Question

In the figure, a semicircle circumscribes a trapezium ABCD. If AB = 2CD. LADC = 120 and the radius of the semicircle is r, then the area of the shaded region is......

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Answer 1

Let us consider a figure.

Here ABCD is the trapezium in the semi-circle with center '0'

∠ ADB=90°

AD=ABcosθ=2Rcosθ

AE=ADcosθ=2Rcos

2

θ

ER=AB−2AC=2R−4Rcos

2

θ

DE=ADsinθ=2Rsinθcosθ

Area of Trapezium =

2

1

(AB+CD)×DE

=

2

1

(2R+2R−4Rcos

2

θ)×2Rsinθ

=

2

1

[4R−4Rcos

2

θ]2Rsinθ

A=4R

2

(1−cos

2

θ)cosθsinθ

A=4R

2

sin

2

θcosθ

For the maximum area, differentiate w.r.t θ

θ

dA

=12R

2

sin

2

θcos

2

θ−4R

2

sin

4

θ=0

⇒sin

2

θ4R

2

[3cos

2

θ−sin

2

θ]=0

⇒sin

2

θ=0or3cos

2

θ−sin

2

θ=0

⇒θ=0°tan

2

θ=3

tanθ=

3

θ=

3

π

Therefore area will be maximum at θ=

3

π

A=4R

2

sin

3

θcosθ

Atθ=

3

π

A=4R

2

sin

3

3

π

cos

3

π

=4R

2

(

2

3

)

3

×

2

1

=2R

2

4

3

3

A=

4

3

3

R

2

Hence option A is the correct answer.

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