ML Aggarwal Mensuration Exe-17.2 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-17.2 Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Mensuration Exe-17.2 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-17 | Mensuration |

Writer / Book | Understanding |

Topics | Solutions of Exe-17.2 |

Academic Session | 2024-2025 |

**Mensuration Exe-17.2**

ML Aggarwal Class 10 ICSE Maths Solutions

**Take π = 22/7 unless stated otherwise.**

**Question 1. ****Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.**

**Answer :**

10 Slant height of a cone (l) = 10 cm

and radius of the base = 7 cm

Curved surface area = πrl

= 22/7 × 7 × 10 = 220 cm^{2}

**Question 2. ****Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.**

**Answer :**

The diameter of the base of a cone = 10.5cm

Its radius (r) = 10.5/7 = 5.25 cm

and slant height (l) = 10 cm

Curved surface area = πrl

= 22/7 × 5.25 × 10 cm^{2}

= 165.0 cm^{2}

**Mensuration Exe-17.2**

ML Aggarwal Class 10 ICSE Maths Solutions

Page 406

**Question 3. ****Curved surface area of a cone is 308 cm**^{2} and its slant height is 14 cm. Find :

^{2}and its slant height is 14 cm. Find :

**(i) radius of the base**

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

**Answer :**

##### (i) Curved surface area of a cone = 308 cm^{2}

Slant height = 14 cm

⇒ πr*l =* 308

⇒ (22/7) × r × 14 = 308

⇒ r = (308×7)/(22×14)

= 7

Hence the radius of the cone is 7 cm.

**(ii)** Total surface area of the cone = Base area + curved surface area

= πr^{2}+ πr*l*

= (22/7) × 7^{2} + 308

= (22/7) × 49 + 308

= 154 + 308

= 462

Hence the total surface area of the cone is 462 cm^{2}.

**Question 4. ****Find the volume of the right circular cone with**

**(i) radius 6 cm and height 7 cm**

**(ii) radius 3.5 cm and height 12 cm.**

**Answer :**

**(i) **Given radius, r = 6 cm

Height, h = 7 cm

Volume of the cone = (1/3)πr^{2}h

= (1/3) × (22/7) × 6^{2 }× 7

= 22 × 12

= 264 cm^{3}

Hence the volume of the cone is 264 cm^{3}.

**(ii)** Given radius, r = 3.5 cm

Height, h = 12 cm

Volume of the cone = (1/3)r^{2}h

= (1/3) × (22/7) × 3.5^{2 }× 12

= (22/7) × 12.25 × 4

= 154 cm^{3}

Hence, the volume of the cone is 154 cm^{3}.

**Question 5. ****Find the capacity in litres of a conical vessel with:**

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

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

**Answer :**

**(i) Radius = 7 cm**

and slant height (l) = 25 cm

We know that *l*^{2} = h^{2 }+ r^{2}

Height of the conical vessel, h = √(*l*^{2}– r^{2})

= √(25^{2 }– 7^{2})

= √(625 – 49)

= √576

= 24 cm

Volume of the cone = (1/3)πr^{2}h

= (1/3) × (22/7) × 7^{2 }× 24

= 22 × 7 × 8

= 1232 cm^{3}

= 1.232 litres **[1 litre = 1000 cm ^{3}]**

Hence the volume of the cone is 1.232 litres.

**(ii)** Given height, h = 12 cm

Slant height, *l* = 13 cm

We know that *l*^{2} = h^{2 }+ r^{2}

Radius of the conical vessel, r = √(*l*^{2 }– h^{2})

= √(13^{2 }– 12^{2})

= √(169 – 144)

= √25

= 5 cm

Volume of the cone = (1/3)πr^{2}h

= (1/3) × (22/7) × 5^{2 }× 12

= (22/7) × 25 × 4

= 2200/7 cm^{3}

= 2.2/7 litres **[1 litre = 1000 cm ^{3}]**

= 0.314 litres

Hence, the volume of the cone is 0.314 litre**s.**

**Question 6. ****A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?**

**Answer 6:**

Given diameter, d = 3.5 m

So radius, r = 3.5/2 = 1.75

Depth, h = 12 m

Volume of the cone = (1/3) πr^{2}h

= (1/3) × (22/7) × 1.75^{2 }× 12

= (22/7) × 1.75^{2 }× 4

= 38.5 m^{3}

= 38.5 kilolitres **[1 kilolitre = 1m ^{3}]**

Hence the volume of the conical pit is 38.5 kilolitres.

