From a circular piece of carboard of radius \( 3 \mathrm{~cm} \) two sectors of \( 90^{\circ} \) have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters (Take \( \pi=22 / 7) \)

Given:

From a circular piece of carboard of radius \( 3 \mathrm{~cm} \) two sectors of \( 90^{\circ} \) have been cut off.

To do:

We have to find the perimeter of the remaining portion nearest the hundredth centimetres.

Solution:

Let $ABCD$ be the cardboard as shown in the figure.

Two sectors of \( 90^{\circ} \) have been cut off.

This implies,

We get a semicular cardboard piece.

Let $ABD$ be the remaining cardboard piece.

Therefore,

Perimeter of arc $ABD=\frac{1}{2}(2 \pi r)$

$=\pi (3)$

$=\frac{22}{7} \times 3$

$=\frac{66}{7}$

$=9.428 \mathrm{~cm}$

The perimeter of the remaining portion nearest hundredth centimeters is $9.428 \mathrm{~cm}$.

Tutorialspoint

Updated on: 10-Oct-2022

97 Views

Kickstart Your Career

Get certified by completing the course

Get Started