# Ncert Solutions For Class 10 Maths Ex 13.4

## Ncert Solutions For Class 10 Maths Chapter 13 Ex 13.4

1: A drinking glass with 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. Calculate the capacity of the glass.

Ans.- Given, R = 2, r = 1 cm and h = 14 cm

We know that, volume of frustum is

=13πh(R2+r2+Rr)$=\frac{1}{3}\pi h\left ( R^{2} +r^{2}+Rr\right )$ =13π×14(22+12+2)$=\frac{1}{3}\pi \times 14\left ( 2^{2}+1^{2}+2 \right )$ =13×227×14×7$=\frac{1}{3}\times \frac{22}{7 }\times 14\times 7$ =10223cm3$=102\frac{2}{3}cm^{3}$

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. Calculate the curved surface area of the frustum.

Ans.- Given, Slant height l = 4 cm, perimeters = 18 cm and 6 cm

Radii can be calculated as follows:

We know that, Radius= Perimeter2π$\frac{Perimeter}{2\pi }$

= 182π=9π$\frac{18}{2\pi }=\frac{9}{\pi }$

And radius = 62π=3π$\frac{6}{2\pi }=\frac{3}{\pi }$

Therefore the curved surface area of frustum is

= π(R+r)l$\pi \left ( R+r \right )l$

= π(9π+3π)4$\pi\left ( \frac{9}{\pi }+\frac{3}{\pi } \right )4$= 48cm2$48cm^{2}$

3: A fez, the cap used by the Turks, is shaped like the frustum of a cone. 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, calculate the area of material used for making it.

Ans.- Given,  R = 10 cm, r = 4 cm, slant height = 15 cm

We know that, Curved surface area of frustum is

= π(R+r)l$\pi \left ( R+r \right )l$

= π(10+4)15$\pi \left ( 10+4 \right )15$

= 227×14×15=660$\frac{22}{7}\times 14\times 15= 660$

Now area of upper base is

= πr2=π×42$\pi r^{2}=\pi \times 4^{2}$

= 16π=5027$16\pi =50\frac{2}{7}$

So the total surface area is

660+5027=71027cm2$660+50\frac{2}{7}=710\frac{2}{7}cm^{2}$

4: A vessel, opened from the top and made up of an aluminium sheet, is in the form of a frustum of a cone having height, radii of its lower and upper ends are 16cm, 8 cm and 20 cm, respectively. Calculate the cost of the milk which can completely fill the vessel, at the rate of Rs 20 per litre. Also calculate the cost of aluminium sheet used to make the vessel, if it costs Rs 8 per 100 cm2.

Ans.- Given,  Height of frustum = 16 cm, R = 20 cm, r = 8 cm

We know that, volume of frustum is

= 13πh(R2+r2+Rr)$\frac{1}{3}\pi h\left ( R^{2}+r^{2}+Rr \right )$

= 13π×16(202+82+160)$\frac{1}{3}\pi \times 16\left ( 20^{2}+8^{2}+160 \right )$

= 13×227×16(400+64+160)$\frac{1}{3}\times \frac{22}{7}\times 16\left ( 400+64+160 \right )$

= 13×227×16×604=10449.92cm3$\frac{1}{3}\times \frac{22}{7}\times 16\times 604= 10449.92cm^{3}$

Cost of milk @ Rs. 20 per 1000 cubic cm

10.44992 x 20 = Rs. 208.99

For calculating surface area, we need to find slant height which can be calculated as follows:

l=h2+(Rr)2$l= \sqrt{h^{2}+\left ( R- r \right )^{2}}$

= 162+(208)2$\sqrt{16^{2}+\left ( 20- 8 \right )^{2}}$

= 256+144=400=20cm$\sqrt{256+144}=\sqrt{400}=20cm$

Now the surface area of frustum is

= π(R+r)l+πr2$\pi \left ( R+r \right )l+\pi r^{2}$

= π[(R+r)l+r2]$\pi \left [ \left ( R+r \right ) l+r^{2}\right ]$

= π[(20+8)20+82]$\pi \left [ \left ( 20+8 \right ) 20+8^{2}\right ]$

= π(560+64)$\pi \left ( 560+64 \right )$

= 227×624=1959.36$\frac{22}{7}\times 624= 1959.36$

Therefore Cost of metal sheet @ Rs. 8 per 100 sq cm = 19.5936 x 8 = Rs. 156.75

5: A metallic right circular cone of  20 cm high and whose vertical angle is 60$60^{\circ}$ 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, calculate the length of the wire.

Ans.-  Here volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.

In the given figure; AO = 20 cm and hence height of frustum LO = 10 cm

In triangle AOC we have angle CAO = 30 (half of vertical angle of cone BAC)

Therefore;

tan30=OCAO$tan 30^{\circ}= \frac{OC}{AO}$

Or, 13=OC20$\frac{1}{\sqrt{3}}=\frac{OC}{20}$

Or, OC= 203$\frac{20}{\sqrt{3}}$

Using similarity cirteria in triangles AOC and ALM it can be shown that LM = 10/√3 (because LM bisects the cone through its height)

Similarly, LO = 10 cm

Volume of frustum can be calculated as follows:

V = 13π×10(4003+1003+2003)$\frac{1}{3}\pi \times 10\left ( \frac{400}{3}+\frac{100}{3} +\frac{200}{3}\right )$

= 70009πcubcm$\frac{7000}{9}\pi \, cub cm$

Volume of cylinder is given as follows:

πr2h$\pi r^{2}h$

Or, π×(132)2×h$\pi \times \left ( \frac{1}{32} \right )^{2}\times h$

= 70009π$\frac{7000}{9}\pi$

Or, h= 70009×1024=796444.44cm$\frac{7000}{9}\times 1024= 796444.44cm$

= 7964.4m