In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Use π = \(\frac{22}{7}\), unless stated otherwise.

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:

We have,

d_{1} = 4 cm

∴ r_{1} = \(\frac{d_{1}}{2}\) = 2 cm

and d_{2} = 2 cm

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

and h = 14 cm

Volume of the glass

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:

We have,

Slant height (l) = 4 cm

Circumference of one end = 2πr_{1} = 18 cm

and Circumference of other end = 2πr_{2} = 6 cm

⇒ πr_{1} = \(\frac{18}{2}\) = 9 cm

and πr_{2} = \(\frac{6}{2}\) = 3 cm

∴ Curved surface area of the frustum of the cone

= π(r_{1} + r_{2}) l = (πr_{1} + πr_{2}) l = (9 + 3 ) × 4 cm^{2}

= 12 × 4 cm^{2} = 48 cm^{2}.

Question 3.

A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). 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, the radius of the open side (r_{1}) = 10 cm

The radius of the upper base (r_{2}) = 4 cm

Slant height (l) = 15 cm

∴ Area of the material required = [Curved surface area of the frustum] + [Area of the top end]

= π(r_{1} + r_{2})l + πr_{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 ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm^{2}. (Take π = 3.14)

Solution:

We have, r_{1} = 20 cm, r_{2} = 8 cm and h = 16 cm

Area of the bottom = πr_{2}^{2}

= (\(\frac{314}{100}\) × 8 × 8) cm^{2} = 200.96 cm^{2}

∴ Total area of metal required

= 1758.4 cm^{2} + 200.96 cm^{2} = 1959.36 cm^{2}

Cost of metal required for 100 cm^{2} = ₹ 8

∴ Cost of metal required for 1959.36 cm^{2}

= ₹ \(\frac{8}{100}\) × 1959.36 = ₹ 156.75

Question 5.

A metallic right circular cone 20 cm high and whose vertical angle is 60° 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 \(\frac{1}{16}\) find the length of the wire

Solution:

Let us consider the frustum DECB of the metallic cone ABC

Thus, the required length of the wire = 7964.44 m