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The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and △CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

The figure shows a kite, in which BCD is in the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and △CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mensuration

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Answer

From figure,

∠ECF = ∠BCD = 90°.

Area of right angle △ECF = 12×\dfrac{1}{2} \times CF × EC

= 12×6×6\dfrac{1}{2} \times 6 \times 6

= 18 cm2.

Area of quadrant BCD = πr24\dfrac{πr^2}{4}

=227×4224=22×42×427×4=3880828=1386 cm2.= \dfrac{\dfrac{22}{7} \times 42^2}{4} \\[1em] = \dfrac{22 \times 42 \times 42}{7 \times 4} \\[1em] = \dfrac{38808}{28} \\[1em] = 1386 \text{ cm}^2.

Area of shaded region = Area of quadrant BCD + Area of right angle △ECF

= 18 + 1386 = 1404 cm2.

Hence, area of shaded region = 1404 cm2.

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