Class 10 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.4

Last Updated : 04 Mar, 2021

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:

Given values:

Height of frustum (h) = 14 cm

Radius of larger circle end (R) = $\frac{4}{2}$ = 2 cm

Radius of smaller circle end (r)= $\frac{2}{2}$ = 1 cm

Capacity of frustum-shaped glass = Volume of Frustum

= $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3}$ × π × 14 ((1 × 1) + (2 × 2) × (2 × 1))

= $\frac{1}{3} × \frac{22}{7}$ × 14 × 7 (taking π= $\frac{22}{7}$ )

= $\frac{(22 × 14)}{3}$

= 102.67 cm³

Hence, the capacity of frustum-shaped glass = 102.67 cm³

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:

Slant height of frustum (l) = 4 cm

Let radius of smaller circle end = r

Let radius of larger circle end = R

Circumference of circle = 2π × (radius of circle)

Circumference of larger circle = 2πR

18 cm² = 2πR

R = $\frac{18}{2π}$

R = $\frac{9}{π}$ cm

Circumference of smaller circle = 2πr

6 cm² = 2πr

r = $\frac{6}{2π}$

r = $\frac{3}{π}$ cm

Now, as curve surface area of frustum = π (r+R) l

= π × ( $\frac{9}{π}+\frac{3}{π}$ ) × 4

= 12 × 4 (Taking π common and canceling it)

= 48cm²

Hence, the curved surface area of the frustum = 48cm²

Question 3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). 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:

Given values:

Slant height of frustum (l)= 15 cm

Let radius of smaller circle end (r) = 4 cm

Let radius of larger circle end (R) = 10 cm

Area of material used for making it = Curve surface area + area of upper base

= (π(r+R)l) + (πr²)

= π ((r+R)l + r²) (Taking π common)

= π ((4+10) × 15 + (4 × 4))

= $\frac{22}{7}$ × (226) (Taking π = $\frac{22}{7}$ )

= 710.286 cm²

Hence, the area of material used for making it = 710.286 cm²

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 cm². (Take π = 3.14)

Solution:

Given values:

Height of frustum (h)= 16 cm

Let radius of smaller circle end (r) = 8 cm

Let radius of larger circle end (R) = 20 cm

The amount of milk to fill the container = Volume of frustum

= $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3}$ × 3.14 × 16 (8×8 + 20×20 + 8×20) (Taking π=3.14)

= $\frac{1}{3}$ × 3.14 × 16 × (624)

= 10449.92 cm³

Cost of 1 litre milk = ₹ 20

And as, 1 m³= 1000 cm³= 1 litre

10449.92 cm³= ( $\frac{1}{1000}$ ) ×10449.92 litres

cost of 10449.92 cm³= ( $\frac{10449.92}{1000}$ ) × 20

= ₹ 208.998

Now, metal sheet used to make the container = Curve surface area + area of lower base

= (π(r+R)l) + (πr²)

= π ((r+R)l + r²) (Taking π common)

= π ((20+8) × (√(16²+(20-8)²)) + (8 × 8)) (Slant height (l) = √(h²+(R-r)²))

= 3.14 × (28 × √400 + 64) (Taking π = 3.14)

= 3.14 × (624)

= 1959.36 cm²

Hence, the metal sheet used to make the container = 1959.36 cm²

As, cost of 100 cm²= ₹ 8

1959.36 cm² = (8/100) × 1959.36

= ₹156.748

Hence, the cost of the milk which can completely fill the container = ₹ 208.998

and, the cost of metal sheet used to make the container = ₹156.748

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 1/16 cm, find the length of the wire.

Solution:

As the angle is cut into two equal parts, the height gets half too.

Let radius of smaller circle end = r

Let radius of larger circle end = R

In ∆PFR and ∆PEB

tan ∝ = $\frac{(opposite \hspace{0.1cm}side)}{(adjacent \hspace{0.1cm}side)}$

tan 30° = $\frac{EB}{PE} = \frac{FR}{PF}$

$\frac{1}{√3} = \frac{r} {10} = \frac{R}{20}$

R = $\frac{20}{√3}$

r = $\frac{10}{√3}$

and as height of frustum = 10 cm

So according to the question,

Frustum is converted to cylindrical wire having diameter $\mathbf{\frac{1}{16}}$ cm

Volume of Frustum = Volume of Cylinder

Volume of Frustum = $\frac{1}{3}$ πh (r² + R² + rR)

= $\frac{1}{3} × π × 10 (\frac{10}{√3} × \frac{10}{√3} + \frac{20}{√3} × \frac{20}{√3} + \frac{20}{√3} × \frac{10}{√3})$

= $\frac{1}{3} × π × 10 × (\frac{100}{3} + \frac{400}{3} + \frac{200}{3})$

= $\frac{1}{3} × π × 10 × \frac{700}{3}$

= $\frac{7000π}{ 9}$ cm³ ………………………..(1)

Volume of Cylinder = $\frac{1}{3}$ π(radius)²H

= $\frac{1}{3}$ π( $\frac{1}{16×2}$ )²H …………………(2)

As (1) = (2) , then

7000π / 9 = 1/3 π(1/(16×2))²H

H = $\frac{7000π × 32 × 32}{9 × π}$ (cancel π from both side)

H = 796444.443 cm

H = 7964.44 m

Hence, the length of the wire = 7964.44 m