. In the figure, ABCD is …

IN THE GIVEN QUESTION, SIDES OF RECTANGLE AB=10 cm
AND AD=5 cm.
THE LARGEST SEMI CIRCLE INSCRIBED IN THIS RECTANGLE WILL BE WITH DIAMETER 10 cm
THEREFORE , ITS RADIUS WILL BE 5 cm.
ITS AREA WILL BE =π
⇒
⇒22*25/7
⇒AREA OF SEMICIRCLE=550/7  consider this as equation (1)
NOW,ΔAPB IS AN ISOSCELES TRIANGLE WITH SIDES AP=AB,
THEREFORE LET AP=AB= X(say)
THEN IN TRIANGLE APB , 
WE GET ∠PAB=∠PBA (ANGLES OPPOSITE TO EQUAL SIDES)=45°(ANGLE SUM PROPERTY IF A TRIANGLE)
BY SIN 45° IN TRIANGLE WE GET,
PB/AB=
⇒
⇒X=10 cm.
therefore AREA OF ΔAPB=1/2*b*h
⇒1/2*x*x
⇒1/2*10*10
⇒AREA OF ΔAPB=50  ...consider this as equation (2)
NOW, AREA OF SHADED REGION = AREA OF SEMICIRCLE-AREA OF TRIANGLE
⇒AREA OF SHADED REGION =550/7-50 
⇒550/7-350/7
⇒200/7
⇒28.571528  
THEREFORE THE AREA OF SHADED REGION IS 28.571528