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 32underroot3 sp cm . find the radius of the circle
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From the
figure, O is centre of the circle and OPQR is a rhombus.
Let the diagonals OQ and PR intersect at S
Given area of rhombus OPQR = 32√3 cm2
Let OP = OQ = OR = r cm
OS = SQ = (r/2) cm and RS = PS
In right ΔOSP, OP2 = OS2 + PS2 (By Pythagoras theorem)
⇒ r2 = (r/2)2 + PS2
⇒ PS2 = r2 – (r/2)2 = 3r2/4
∴ PS = (√3r/2)
⇒ PR = 2PS = √3r
⇒ r2 = 64
∴ r = 8
Area of circle = πr2 sq units
Let the diagonals OQ and PR intersect at S
Given area of rhombus OPQR = 32√3 cm2
Let OP = OQ = OR = r cm
OS = SQ = (r/2) cm and RS = PS
In right ΔOSP, OP2 = OS2 + PS2 (By Pythagoras theorem)
⇒ r2 = (r/2)2 + PS2
⇒ PS2 = r2 – (r/2)2 = 3r2/4
∴ PS = (√3r/2)
⇒ PR = 2PS = √3r
⇒ r2 = 64
∴ r = 8
Area of circle = πr2 sq units
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Answer:- Radius of the circle is 8 cm.
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