### 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) = 4/2 = 2 cm

Radius of smaller circle end (r)= 2/2 = 1 cm

Capacity of frustum-shaped glass = Volume of Frustum

= 1/3 πh (r^{2}+ R^{2}+ rR)= 1/3 × π × 14 ((1 × 1) + (2 × 2) × (2 × 1))

= 1/3 × 22/7 × 14 × 7 (taking π=22/7)

= (22 × 14)/3

= 102.67 cm^{3}Hence, the capacity of frustum-shaped glass = 102.67 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:**

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 cm

^{2}= 2πRR = (18)/2π

R = 9/π cm

Circumference of smaller circle = 2πr

6 cm

^{2}= 2πrr = (6)/2π

r = 3/π cm

Now, as curve surface area of frustum = π (r+R) l= π × (3/π + 9/π) × 4

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

= 48cm^{2}Hence, the curved surface area of the frustum = 48cm

^{2}

### 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) + (πr^{2})= π ((r+R)l + r

^{2})(Taking π common)

=π ((4+10) × 15 + (4 × 4))= 22/7 × (226) (Taking π = 22/7)

= 710.286 cm^{2}Hence, the area of material used for making it = 710.286 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)

**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

= 1/3 πh (r^{2}+ R^{2}+ rR)= 1/3 × 3.14 × 16 (8×8 + 20×20 + 8×20) (Taking π=3.14)

= 1/3 × 3.14 × 16 × (624)

= 10449.92 cm

^{3}Cost of 1 litre milk = ₹ 20

And as, 1 m^{3 }= 1000 cm^{3 }= 1 litre10449.92 cm

^{3 }= (1/1000) ×10449.92 litrescost of 10449.92 cm

^{3 }= (10449.92/1000) × 20

= ₹ 208.998Now, metal sheet used to make the container = Curve surface area + area of lower base

= (π(r+R)l) + (πr^{2})= π ((r+R)l + r

^{2}) (Taking π common)= π ((20+8) × (√(16

^{2}+(20-8)^{2})) + (8 × 8)) (Slant height (l) = √(h^{2}+(R-r)^{2}))= 3.14 × (28 × √400 + 64) (Taking π = 3.14)

= 3.14 × (624)

= 1959.36 cm

^{2}Hence, the metal sheet used to make the container = 1959.36 cm

^{2}As, cost of 100 cm

^{2}= ₹ 81959.36 cm

^{2}= (8/100) × 1959.36

= ₹156.748Hence, 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 ∝ = (opposite side) / (adjacent side)

tan 30° = EB/PE = FR/PF

1/√3 = r/ 10 = R/20

R = 20/√3

r = 10/√3

and as height of frustum = 10 cm

So according to the question,

Frustum is converted to cylindrical wire having diameter 1/16 cm

Volume of Frustum = Volume of Cylinder

Volume of Frustum = 1/3 πh (r^{2}+ R^{2}+ rR)= 1/3 × π × 10 (10/√3×10/√3 + 20/√3×20/√3 + 20/√3×10/√3)

= 1/3 × π × 10 × (100/3 + 400/3 + 200/3)

= 1/3 × π × 10 × (700/3)

= 7000π / 9 cm

^{3}………………………..(1)

Volume of Cylinder = 1/3 π(radius)^{2}H

=1/3 π(1/(16×2))^{2}H …………………(2)As (1) = (2) , then

7000π / 9 = 1/3 π(1/(16×2))

^{2}HH = (7000π × 32 × 32)/(9 × π) (cancel π from both side)

H = 796444.443 cm

H =7964.44 mHence, the length of the wire = 7964.44 m