Question

In the follwing figure ABCD is a rectangle with AD and DC equal to 1 and 2 units respectively .Two quarter circle are drawn with centres at B and A respectively. Now a circle is drawn touching both the quarter circle and done of the sides of the rectangle. Find the area of the shaded region:​

Answer 1

Answer:

13/56

Step-by-step explanation:

sure my answer is right 100%

Answer 2

Answer: 13/56 units

Step-by-step explanation:

Given : ABCD is a rectangle with AD = 1 units and DC = 2 units

To find : Area of shaded region

As it is given that ABCD is a rectangle hence AB = DC and AD = BC

Hence AB = 2 units as DC = 2 units

Assuming P as mid point of AB then AP = 1 units

Taking the mid point of circle in shaded region as O and assuming AOP to be a right angle triangle.

Then, OA = 1 - r   OP = 1-r    AP = 1

Hence  $OA^{2}$ - $AP^{2}$ = OP

Putting values of OA, AP and OP we get

( 1 +$r^{2}$) - 1 = (1 - $r^{2}$)

( 1 +$r^{2}$) - (1 - $r^{2}$) = 1

Applying the formula of  $(a^{2} + b^{2} ) and (a^{2} - b^{2} )$

4r = 1

Hence r = 1/4

Area of shaded region = 2 - $\pi /2$ (1) - $\pi$ ($1/4^{2}$)

= 13/ 56 units

Answer = 13/56 units