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Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.
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Solution
Given that,
Perimeter of rhombus = 80m
Perimeter of rhombus = 4 × side
⇒ 4a = 80
⇒ a = 20m
Let AC = 24 m
∴ OA =`1/2AC=1/2xx24=12m`
In ΔAOB
`OB^2=AB^2-OA^2`
`⇒OB=sqrt(20^2-12^2)`
`sqrt(400-144)`
`sqrt(256)=16m`
Also BO = OD
[Diagonal of rhombus bisect each other at 90°]
∴ BD = 20B = 2 ×16 = 32 m
∴Area of rhombus = `1/2`×32×24=`384𝑚^2`
[∵Area of rhombus = 12×𝐵𝐷×𝐴𝐶]
Concept: Application of Heron’s Formula in Finding Areas of Quadrilaterals
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