**Maths Class 10
Exercise 13.4**

**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, *r*_{1} = 4/2 = 2 cm, *r*_{2} = 2/2 = 1 cm and *h* = 14 m

*h*(

*r*

_{1}

^{2}+

*r*

_{2}

^{2}+

*r*

_{1}

*r*

_{2})

=
1/3 × 22/7 × 14(2^{2} + 1^{2} + 2 × 1)

=
1/3 × 22/7 × 14 × 7

**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: **Let *r*_{1} cm and *r*_{2} cm
be the radii of the ends (*r*_{1} >
*r*_{2}) of the frustum of
the cone.

Slant
height of the frustum (*l*) = 4 cm

We
have, 2Ï€*r*_{1} = 18 cm

Ï€*r*_{1} = 9 cm ……. (i)

And
2Ï€*r*_{2} = 6 cm

Ï€*r*_{2} = 3 cm ……. (ii)

Curved
surface area of the frustum = Ï€(*r*_{1} +
*r*_{2})*l*

=
(Ï€*r*_{1} + Ï€*r*_{2})*l*

=
(9 + 3) × 4 [Putting the values from equations (i) and (ii)]

=
48 cm^{2}

**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, *r*_{1} = 10 cm, *r*_{2} =
4 cm and *l* = 15 cm

Now,
total surface area of frustum = Ï€(*r*_{1} +
*r*_{2})*l *+ Ï€*r*_{2}^{2}

=
22/7 × (10 + 4) × 15 + 22/7 × (4)^{2}

=
660 + 352/7

**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
total cost of milk which can completely fill the container at the rate of Rs.
20 per liter. Also find the cost of metal sheet used to make the container, if
it costs Rs. 8 per 100 cm ^{2}. (Take Ï€ = 3.14)**

**Solution: **We have, *r*_{1} = 20 cm, *r*_{2} =
8 cm and *h* = 16 cm.

*h*(

*r*

_{1}

^{2}+

*r*

_{2}

^{2}+

*r*

_{1}

*r*

_{2})

=
1/3 × 3.14 × 16(20^{2} + 8^{2} + 20 × 8)

=
1/3 × 3.14 × 16(400 + 64 + 160)

=
1/3 × 3.14 × 16 × 624

=
10449.92 cm^{3}

=
10.44992 liters [Since 1000 cm^{3}
= 1 litre]

Cost
of the milk = 10.44992 × 20

=
Rs. 208.9984

=
Rs. 209

Now,
total surface area of the frustum = Ï€(*r*_{1} +
*r*_{2})*l *+ Ï€*r*_{2}^{2}

=
3.14 × 28 × 20 + 3.14 × 64

=
1758.4 + 200.96

= 1959.36
cm^{2}

Area
of the metal sheet used = 1959.36 cm^{2}

Cost
of the metal sheet = 1959.36 × 8/100

=
156.7488

=
Rs. 156.75

**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: **

We have, tan 30˚ =
*r*_{2}/10

1/√3 = *r*_{2}/10

*r*_{2} =
10/√3 cm

Again,
tan 30˚ = *r*_{1}/20

1/√3 = *r*_{1}/20

*r*_{1} =
20/√3 cm

And *h* = 10 cm

Volume
of the frustum of the cone = 1/3 × Ï€*h*(*r*_{1}^{2} + *r*_{2}^{2} + *r*_{1}*r*_{2})

Diameter
of the wire = 1/16 cm

Radius
of the wire = 1/32 cm

Let
the length of the wire be *H* cm.

Then,
volume of the wire = Ï€*r*^{2}*H*

=
22/7 × (1/32)^{2} × *H*

=
11 *H*/3584 cm^{3}

According
to the question, 11 *H*/3584 = 22000/9

*H* = (22000 × 3584)/(11 × 9)

*H* = (2000 × 3584)/9

*H* = 796444.44 cm

*H* = 7964.4 m

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