Question

In the figure ABCD is a rectangular of length 10root over 2 cm nd breadth 5root over 2cm. if APB is an isosceles triangle inscribed ih the semicircle with diameter AB, find the area of the shaded region?

Answer 1

Hi friend ,

Area of the shaded region= area of semicircle- area of triangle 
                             =$\frac{ \pi r^{2} }{2}$-$\frac{1}{2}$ x b x h
                             =$\frac{22}{7} X \frac{ \frac{10 \sqrt{2} }{2} X \frac{10 \sqrt{2} }{2} }{2}$-$\frac{1}{2} X10 \sqrt{2} X 5\sqrt{2}$

                              =[tex] \frac{22}{7} X 25 - 2 X 2 X 5
                              =78.57-50
                              =28.57 cm²
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Answer 2

IN THE GIVEN QUESTION, SIDES OF RECTANGLE AB=10$\sqrt{2}$ cm
AND AD=5$\sqrt{2}$ cm.
THE LARGEST SEMI CIRCLE INSCRIBED IN THIS RECTANGLE WILL BE WITH DIAMETER 10$\sqrt{2}$ cm
THEREFORE , ITS RADIUS WILL BE 5$\sqrt{2}$ cm.
ITS AREA WILL BE =π[tex] r^{2} /2 [/tex]
⇒$22/7*5 \sqrt{2} *5 \sqrt{2} /2$
⇒22*25/7
⇒AREA OF SEMICIRCLE=550/7  $cm^{2}$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=$1/2 \sqrt{2} =$
⇒$x/10 \sqrt{2} =1/ \sqrt{2}$
⇒X=10 cm.
therefore AREA OF ΔAPB=1/2*b*h
⇒1/2*x*x
⇒1/2*10*10
⇒AREA OF ΔAPB=50 $cm^{2}$ ...consider this as equation (2)
NOW, AREA OF SHADED REGION = AREA OF SEMICIRCLE-AREA OF TRIANGLE
⇒AREA OF SHADED REGION =550/7-50$cm^{2}$ 
⇒550/7-350/7
⇒200/7
⇒28.571528 $cm^{2}$ 
THEREFORE THE AREA OF SHADED REGION IS 28.571528 $cm^{2}$