Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.4

Question 1.

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution:

Here

d_{1} = 4 cm

r_{1} = \(\frac {4}{2}\) = 2 cm

and d_{2} = 2 cm

r_{2} = 1 cm

Height h = 14

Hence, capacity of glass

= \(\frac {1}{3}\) Ï€ (r^{2}_{1} + r^{2}_{1} + r_{1}r_{2}) x h

= \(\frac {1}{3}\) x \(\frac {22}{7}\) [2^{2} + 1^{2} + 2 x 1] x 14

= \(\frac {1}{3}\) x \(\frac {22}{7}\) x [4 + 1 + 2] x 14

= \(\frac {22}{3 x 7}\) x 7 x 14 = \(\frac {308}{3}\) cm^{3} = 102 \(\frac {2}{3}\) cm^{2}

Question 2.

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution:

Here,

Slant height l = 4 cm

and let r_{1} r_{2} be the radii of frustum and r_{1} > r_{2
}

Perimeter of circular ends = 2Ï€r_{1}

2Ï€r_{1} = 18 cm

â‡’Â Ï€r_{1} = 9 cm ……(1)

and 2Ï€r_{2} = 6 cm

Ï€r_{2} = 3 cm …….(2)

Curved surface area of frustum

= Ï€(r_{1} + r_{2}) l = (Ï€r_{1} + Ï€r_{2})l ……(3)

Putting value of eqn. (1) and (2) in eqn. (3),

Curved surface area of frustum = (9 + 3) x 4 = 12 x 4 = 48 cm^{2}

Question 3.

A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:

Here

r_{1} = 10 cm (Radius on the open side)

and r_{1} = 4 cm (Radius of the upper base)

Surface area of cap = CSA of frustum + Area of upper circular end

= Ï€(r_{1} + r_{2}) l + Ï€r^{2}_{2
}

Question 4.

A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of â‚¹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costsÂ â‚¹8 per 100 cm^{2}. (Take Ï€ = 3.14)

Solution:

Here

h = 16 cm

r_{1} = 20 cm and r_{2} = 8 cm

Volume of frustum shaped container

= \(\frac {1}{3}\)Ï€h (r^{2}_{1} + r^{2}_{2} + r_{1}_{2})

= \(\frac {1}{3}\) x 3.14 x 16(20^{2} + 8^{2} + 20 x 8)nh

= \(\frac {1}{3}\) x 3.14 x 16 [400 + 64 + 160]

= \(\frac {1}{3}\) x 3.14 x 16 x 624

= 3.14 x 16 x 208

= 10449.92 cm^{3} (1000 cm^{3} = 1 litre)

= \(\frac{10449.92}{1000}\) = 10.44992 litre

Cost of milk = 10.44992 x 20

= â‚¹ 208.9984 = â‚¹ 209

Surface area of frustum

= Ï€ (r_{1} + r_{2})l + Ï€r^{2}_{2} ……….(1)

l = \(\frac{10449.92}{1000}\)

= \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)

= \(\sqrt{(20-8)^{2}+16^{2}}\)

= \(\sqrt{12^{2}+16^{2}}\)

= \(\sqrt{144+256}\)

= \(\sqrt{400}\)

l = 20 cm

Area of metal sheet = Surface area of frustum

= Ï€(r_{1} + r_{2})l + Ï€r^{2}_{2} [From (1)]

= 3.14 (20 + 8) x 20 + 3.14 x 8^{2}

= 3.14 x 28 x 20 + 3.14 x 64

= 1758.4 + 200.96

= 1959.36 cm^{2}

Cost of metal sheet

= 1959.36 x \(\frac{8}{100}\) = ? 156.75

Question 5.

A metallic right circular cone 20 cm high and whose vertical angle is 60Â° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm, find the length of the wire.

Solution:

Here height h of cone = 20 cm

Cone is cut into the middle then height of frustum

= \(\frac{20}{2}\) = 20 cm

Hence AC = 10 cm

In rt. Î”ACE

tan 300 = \(\frac{CE}{AC}\)

â‡’ \(\frac{1}{\sqrt{3}}\) = \(\frac{r_{2}}{10}\)

â‡’ r_{2} = \(\frac{10}{\sqrt{3}}\) cm

Now rt. Î”AOB

tan 300 = \(\frac{OC}{OA}\)

\(\frac{1}{\sqrt{3}}\) = \(\frac{r_{1}}{20}\)

â‡’ r_{1} = \(\frac{20}{\sqrt{3}}\) cm

Volume of frustum

Diameter of wire which is drawn = \(\frac{1}{16}\)cm

Radius of wire r = \(\frac{1}{2}\) x \(\frac{1}{16}\) = \(\frac{1}{32}\)cm

Volume of wire drawn = Volume of frustum

Ï€r^{2}h = \(\frac{22000}{9}\)

= 7964.4 m

Hence, the length of wire is 7964.4 in.