Question

the area of the shaded portion in the figure
A
10 m
B.
A
6m
D
Fig. 14.25​

Answer 1

Let the center of circle be at O. AC is diagonal of square ABCD

AC=42cm

we have AE=CF=1cm each

EF=AC−AE−CF

EF=42−2

radius of circle =21EF=22−1 

Area of shaded region = area (ABCD) − area of circel − 4\times area of each quadrant

=4×4−π×(22−1)2−4×41π(1)2

=16−π{(22−1)2+1}

=16−π{8+1−42+1}

=16−(10−42)π

=16−13.64439

Area =2.36cm2

I HOPE IT'S HELPFUL for you

Answer 2

$\sf \Large Solution!!$

We are given the length and breadth of the rectangle. The breadth of the rectangle is the diameter of the circle. So, we have to first find out the area of the rectangle. Then, we'll find out the area of the square and subtract it from the area of the rectangle to get the area of the shaded portion.

$\sf Area_{(rectangle)}=Length\times Breadth$

$\sf Area_{(rectangle)}=10\:m\times 6\:m=60\:m^{2}$

$\sf Diameter_{(circle)}=6\:m$

$\sf Radius_{(circle)}=\dfrac{6}{2}=3\:m$

$\sf Area_{(circle)}=\pi r^{2}$

$\sf Area_{(circle)}=\dfrac{22}{7}\times (3\:m)^{2}$

$\sf Area_{(circle)}=28.28\:m^{2}$

$\sf Area_{(shaded\:portion)}=Area_{(rectangle)}-Area_{(circle)}$

$\sf Area_{(shaded\:portion)}=60\:m^{2}-28.28\:m^{2}$

$\sf Area_{(shaded\:portion)}=31.72\:m^{2}$