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Standard X

Mathematics

Areas of Combination of Plane Figures

From a thin m...

Question

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

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Solution

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

= 0.5{ AB + DC } × BC - π(BC)²

= 0.5{3.5 cm + 5.5cm } × 3.5cm - π(3.5cm)²
= 4.5cm × 3.5cm - $fraction numerator 22 over denominator 7 cross times 4 end fraction open parentheses 3.5 close parentheses squared$ cm²
= 6.125 cm²

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From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠ BCD = 90∘, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.

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