**Question 7. ****If the volume of a right circular cone of height 9 cm is 48π cm**^{3}, find the diameter of its base.

^{3}, find the diameter of its base.

**Answer :**

Given height of a cone, h = 9 cm

Volume of the cone = 48π

(1/3)πr^{2}h = 48π

⇒ (1/3)πr^{2 }× 9 = 48π

⇒ 3r^{2} = 48

⇒ r^{2} = 48/3 = 16

⇒ r = 4

So diameter = 2 × radius

= 2 × 4

= 8 cm

Hence, the diameter of the cone is 8 cm.

**Question 8. ****The height of a cone is 15 cm. If its volume is 1570 cm**^{3}, find the radius of the base. (Use π = 3.14)

^{3}, find the radius of the base. (Use π = 3.14)

**Answer :**

Height of cone (h) = 15 cm

Volume = 1570 cm^{3}

(1/3)πr^{2}h = 1570

⇒ (1/3)3.14 × r^{2 }× 15 = 1570

⇒ 5 × 3.14 × r^{2} = 1570

⇒ r^{2} = 1570/5×3.14 = 314/3.14 = 100

⇒ r = 10

Hence, the radius of the cone is 10 cm.

**Question 9. ****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 area at the rate of Rs 210 per 100 m**^{2}.

^{2}.

**Answer :**

Slant height of conical tomb (l) = 25 m

and base diameter = 14 m

So radius, r = 14/2 = 7 m

Curved surface area = πr*l*

= (22/7) × 7 × 25 = 550 m^{2}

Hence, the curved surface area of the cone is 550 m^{2}.

Rate of washing its curved surface area per 100 m^{2} = Rs. 210

So total cost = (550/100) × 210 = Rs. 1155

Hence the total cost of washing its curved surface area is Rs. 1155.

**Question 10. ****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 m ^{2} canvas is Rs 70.**

**Answer :**

Height of a conical tent (h) = 10 m

and radius (r) = 24 m

We know that *l*^{2} = h^{2 }+ r^{2}

*⇒ **l*^{2} = 10^{2 }+ 24^{2}

*⇒ **l*^{2} = 100 + 576

*⇒ **l*^{2} = 676

*⇒ **l *= √676

*⇒ **l *= 26

⇒ Curved surface area = πr*l*

= (22/7) × 24 × 26 = 13728/7 m^{2}

Cost of 1 m^{2} canvas = Rs. 70

Total cost = (13728/7) × 70

= Rs. 137280

Hence the cost of the canvas required to make the tent is Rs. 137280.

**Question 11. ****A Jocker’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 cloth required to make 10 such caps.**

**Answer :**

Base radius of a conical cap = 7 cm

and height (h) = 24 cm

We know that *l*^{2} = h^{2}+r^{2}

*l*^{2} = 24^{2 }+ 7^{2}

*⇒ **l*^{2} = 576 + 49

*⇒ **l*^{2} = 625

*⇒ **l*= √625

*⇒ **l*= 25

Curved surface area = πr*l*

= (22/7) × 7 × 25

= 22 × 25

= 550 cm^{2}

So the area of the cloth required to make 10 such caps = 10 × 550 = 5500

Hence the area of the cloth required to make 10 caps is 5500 cm^{2}.

