Question

From a piece of cardbord, in the
shape of a trapezium ABCD, and
AB||CD and angleBCD= 90°
quarter circle is removed. Given
AB = BC = 3.5 cm and DE = 2
cm. Calcualte the area of the
remaining piece of the
cardboard. (Take it to be 22divided by 7


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Answer 1

Answer:

6.125 cm²

Explanation:

A thin metallic 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 - an area of quarter circle BFEC

= 1/2{ AB + DC } × BC - π(BC)²/4

= 1/2{3.5 cm + 5.5cm } × 3.5cm - π(3.5cm)²/4

= 4.5cm × 3.5cm - 22/7 ×3.5×3.5/4 cm²

= 6.125 cm²