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Question
A Rangoli has been drawn on a flor of a house. ABCD and PQRS both are in the shape of a rhombus. Find the radius of semicircle drawn on each side of rhombus ABCD.
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Solution
In rhombus ABCD,
AO = OP + PA
= 2 + 2
= 4 units
And OB = OQ + QB
= 2 + 1
= 3 units
We know that, diagonals of rhombus bisect each other at 90°.
Now, In ΔOAB,
(AB)2 = (OA)2 + (OB)2 ...[By Pythagoras theorem]
⇒ (AB)2 = (4)2 + (3)2
⇒ (AB)2 = 16 + 9
⇒ (AB)2 = 25
⇒ AB = `sqrt(25)`
⇒ AB = 5 units
Since, AB is diameter of semi-circle.
∴ Radius = `"Diameter"/2`
= `(AB)/2`
= `5/2`
= 2.5 units
Hence, radius of the semi-circle is 2.5 units.
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