Question

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.​

Answer 1

Answer :-

Capacity of glass = 102.67 cm³

$\rule{100}2$

A drinking glass is in the shape of the frustum and it's height is 14 cm.

So, height of frustum = h = 14 cm

Diameter of two circular ends is 4 cm and 2 cm.

Let the -

• diamter of smaller circular end be d1 = 2cm
• diamter of bigger circular end be d2 = 4 cm

So,

• radius of smaller circular end = r1 = 1 cm
• radius bigger circular end = r2 = 2 cm

Capacity of glass = Volume of frustum

Volume of frustum = 1/3πh(r1² + r2² + r1r2)

Substitute the known values in above formula

⇒ 1/3 × 22/7 × 14 [(1)² + (2)² + (1)(2)]

⇒ 1/3 × 22 × 2 (1 + 4 + 2)

⇒ 1/3 × 44(7)

⇒ 308/3

⇒ 102.67 cm³

Answer 2

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Volume of the frustrum shaped drinking glass is 102.67 cm³

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• Frustrum shaped glass

1. Height = 14 cm
2. Diameter of its two circular ends are 4 cm and 2 cm
3. $\sf{D_1\:=\:4\:cm}$
4. $\sf{D_2\:=\:2\:cm}$

To FiNd :

• Capacity of the glass (Volume)

SoLuTiOn :

First let's calculate the radius $\sf{r_1\:and\:r_2}$

Let's begin with the calculation of the first radius.

$\sf{r_1\:=\:{\dfrac{D_1}{2}}}$

$\sf{r_1\:=\:{\dfrac{4}{2}}}$

$\sf{\therefore{r_1\:=\:2\:cm}}$

Now, let's calculate the second radius.

$\sf{r_2\:=\:{\dfrac{D_2}{2}}}$

$\sf{r_2\:=\:{\dfrac{2}{2}}}$

$\sf{\therefore{r_2\:=\:1\:cm}}$

We have the height of the frustrum given in the question.

So now substituting the available values in the formula we can calculate the volume of the frustrum shaped drinking glass.

Formula :

$\bold{\large{\purple{\boxed{\sf{Volume\:=\:{\dfrac{1}{3}\:\pi\:h(r_1^2\:+\:r_2^2\:+\:r_1\:\times\:r_2}}}}}}$

Block in the values,

$\leadsto$ $\sf{\dfrac{1}{3}\:\times\:{\dfrac{22}{7}\:\times\:14\:(2^2\:+\:1^2\:+\:2\:\times\:1)}}$

$\leadsto$ $\sf{\dfrac{1}{3}\:\times\:{\dfrac{22}{7}\:\times\:14\:(4\:+\:1\:+\:2)}}$

$\leadsto$ $\sf{\dfrac{22}{21}\:14\:(5\:+\:2)}$

$\leadsto$ $\sf{\dfrac{308}{21}\:\times\:7}$

$\leadsto$ $\sf{14.6666666667\:\times\:7}$

$\leadsto$ $\sf{102.67\:cm^3}$