These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.7 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.7

Question 1.

Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm.

(ii) radius 3.5 cm, height 12 cm.

Solution:

(i) We have given

radius of right circular cone = 6 cm

and height of right circular cone = 7 cm.

Volume of right circular cone = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7} 6 \times 6 \times 7\)

= 264 cm^{3}

(ii) We have given

Radius of cone = 3.5 cm

and height of cone = 12 cm.

Volume of cone = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7}\) × 3.5 × 3.5 × 12

= 154 cm^{3}

Question 2.

Find the capacity in litres of a conical vessel with

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

(ii) height 12 cm and slant height 13 cm.

Solution:

(i) We have given that,

Radius of conical vessel = 7 cm

and slant height = 25 cm

By Pythagoras theorem,

∴ Volume of conical vessel = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7}\) × 7 × 7 × 24

= 1232 cm^{3}

We know that

1000 cm^{3} = 1 l

∴ 1232 cm^{3} = 1.232 l

Therefore, capacity of conical versel is 1.232 litres.

(ii) We have given that

Height of conical vessel (h) = 12 cm

Slant height of conical vessel (l) = 13 cm

By Pythagoras theorem

∴ Volume of conical vessel = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7}\) × 5 × 5 × 12

= \(\frac{2200}{7}\) cm^{3}

Since, 1000 cm^{3} = 1 l

\(\frac{2200}{7}\) cm^{3} = \(\frac{2.2}{7}\) l

Therefore, the capacity of conical vessel is \(\frac{2.2}{7}\) litres.

Question 3.

The height of a cone is 15 cm. If its volume is 1570 cm^{3}, find the radius of the base.

Solution:

We have given

Height of cone (h) = 15 cm

and volume of cone = 1570 cm^{3}

We know that

Volume of cone = \(\frac{1}{3} \pi r^{2} h\)

Therfore, radius of required cone = 10 cm.

Question 4.

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

Solution:

We have given that

Height of cone = 9 cm

and volume of cone = 48π

But we know that volume of cone = \(\frac{1}{3} \pi r^{2} h\)

Therefore diameter of required right circular cone is 2 × 4 = 8 cm.

Question 5.

A conical pit top diameter 3.5 is 12 m deep. What is its capacity in kilo liters.

Solution:

We have given that

diameter of conical pit = 3.5m.

radius of conical pit = \(\frac{3.5}{2}\) m = 1.75 m

and height of conical pit = 12 m

∴ Volume of conical pit = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7}\) × 1.75 × 1.75 × 12

= 38.5 m^{3}

We know that

1 m^{3} = 10001

38.5 m^{3} = 1000 × 38.5

= 38500 litres

= 38.5 kilolitres.

Question 6.

The volume of a right circular cone is 9856 cm^{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:

(i) We have given that

diameter of cone = 28 cm

Radius of cone = 14 cm

and volume of cone = 9856 cm^{3}

We know that,

Volume of cone = \(\frac{1}{3} \pi r^{2} h\)

9856 = \(\frac{1}{3} \times \frac{22}{7}\) × 14 × 14 × h

h = \(\frac{9856 \times 3 \times 7}{22 \times 14 \times 14}\) = 48 cm

(ii) We have,

height of the cone = 48 cm

and radius of cone = 14 cm

By Pathagoras theorem,

AC = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}\)

= \(\sqrt{(48)^{2}+(14)^{2}}\)

= \(\sqrt{2304+196}\)

= 50 cm.

Therefore slant height of the cone = 50 cm

(iii) We know that,

Curved surface area of cone = πrl

= \(\frac {22}{7}\) × 14 × 50

= 2200 cm^{2}

Hence, curved surface area of the cone is 2200 cm^{2}

Question 7.

A right triangle ABC with side 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Solution:

When we revolved the right ∆ABC about the side 12 cm, then it form a right circular cone of radius 6 cm and height 12 cm.

So, volume of such cone = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac {1}{3}\) × π × 5 × 5 × 12

= 100π cm^{3}

Therefore, volume of the solid formed by the revolved a right angle triangle about the side 12 cm is 100π cm^{3}.

Question 8.

If the triangle ABC in the question 7 is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solid obtained.

Solution:

If we revolved the right angle triangle ABC about the side 5 cm is formed a right circular cone.

So, volume of the such cone = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac {1}{3}\) × π × 12 × 12 × 5

= 240π

Therefore, volume of the solid formed by the revolved a right angle triangle about the side 5 cm is 240π cm^{3}.

Ratio of the volumes of the two solid obtained by the revolved a right angle triangle is = \(\frac{100 \pi}{240 \pi}\) = 5 : 12

Question 9.

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it form rain. Find the area of the cavas required.

Solution:

We have given that

Diameter of right circular cone = 10.5 cm

Radius of right circular cone = 3.25 m

and height of the cone = 3m

∴ Volume of the cone = \(\frac{1}{3} \pi r^{2} h\)

= \(\frac{1}{3} \times \frac{22}{7}\) × 5.25 × 5.25 × 3

= 86.625 m^{3}

Again,

By Pythagoars theorem.

AC = \(\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}\)

= \(\sqrt{(3)^{3}+(5.25)^{2}}\)

= \(\sqrt{9+27.5625}\)

= 6.04 m (approx)

Therefore,

Canvas required to covers it = Curved surface area of cone

Now, Curved surface area of cone = πrl

= \(\frac {22}{7}\) × 5.25 × 6.04

= 99.66 m^{2}

Hence, canvas required to cover the conical shape heap of wheat is 99.66 m^{2}.