**Question 12. **

**(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.**

**(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.**

**(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find:**

**(i) the ratio of their volumes.**

**(ii) the ratio of their lateral surface areas.**

**Answer :**

**(a)** Let r_{1} and r_{2 }be the radius of the given cones and h be their height.

Ratio of radii, r_{1}:r_{2} = 3:4

Volume of cone, V_{1} = (1/3)πr_{1}^{2}h

Volume of cone, V_{2} = (1/3)πr_{2}^{2}h

V_{1} /V_{1} = (1/3)πr_{1}^{2}h/(1/3)πr_{2}^{2}h

= r_{1}^{2}/ r_{2}^{2}

= 3^{2}/4^{2}

= 9/16

Hence the ratio of the volumes is 9 :16.

**(b)** Let h_{1} and h_{2 }be the heights of the given cones and r_{1} and r_{2 }be their radii.

Ratio of heights, h_{1}: h_{2} = 5:2

Ratio of radii, r_{1}:r_{2} = 2:5

Volume of cone, V_{1} = (1/3)πr_{1}^{2}h_{1}

Volume of cone, V_{2} = (1/3)πr_{2}^{2}h_{2}

V_{1} /V_{1} = (1/3)πr_{1}^{2}h_{1}/ (1/3)πr_{2}^{2}h_{2}

= r_{1}^{2}h_{1}/r_{2}^{2}h_{2}

= 2^{2 }× 5/5^{2 }× 2

= 4 × 5/25 × 2

= 20/50

= 2/5

Hence

the ratio of the volumes is 2:5.

**(c)** Let the height of bigger cone = h and radiu**s **= r

∴ Volume = 1/3πr^{2}h

And height of the smaller cone = h/2

And radius = r/2

∴ Volume = 1/3π (r/2)^{2} (h/2)

= 1/3π **× **r^{2}/4 × h/2

= 1/24πr^{2}h

And ratio in their values = (1/24pr^{2}h) : (1/3pr^{2}h)

= 1/8 : 1

= 1 : 4

**Question 13. ****Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.**

**Answer :**

Diameter of the base of the conical tent = 20 m

Radius (r) = 20/2 = 10 m

and slant height (h) = 42 m

Curved surface area of the conical tent = πr*l*

= (22/7) × 10 × 42

= 22 × 10 × 6

= 1320 m^{2}

So, the area of canvas required is 1320 m^{2}.

Since 10% of this area is used for folds and stitches, actual cloth needed = 1320 + 10% of 1320

= 1320 + (10/100) × 1320

= 1320 + 132

= 1452 m^{2}

Width of the cloth = 2 m

Length of the cloth = Area/width = 1452/2 = 726 m

Cost of canvas = Rs.80 per metre.

Total cost = 80 × 726 = Rs. 58080

Hence the total cost of the canvas is Rs. 58080.

**Question 14. ****The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.**

**Answer :**

Perimeter of the base of a cone = 44 cm

2πr = 44

⇒ 2 × 22/7 × r = 44

⇒ r = 44 × 7/(2 × 22)

⇒ r = 7 cm

Slant height, *l* = 25

height, h = √(*l*^{2}– r^{2})

⇒ h = √(25^{2 }– 7^{2})

⇒ h = √(625 – 49)

⇒ h = √576

⇒ h = 24 cm

Volume of the cone, V = (1/3)πr^{2}h

⇒ V = (1/3) × (22/7) × 7^{2 }× 24

⇒ V = (22/7) × 49 × 8

⇒ V = 22 × 7 × 8

⇒ V = 1232

Hence the volume of the cone is 1232 cm^{3}.

Curved surface area of the cone = πr*l*

*= *(22/7) × 7 × 25

= 22 × 25

= 550 cm^{2}

Hence, the curved surface area of the cone is 550 cm^{2}.

