Question

In Fig. 2, ABCD is a rectangle. The radii of the semicircles drawn on AD and BC and the radius of the circle drawn in between are same. Given AD-7 cm, calculate the area of the shaded region.​

Answer 1

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Given AB=14cm,BC=AD=7cm

✏we know that the area of rectangle=l×b

↣area of the semicircle=

$π× {r}^{2}$

↣Here the radius of the big ssemicircle

$({r}^{2} ) = \frac{14}{2} = 7cm$

✏The radius of the small semicircle

$( {r}^{2} ) = \frac{7}{2} cm$

✏Area of shaded region = Area ABCD + 2 × area small semi-circle - area big semi-circle

$(l \times b) + 2. \frac{1}{2} \pi {r1}^{2} - \frac{1}{2} \pi {r2}^{2}$

$(14 \times 7) + 2 \times \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} - \frac{1}{2} \times 7 \times 7$

$98 + 38.5 - 77 = 59.5 {cm}^{2}$

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Answer 2

Given AB=14cm,BC=AD=7cm

we know that the area of rectangle=l×b

area of the semicircle= 2 π×r 2

Here the radius of the big semicircle(r 1 )= 214

=7cm

the radius of the small semicircle(r

2

)=

2

7

cm

Area of shaded region = Area ABCD + 2 × area small semi-circle - area big semi-circle

=(l×b)+2⋅

2

1

πr

1

2

−

2

1

πr

2

2

=(14×7)+2×

2

1

×

7

22

×

2

7

×

2

7

−

2

1

×

7

22

×7×7

=98+38.5−77=59.5cm

2