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Question
Find the perimeter of a rhombus whose diagonals are 24cm and 10cm.
Solution
In the given Rhombus diagonal AC = 24cm and diagonal BD = 10cm
We Know that, diagonals of a Rhombus bisect at right angles. In Triangle AOB,
∠AOB = 90°, AB is the hypotenuse
OB = `12"cm"(1/2 (24"cm"))` and
AO = `5"cm"(1/2 (10"cm"))`
AB = `sqrt("OB"^2 + "OA"^2)`
= `sqrt(12^2 + 5^2)`
= `sqrt(144 + 25)`
= `sqrt(169)`
= 13
Further all sides of a Rhombus are equal by definition
So, AB = BC = CD = AD = 13cm
Perimeter
= 4(13)
= 52cm.
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