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

## Gujarat Board Textbook Solutions Class 9 Maths 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) r = 6 cm

h = l cm

∴ Volume of the right circular cone

= \(\frac{1}{3}\) πr^{2}h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (6)^{2} x 7

= 264 cm^{3}

(ii) r = 3.5 cm

h = 12 cm

∴ Volume of the right circular cone

= \(\frac{1}{3}\) πr^{2}h = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (3.5)^{2} x 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, slant height 13 cm

Solution:

(i) r = 7 cm

l = 25 cm

r^{2} + h^{2} = l^{2}

⇒ (7)^{2} + h^{2} = (25)^{2}

⇒ h^{2} = (25)^{2} – (7)^{2}

⇒ h^{2} = 625 – 49

⇒ h^{2}= 576

⇒ h = \(\sqrt{576}\)

⇒ h = 24 cm

∴ Capacity = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (7)^{2} x 24

= 1232 cm^{3} = 1.232 L

(ii) h = 12 cm

l = 13 cm

r^{2} + h^{2} = l^{2}

⇒ r^{2} + (12)^{2} = (13)^{2}

⇒ r^{2} + 144 = 169

⇒ r^{2} = 169 – 144

⇒ r^{2} = 25

⇒ r = \(\sqrt{25}\)

⇒ r = 5 cm

∴ Capacity = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (5)^{2} x 12

= \(\frac{2200}{7}\) cm^{3} = \(\frac{2200}{7000}\)L = \(\frac{11}{35}\) L

Question 3.

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)

Solution:

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

h = 15 cm

Volume = 1570 cm^{3}

⇒ \(\frac{1}{3}\) πr^{2}h x 1570

\(\frac{1}{3}\) x 3.14 x r^{2} x 15 = 1570

⇒ r^{2} = \(\frac{1570 \times 3}{3.14 \times 15}\)

⇒ r^{2} = 100

⇒ r^{2} = 100

⇒ r = \(\sqrt{100}\)

⇒ r = 10 cm

Hence, the radius of the base of the cone is 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:

Let the radius of the base of the right circular

cone be r cm.

h = 9 cm

Volume = 48π cm^{3}

⇒ \(\frac{1}{3}\)πr^{2}h = 48π

⇒ \(\frac{1}{3}\)πr^{2}h = 48

⇒ \(\frac{1}{3}\) x r^{2} x 9 = 48

⇒ r^{2} = \(\frac{48 \times 3}{9}\)

⇒ r^{2} = 16

⇒ r^{2} = \(\sqrt{16}\) = 4 cm

⇒ 2r = 2(4) = 8 cm

Hence, the diameter of the base of the right circular cone is 8 cm.

Question 5.

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

Solution:

For conical pit

Diameter = 3.5 cm

∴ Radius (r) = \(\frac{3.5}{2}\) m = 1.75 m

Depth (h) = 12 m

∴ Capacity of the conical pit

\(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (1.75)m^{2} x 12 m^{3}

= 38.5 m^{3} = 38.5 x 1000L = 38.5 KL

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) Diameter of the base = 28 cm

∴ Radius of the base (r) = \(\frac{28}{2}\) cm = 14 cm

Let the height of the cone be h cm.

Volume = 9856 cm^{3}

⇒ \(\frac{1}{3}\) πr^{2}h = 9856

⇒ \(\frac{1}{3}\) x \(\frac{22}{7}\) x (14)^{2} x h = 9856

⇒ h = \(\frac{9856 \times 3 \times 7}{22 \times(14)^{2}}\)

⇒ h = 48 cm

Hence, the height of the cone is 48 cm.

(ii) r = 14 cm

h = 48 cm

∴ I = \(\sqrt{r^{2}+h^{2}}\) = \(\sqrt{(14)^{2}+(48)^{2}}\)

= \(\sqrt{196+2304}\) = \(\sqrt{2500}\) = 50 cm

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

(iii) r = 14 cm

l = 50 cm

∴ Curved surface area = πrl

= \(\frac{22}{7}\) x 14 x 50 = 2200 cm^{2}

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

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:

The solid obtained will be a right circular cone whose radius of the base is 5 cm and height is 12 cm.

∴ r = 5 cm

h = 12 cm

∴ Volume = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) x (5)^{2} x 12 cm^{3}

= 100 π cm^{3}

Hence, the volume of the solid so obtained is 100 π cm^{3}

Question 8.

If the triangle ABC in 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 Questions 7 and 8.

Solution:

The solid obtained will be a right circular cone whose radius of the base is 12 cm and the height is 5 cm.

∴ r = 12 cm

h = 5 cm

Volume = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) x π x (12)^{2} x 5 cm^{3}= 24071 cm^{3}

The ratio of the volumes of the two solids obtained = 100π : 240 π = 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:

For heap of wheat

Diameter = 10.5 m

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

Height (A) = 3m

∴ Volume = \(\frac{1}{3}\)πr^{2}h

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x (5.25)^{2} x 3

= 86.625 m^{2}

Slant height (l) = \(\sqrt{r^{2}+h^{2}}\)

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

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

= \(\sqrt{36.5629}\)

= 6.05 m (approx.)

∴ Curved surface area = πrl 22

= \(\frac{22}{7}\) x 5.25 x 6.05

= 99.825 m^{2}

Hence, the area of the canvas required is 99.825 m^{2}.