Question

the area of a rhombus is 100 cm² one of its diagonals is twice the other what is the perimeter of the rhombus​

Answer 1

Answer:

Question:-

The area of a rhombus is 100 cm². If one of the diagonals is twice the other, find the perimeter of the rhombus.

Solution:-

Diagonal 1 (d₁) = x

Diagonal 2 (d₂) = 2x [given]

Area = 100 cm² [given]

Area = $\frac{1}{2} * diagonal * diagonal$

$\frac{1}{2} * x * 2x = 100$

$x^{2} = 100$

$x = \sqrt{100}$

$x = 10$

d₁ = 10 cm = x

d₂ = 20 cm = 2x

We know that the diagonals of a rhombus bisect each other perpendicularly, i.e. forming an angle of 90 degrees.

We can find out the measure of a side using Pythagoras Theorem.

First side = 5 cm [after being bisected]

Second side = 10 cm [after being bisected]

Third side = x [side of rhombus]

Pythagoras Theorem states:-

(P)² + (B)² = (H)²

(10)² + (5)² = (H)²

100 + 25 = (H)²

125 = (H)²

√125 = H

11.18 = H = side of rhombus

Perimeter of Rhombus = 4s

= 4(11.18)

= 44.72 cm OR 45 cm [after round off]

Hope it helps

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