Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.7 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 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) Here, r = 6 cm and h = 7 cm

(ii) Here, r = 3.5 cm and h = 12 cm

Question 2.

Find the capacity in litres with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

Solution:

(i) Here, r = 7 cm and l = 25 cm.

Let the height of the cone be h cm. Then,

h² = l² – r² = 252 – 72

= 625 – 49 = 576

⇒ h = \(\sqrt{576}\)

= 24 cm

Volume of the conical vessel

∴ Capacity of the vessel in litres = \(\left(\frac{1232}{1000}\right)\)l = 1.2321.

(ii) Here, h = 12 cm and l = 13 cm Let the radius of the base of the cone be r cm. Then,

r² = l² – h² = 13² – 12²

= 169 – 144 = 25

⇒ r = \(\sqrt{25}\) = 5 cm

Volume of the conical vessel

Question 3.

The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Solution:

Here, h = 15 cm and volume = 1570 cm³ Let the radius of the base of cone be r cm.

Volume = 1570 cm³

⇒ \(\frac { 1 }{ 3 }\)πr²h = 1570

⇒ \(\frac { 1 }{ 3 }\) x 3.14 x r² x 15 = 1570

⇒ r² = \(\frac{1570}{3.14 \times 5}\) = 100

⇒ r = \(\sqrt{100}\) = 10

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

Question 4.

If the volume of a right circular cone of height 9 cm is 48 n cm³, find the diameter of its base.

Solution:

Here, h = 9 cm and volume = 48π cm³

Let the radius of the base of the cone be r cm

∴ Volume = 48n cm³

⇒ \(\frac { 1 }{ 3 }\) πr²h = 48π

⇒ \(\frac { 1 }{ 3 }\) x r² x 9 = 48π

⇒ 3r² = 48

⇒ r² = \(\frac{48}{3}\) = 16

⇒ r = \(\sqrt{16}\) = 4

Thus, the diameter of the base of the cone = 8 cm.

Question 5.

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Solution:

Diameter of the top of the conical pit = 3.5 m

Question 6.

The volume of a right circular cone is 9856 cm³. 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) Diameter of the base of the cone = 28 cm

∴ Radius = r = ( \(\frac{28}{2}\) ) cm = 14 cm.

Volume of the cone = 9856 cm³

Let the height of the cone be h cm.

Now, volume = 9856 cm³

⇒ \(\frac { 1 }{ 3 }\) πr²h = 9856

⇒ \(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 14 x 14 x h = 9856

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

Thus, the height of the cone = 48 cm.

(ii) Here, r = 14 m and h = 48 cm

Let l be the slant height of the cone. Then,

l² = h² + r² = 48² + 14²

= 2304+ 196 = 2500

⇒ l = \(\sqrt{2500}\) = 50

Thus, the slant height of the cone = 50 cm.

(iii) Here, r = 14 m and l = 50 cm.

Curved surface area = πrl

= (\(\frac { 22 }{ 7 }\) x 14 x 50) cm²

= 2200 cm².

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:

On revolving the right ∆ ABC about the side AB (= 12 cm), we get a cone as shown in the figure.

Question 8.

If the triangle ABC in the question 7 above 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 solids obtained in question 7 and 8.

Solution:

On revolving the right ∆ ABC about the side BC (= 5 cm), we get a cone as shown in the figure,

Volume of solid so obtained = \(\frac { 1 }{ 3 }\) πr²h

= \(\frac { 1 }{ 3 }\) x π x 12 x 12 x 5 cm³

= 240π cm³

∴ Ratio of their volumes = 100π : 240π

(i.e., of Q 7 Q 8) = 5 : 12.

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:

Diameter of the base of the cone = 10.5 m

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

Height of the cone = 3m

∴ Volume of the cone (heap) = \(\frac { 1 }{ 3 }\)πr²h

= (\(\frac { 1 }{ 3 }\) x \(\frac { 22 }{ 7 }\) x 5.25 x 5.25 x 3) m³

= 86.625 m³

To find the slant height l :

We have, l² = h² + r² = 3² + (5.25)²

= 9 + 27.5625 = 36.5625

⇒ l = \(\sqrt{36.5625}\) = 6.0467 (approx.)

Canvas required to protect wheat from rain

= Curved surface area

= πrl = (\(\frac { 22 }{ 7 }\) x 5.25 x 6.0467) m²

= 99.77 m² (approx.)