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Question
Find the area of a rhombus whose perimeter is 260cm and the length of one of its diagonal is 66cm.
Solution
The perimeter of the Rhombus = 260cm
Each side of the Rhombus = `(1)/(4)(260)` = 65cm
In the given, Rhombus, AB = 65cm, diagonal AC = 66cm
We know that, diagonals of a Rhombus bisect at right angles.
In Triangle AOB,
∠AOB = 90°, AB is the hypotensue
AO = `33"cm"(1/2 (66"cm"))`
AB2 = OB2 + OA2
⇒ OB
= `sqrt("AB"^2 - "OA"^2)`
= `sqrt(65^2 - 33^2)`
= `sqrt(4225 - 1089)`
= `sqrt(3136)`
= 56
⇒ AB = 112cm
We know that the area of a rhombus whose diagonals are d1 and d2, is
A = `(1)/(2) xx "d"_1 xx "d"_2`
∴ the area of a rhombus whose diagonals are 112 and 66, is
A = `(1)/(2) xx 112 xx 66`
= 3696cm2.
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