Question

From a thin metallic piece, in the shape of a trapezium ABCD, in which $AB\parallel CD$ and $\angle BCD={90}^{\circ }$, a quarter circle BEFC is removed. Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of the remaining piece of the metal sheet.

Answer 1

A thin metalic piece in the shape of trapezium ABCD in which AB || CD and BCD = 90° a quarter circle BFEC is removed as shown in figure .

Given , AB = BC =3.5 cm and DE = 2cm

here you can see that , CE = CB = 3.5 cm { because CE and BC are the radii of quarter circle BFEC }
so,DC = DE + EC = 2cm + 3.5 cm = 5.5 cm

now, area of remaining part of trapezium ABCD = area of trapezium ABCD - area of quarter circle BFEC

$=\frac{1}{2}(AB+DC)\times BC-\frac{\pi (BC{)}^2}{4}$

$=\frac{1}{2}(3.5+5.5)\times 3.5-\frac{\pi (3.5{)}^2}{4}$

$=4.5\times 3.5-\frac{22}{7}\times \frac{3.5\times 3.5}{4}$

= 6.125 $c{m}^2$