^{2}

**Question 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 ₹70.**

**Solution:**

We have, h = 10 m

Hence, the slant height of the canvas tent is 26 m.

(ii) Canvas required to make the tent = Curved surface area of tent

= πrl

= π x 24 x 26 = 624π m^{2}

∵ Cost of 1 m2 canvas = ₹ 70

Cost of 624n m2 canvas = ₹ 70 x 624π

= ₹ 70x 624 x \(\frac { 22 }{ 7 }\)

= ₹ 10 x 624 x 22

= ₹ 137280

Hence, the cost of the canvas is ₹ 137280.

**Question 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:**

Let r, h and l be the radius, height and slant height of the tent, respectively.

The extra material required for stitching margins and cutting = 20 cm = Q^{2} m Hence, the total length of tarpaulin required = 62.8 + Q^{2} = 63 m

**Question 6.**

**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 ₹ 210 per 100 m ^{2}.**

**Solution:**

We have, slant height, l = 25m

and diameter = 14m

∴ Radius, r= 7m

**Question 7.**

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

**Solution:**

∵ The sheet required to make 1 cap = 550 cm^{2}

∴ The sheet required to make 10 caps = 550 x 10= 5500 cm^{2}

**Question 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 ₹12 per m ^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take \( \sqrt{104} \) = 102)**

**Solution:**

Curved surface area of a cone = πrl = 3.14 x 0.2 x 1.02 = 0.64056 m

^{2}

Cost of painting per m2 = ₹12

Cost of painting 0.64056 m2 = ₹12 x 0.64056= ₹ 7.68672

Cost of painting for 1 cone = ₹ 7.68672

Cost of painting 50 cones = ₹ 7.68672x 50= ₹ 384.336= ₹ 384.34

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