**NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.7**

NCERT Solutions for Class 9 Maths Chapter 13 Surface
Areas and Volumes Ex 13.7 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 13 Surface
Areas and Volumes Ex 13.7.

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

**Ex 13.7 Class 9 Maths 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) Here, radius of the cone (r) = 6
cm

Height of the cone (h) = 7 cm

**Ex 13.7 Class 9 Maths Question 2.**

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

**Solution:
**(i) Here, radius (r) = 7 cm and slant height (l) = 25
cm

Thus, the required capacity of the conical vessel is 1.232 litres.

(ii) Here, height (h) = 12 cm and slant height (l) =
13 cm

∴ Capacity of the conical vessel

Thus, the required capacity of the conical vessel is 11/35 litres.

**Ex 13.7 Class 9 Maths Question 3.
**The height of a cone is 15 cm. If its volume is 1570
cm³, find the radius of the base. (Use Ï€ = 3.14)

**Solution:
**Here, height of the cone (h) = 15 cm

Volume of the cone = 1570 cm

^{3}

Let the radius of the base be ‘r’ cm.

Thus, the required radius of the base is 10 cm.

**Ex 13.7 Class 9 Maths Question 4.
**If the volume of a right circular cone of height 9
cm is 48 cm³, find the diameter of its base.

**Solution:
**Volume of the cone = 48 cm

^{3}

Height of the cone (h) = 9 cm

Let the base radius be r cm.

Diameter = 2 × radius

∴ Diameter of the base of the cone = (2 × 4) cm = 8 cm

**Ex
13.7 Class 9 Maths Question 5.
**A conical pit of top diameter 3.5 m is 12 m deep.
What is its capacity in kilolitres?

**Solution:
**Here, diameter of the conical pit = 3.5 m

Radius of the conical pit (r) = 3.5/2 m = 35/20 m,

Depth (h) = 12 m

⇒ Volume (capacity) = 1/3 × Ï€r^{2}h

Thus, the capacity of the conical pit is 38.5 kl.

**Ex
13.7 Class 9 Maths 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:
**Volume of the cone = 9856 cm

^{3}

Diameter of the base = 28 cm

Radius of the base (r) = 28/2 = 14 cm

(i) Let the height of the cone be h cm.

Thus, the required height is 48 cm.

(ii) Let the slant height be l cm.

⇒ l^{2} = r^{2} + h^{2}

⇒ l^{2} = 14^{2} + 48^{2} =
196 + 2304 = 2500

∴ l = 50

Thus, the required slant height is 50 cm.

(iii)
The curved surface area of a cone = Ï€rl

∴ Curved surface area = 22/7 × 14 × 50 cm^{2}

= 2200 cm^{2 }

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

**Ex 13.7 Class 9 Maths Question 7.
**A right triangle ABC with sides 5 cm, 12 cm and 13
cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

**Solution:
**Sides of the right triangle ABC are 5 cm, 12 cm and
13 cm.

The right-angled triangle is revolved about the side 12 cm.

Thus,
we have radius of the base of the cone so formed (r) = 5 cm

Height (h) = 12 cm

∴ Volume of the cone so obtained = 1/3 × Ï€r^{2}h

= 1/3 × Ï€ × (5)^{2} ×
12 cm^{3}

= 100 Ï€ cm^{3}

Thus, the required volume of the cone is 100Ï€ cm^{3}.

**Ex 13.7 Class 9 Maths Question 8.
**If the triangle ABC in Question 7 above is revolved
around the side 5 cm, then find the volume of the solid so obtained. Find also
the ratio of the volumes of the two solids obtained in Questions 7 and 8.

**Solution:
**Since the right triangle is revolved around the side
5 cm.

∴ Height of the cone so obtained (h) = 5 cm

Radius of the cone (r) = 12 cm

Thus, the required ratio is 5 : 12.

**Ex 13.7 Class 9 Maths Question 9.
**A heap of wheat is in the form of a cone whose
diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be
covered by canvas to protect it from rain. Find the area of the canvas
required.

**Solution:
**Here, the heap of wheat is in the shape of a cone
with base diameter = 10.5 m

Thus, the required volume = 86.625 m

^{3}

Now,
the area of the canvas to cover the heap must be equal to the curved surface
area of the conical heap.

^{2}.

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