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प्रश्न
If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.
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उत्तर
Let ABCD be the rhombus.
Diagonals AC and BD intersect at point E.
l(AC) = 30 cm …(i)
and Area of a rhombus = 240 sq. cm …(ii)
Area of a rhombus = `1/2` (product of diagnols)
⇒ 240 = `1/2` × (30 × DB)
⇒ DB = `(240 xx 2)/30` = 16 …(iii)
Diagonals of a rhombus bisect each other.
∴ `l(AE) = 1/2 l(AC)`
= `1/2 xx 30`
= 15 cm …(iv)
∴ `l(DE) = 1/2 l(DB)`
= `1/2 xx 16`
= 8 cm …(v)
In ΔADE,
∠AED = 90° …[Diagonals of a rhombus are perpendicular to each other]
AE2 + DE2 = AD2 …[Pythagoras theorem]
⇒ 152 + 82 = AD2 ...[From (iv) and (v)]
⇒ AD2 = 225 + 64 = 289
⇒ AD = `sqrt289` …[Taking square root of both sides]
= 17 cm
Thus, the side of the rhombus = 17 cm
Perimeter of rhombus = 4 × side
= `4 xx 17`
= 68 cm
∴ The perimeter of the rhombus is 68 cm.
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