Question

If the area of the rhombus is 96cm² and one of its diagonals is 16cm, find its perimeter.​

Answer 1

Area = 96cm²

given that it is a rhombus.

We know that,

area of rhombus = $\frac{1}{2}$ × diagonal 1 × diagonal 2

diagonal 1 = 16cm

let diagonal 2 be x cm

96 = $\frac{1}{2}$ × 16 × x

96 = 8 × x

$\frac{96}{8}$ = x

12 cm = x

In a rhombus, the diagonals bisect each other at 90°

therefore, if we divide the rhombus into 4 triangles, the meeting point of the diagonals will be 90°.

And as the diagonals bisect each other, to find the side we have to apply the Pythagoras theorem

(half of diagonal 1)² + (half of diagonal 2)² = (side)²

(8)² + (6)² = (side)²

64 + 36 = (side)²

100 = (side)²

√100 = side

10 cm = side

perimeter = 10 + 10 + 10 + 10 = 40cm (as all sides of a rhombus are equal)

Hope it helps

have a great day