Question

Find the area of a rhombus of perimeter 60cm with one of its diagonal as 24cm.

Answer 1

Answer:

Given.

abcd= rhombus

o= point of intersection of the two diagnols

Perimeter of the rhombus = 60 cm

4a = 60........ (where a= length of one side)

a = 60/4 = 15 cm

therefore, Side = 15 cm

using Pythagorean Theorem to find the length of the second diagnol

H² = B² + P²

(AB)² =  (OA)² + (OB)²..........( o divides the diagnols in two equal parts)

(15)² = (OA)² + (12)²

225 = (OA)² + 144

225 - 144 = (OA)²

81 = (OA)²

9 = OA

Then

AC = 2 x OA

AC = 2 x 9 = 18 cm

Now, Area = 1/2 * product of its diagonals

                = 1/2 * 24 * 18

                = 216 cm^2

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