Calculate the blue area

Blue Area = Red Area - Yellow Area

let $S_{c}$ = Area of Sector (CEA)

$S_{b}$ = Area of Sector (BAE)

k = Area of Kite (ABEC)

$x=\left| DB \right| \\ y=\left| AD \right| \\ \triangle BAD\sim \triangle ACD\Rightarrow \frac { x }{ y } =\frac { 5 }{ 10 } \Rightarrow x=\frac { y }{ 2 }$

$\triangle BAD$ is right angled triangle $\Rightarrow y^{ 2 }+\frac { y^{ 2 } }{ 4 } =25\Rightarrow y=2\sqrt { 5 } \Rightarrow \left| AE \right| =4\sqrt { 5 }$

$\left| BC \right| =\sqrt { 100 + 25 }=5\sqrt{5}$

$\angle C=\cos ^{ -1 }{ \left( \frac { 10^{ 2 }+10^{ 2 }-\left( 4\sqrt { 5 } \right) ^{ 2 } }{ 2\times 10\times 10 } \right) } =\cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } \\ \angle B=\cos ^{ -1 }{ \left( \frac { 5^{ 2 }+5^{ 2 }-\left( 4\sqrt { 5 } \right) ^{ 2 } }{ 2\times 5\times 5 } \right) } =\cos ^{ -1 }{ \left( \frac { -3 }{ 5 } \right) } =180-\cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } \\ S_{ c }=100\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } \\ S_{ b }=25\pi \frac { 180-\cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } =\frac { 25\pi }{ 2 } -25\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } \\ \\ k=\frac { 1 }{ 2 } \left( 4\sqrt { 5 } \right) \left( 5\sqrt { 5 } \right) =50$

Red Area = $S_{c} + S_{b} - k =100\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } +\frac { 25\pi }{ 2 } -25\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } -50=\frac { 25\pi }{ 2 } +75\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } -50$

Yellow Area= 2* Area of quarter circle - Area of square

Yellow Area= $2\left( \frac { 25\pi }{ 4 } \right) -{ 5 }^{ 2 }=\frac { 25\pi }{ 2 } -25$

Blue Area = $\frac { 25\pi }{ 2 } +75\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } -50-\left( \frac { 25\pi }{ 2 } -25 \right) =75\pi \frac { \cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } }{ 360 } -25\approx 9.77357$

Take O as the origin, OA as the x-axis, OB - the y-axis; then G:(4,8) Yellow Area =(25-2(25-25Pi/4)), Red Area (By integration) =15.226 cm² ::: Integral of [(25 -x^2)^(1/2)+5 - (100 -(x-10)^2)^(1/2)] from 0 to 4. Then the required blue area = Semi-circle on OB - Yellow Area - Red Area =25Pi/2 - (15.226) - (25-2(25-25Pi/4)) = 9.774 cm²