Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Textbook Exercise Questions and Answers.

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

Take 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,

Radius of glass of upper end (r_{1}) = \(\frac{4}{2}\) = 2 cm

Radius of glass of lower end (r_{2}) = \(\frac{2}{2}\) = 1 cm

Height of the glass (h) = 14cm

Capacity (volume) of the glass = \(\frac{\pi h}{3}\) (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= \(\frac{22 \times 14}{7 \times 3}\) (2^{2} + 1^{2} + 2 × 1)

= \(\frac{44}{3}\) (4 + 1 + 2)

= \(\frac{44 \times 7}{3}\)

= 10267 cm^{3}

Hence, capacity of the glass = 10267 cm^{3}.

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,

Perimeter (circumference) of the upper end = 18 cm

2πr_{1} = 18

r_{1} = \(\frac{18}{2 \pi}=\frac{9}{\pi}\) cm

Perimeter (circumference) of the lower end = 6 cm

2πr_{2} = 6

r_{2} = \(\frac{6}{2 \pi}=\frac{3}{\pi}\)

Slant height of frustum of cone (l) = 4 cm

Curved surface area of frustum of cone = πl (r_{1} + r_{2})

= π × 4\(\left(\frac{9}{\pi}+\frac{3}{\pi}\right)\)

= π × 4 × \(\frac{12}{\pi}\)

= 4 × 12 = 48 cm^{2}.

Hence, curved surface area of frustum of cone = 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:

We have,

Radius of fez at the upper base (closed base) = r_{2} = 4 cm

Radius of fez at lower base (open base) = r_{1} = 10 cm

Slant height of the fez (l) = 15 cm

∴ Area of material used = C.S.A. + Area of closed base

= π l (r_{1} + r_{2}) + π r_{2}^{2}

= \(\frac{22}{7}\) × 15 (10 + 4) + \(\frac{22}{7}\) × 4

= \(\frac{22}{7}\) × 15(14) + \(\frac{22}{7}\) × 16

= \(\frac{22}{7}\) × 15 × 14 + 50\(\frac{2}{7}\)

= 660 + 50\(\frac{2}{7}\)

= 710\(\frac{2}{7}\) cm^{2}

Hencte, area of material used = 710\(\frac{2}{7}\) cm^{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).

OR

A bucket is in the form of a frustum of cone of height 16 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the bucket, at the rate of ₹ 40 per liotre. (use π = 3.14)

Solution:

We have,

Radius of container of upper end (r_{1}) = 20 cm

Radius of container of lower end (r_{2}) = 8 cm

Height of container (h) = 16 cm

Volume of container = \(\frac{\pi h}{3}\) (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

= \(\frac{3.14 \times 16}{3}\) (20^{2} + 8^{2} + 20 × 8)

= \(\frac{50 \cdot 24}{3}\) × 624

= 50.24 × 208

= 10449.92 cm^{3}.

Milk in the container = \(\frac{10449.92}{1000}\) litres

= 10.45 litres (approx)

Cost of the milk = 10.45 × 20 = ₹ 209.

Area of the metal sheet used = πl (r_{1} + r_{2}) + πr_{2}^{2}

= 3.14 × 20(20 + 8) + 3.14 × 8^{2}

= 62.8 × 28 + 200.96

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

Cost of metal sheet used = ₹ \(\frac{1959 \cdot 36 \times 8}{100}\)

= ₹ 156.75

Hence, cost of milk = ₹ 209

and cost of metal sheet used = ₹ 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 \(\frac{1}{16}\) cm, find the length of the wire.

Solution:

We have,

Height of the cone = 20 cm

Since cone is cut into parts at the middle.

So, height of the frustum on , (ACDB) = h = \(\frac{20}{2}\) = 10 cm

In right ∆VOB tan 30° = \(\frac{\mathrm{BO}}{\mathrm{VO}}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{r_1}{20}\)

⇒ r_{1} = \(\frac{20}{\sqrt{3}}\) cm

and in right triangle VO’D

tan 30° = \(\frac{\mathrm{O}^{\prime} \mathrm{D}}{\mathrm{VO}^{\prime}}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{r_2}{10}\)

⇒ r_{2} = \(\frac{10}{\sqrt{3}}\) cm

Radius of wire = \(\frac{1}{2 \times 16} \Rightarrow \frac{1}{32}\) cm

Let length of wire be x cm.

Volume of wire drawn = π × \(\left(\frac{1}{32}\right)^2\) × x

According to question,

Volume of wire drawn = Volume of metal used in the frustum

⇒ \(\pi \times\left(\frac{1}{32}\right)^2 \times x=\frac{7000 \pi}{9}\)

⇒ \(\pi \times \frac{1}{1024} \times x=\frac{7000 \pi}{9}\)

⇒ x = \(\frac{7000 \pi \times 1024}{\pi \times 9}\)

⇒ x = 796444.44 cm

⇒ x = \(\frac{796444 \cdot 44}{100}\) m

⇒ x = 7964.4 m.

Hence, length of wire = 7964.4 m.