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Question
Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.
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Solution
Let `square`ABCD be the rhombus.
AC = 20 cm, BD = 21 cm
AO = `1/2` AC ...[Diagonals of a rhombus bisect each other]
= `1/2 xx 20`
= 10 cm ...(i)
BO = `1/2` BD ...[Diagonals of a rhombus bisect each other]
= `1/2 xx21`
= `21/2`
= 10.5 cm ...(ii)
In ∆AOB,
∠AOB = 90° ...[Diagonals of a rhombus are perpendicular to each other]
By Pythagoras theorem,
∴ AB2 = AO2 + BO2
∴ AB2 = 102 + 10.52
∴ AB2 = 100 + 110.25
∴ AB2 = 210.25
∴ AB = `sqrt(210.25)`
∴ AB = 14.5 cm
Perimeter of `square`ABCD = 4 × AB
= 4 × 14.5
= 58 cm
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