OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is 32 (under-root)3 cm2, find the radius of the circle.
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O is centre of the circle
OPQR is a rhombus.
Let the diagonals OQ and PR intersect at S
area of rhombus OPQR = 32√3 cm2
Let
OP = OQ = OR = r
OS = SQ = r/2
RS = PS
In right ΔOSP
By Pythagoras theorem
OP² = OS² + PS²
r² = (r/2)² + PS²
PS² = r² – (r/2)²
= 3r²/4
so, PS = (√3r/2)
PR = 2PS = √3r
area of rhombus = 1/2 x d₁ x d₂
area of rhombus OPQR = 1/2 x OQ x PR
32√3 = 1/2 x r x √3r
32 = 1/2 r²
r² = 64
r = 8
Area of circle = πr²
= 22/7 x 8²
= 201. 14 sq cm
OPQR is a rhombus.
Let the diagonals OQ and PR intersect at S
area of rhombus OPQR = 32√3 cm2
Let
OP = OQ = OR = r
OS = SQ = r/2
RS = PS
In right ΔOSP
By Pythagoras theorem
OP² = OS² + PS²
r² = (r/2)² + PS²
PS² = r² – (r/2)²
= 3r²/4
so, PS = (√3r/2)
PR = 2PS = √3r
area of rhombus = 1/2 x d₁ x d₂
area of rhombus OPQR = 1/2 x OQ x PR
32√3 = 1/2 x r x √3r
32 = 1/2 r²
r² = 64
r = 8
Area of circle = πr²
= 22/7 x 8²
= 201. 14 sq cm
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