In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is 32 √cm2, find the radius of the circle.
From the figure, O is the centre of the circle and OPQR is a rhombus.
Let the diagonals OQ and PR intersect at S
Given area of rhombus OPQR =32√ cm2
Let OP=OQ=OR=r cmOS=SQ=(r2) cm and RS=PS
In right ΔOSP,OP2=OS2+PS2(By Pythagoras theorem)⇒r2=(r2)2+PS2⇒PS2=r2–(r2)2=3r24∴PS=√r2⇒PR=2PS=√r
Area of rhombus OPQR =12×d1×d2=12×OQ×PR⇒32√=12×r×√r⇒32=12×r2r2=64r=8 cm
Radius of circle = 8 cm