**Question 15. ****The volume of a right circular cone is 9856 cm**^{3} and the area of its base is 616 cm^{2}. Find:

^{3}and the area of its base is 616 cm

^{2}. Find:

**(i) the slant height of the cone.**

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

**Answer :**

Given base area of the cone = 616 cm^{2}

πr^{2} = 616

⇒ (22/7) × r^{2} = 616

⇒ r^{2} = 616 × 7/22

⇒ r^{2} = 196

⇒ r = 14

Given volume of the cone = 9856 cm^{3}

(1/3)πr^{2}h = 9856

⇒ (1/3) × (22/7) × 14^{2 }× h = 9856

⇒ h = (9856 × 3 × 7)/(22 × 14^{2})

⇒ h = (9856 × 3 × 7)/(22 × 196)

⇒ h = 48

**(i)** Slant height, *l *= √(h^{2 }+ r^{2})

*⇒ **l *= √(48^{2 }+ 14^{2})

*⇒ **l *= √(2304 + 196)

*⇒ **l *= √(2500)

*⇒ **l *= 50

Hence, the slant height of the cone is 50 cm.

**(ii)** Total surface area of the cone = πr(*l *+ r)

= (22/7) × 14 × (50 + 14)

= 22 × 2 × 64

= 2816 cm^{2}

Hence the total surface area of the cone is 2816 cm^{2}.

**Question 16. ****A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)**

**Answer :**

Sides of a right triangle are 6 cm and 8 cm

It is revolved around 8 cm side

Radius (r) = 6 cm

Height (h) = 8 cm

Slant height (l) = 10 cm

Volume of the cone, V = (1/3)πr^{2}h

⇒ V = (1/3) × 3.14 × 6^{2 }× 8

⇒ V = (1/3) × 3.14 × 36 × 8

⇒ V = 3.14 × 12 × 8

⇒ V = 301.44 cm^{3}

Hence, the volume of the cone is 301.44 cm^{3}.

Curved surface area of the cone = πr*l*

= 3.14 × 6 × 10

= 188.4

Hence, the curved surface area of the cone is 188.4 cm^{2}.

**Question 17. ****The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be 1/27 of the volume of the given cone, at what height above the base is the section cut?**

**Answer :**

Height of a cone (H) = 30 cm

A small cone is cut off from the top of the cone given

Volume of the given cone = (1/3)πR^{2}H

Volume of the small cone = 1/27 th of the volume of the given cone.

(1/3)πr^{2}h = (1/27) × (1/3)πR^{2}H

Substitute H = 30

(1/3)πr^{2}h = (1/27) × (1/3)πR^{2 }× 30

⇒ r^{2}h/R^{2} = 30/27

⇒ r^{2}h/R^{2} = 10/9 **….(i)**

From figure, r/R = h/H

r/R = h/30 **….(ii)**

Substitute (ii) in (i)

(h/30)^{2 }× h = 10/9

⇒ h^{3}/900 = 10/9

⇒ h^{3} = (900 × 10)/9 = 1000

Taking cube root on both sides.

h = 10 cm

H – h = 30 – 10 = 20

The small cone is cut at a height of 20 cm above the base.

**Question 18. ****A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find:**

**(i) the radius of the cone.**

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

**Answer :**

**(i)** Given radius of the semi circular lamina, r = 35 cm

A cone is formed by folding it.

So the slant height of the cone, *l *= 35 cm

Let r_{1} be radius of cone.

Semicircular perimeter of lamina becomes the base of the cone.

⇒ πr = 2πr_{1}

⇒ r = 2r_{1}

⇒ 35 = 2 r_{1}

⇒ r_{1} = 35/2 = 17.5 cm

Hence the radius of the cone is 17.5 cm.

**(ii)** Curved surface area of the cone = πr_{1}*l*

= (22/7) × 17.5 × 35

= 22 × 17.5 × 5

= 1925 cm^{2}

Hence the lateral surface area of the cone is 1925 cm^{2}.

-: End of ML Aggarwal Mensuration Exe-17.2 Class 10 ICSE Maths Solutions :–

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