Question

The circumferences of circular faces of a frustum are 132cm and 88cm and it's height is 24cm. To find the curved surface area of the frustum complete the following activity.(π=22 upon 7)

Answer 1

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Answer 2

The curved surface area of the frustum 2750 $cm^{2}$.

Step-by-step explanation:

We have,

Height(h) = 24 cm and the circumferences of circular faces of a frustum = 132 cm and 88 cm.

We have,

$2\pi R_{1} =132$

⇒ $2\times \dfrac{22}{7} \times R_{1} =132$

⇒$2\times \dfrac{1}{7} \times R_{1} =6$

⇒  $R_{1} =21$ cm

And,$2\times \dfrac{22}{7} \times R_{2} =88$

⇒$\dfrac{1}{7} \times R_{2} =2$

⇒ $R_{2} =14$ cm

Slant height, l = $\sqrt{h^{2}+(R_{1}-R_{2} )^{2}}$

$= \sqrt{24^{2}+(21-14 )^{2} }=\sqrt{576+49}=25$ cm

∴  The curved surface area of the frustum $=\pi(R_{1}+R_{2} )l$

$=\dfrac{22}{7}\times (21+14)\times 25$

$=\dfrac{22}{7}\times 35\times 25=110\times 25=2750cm^{2}$

Hence, the curved surface area of the frustum 2750 $cm^{2}$.