Question

The diagonal of a rhombus measure 14 cm and 48 cm . Find its perimeter.

Answer 1

Hers is your answer.

We know that diagonals of rhombus bisect ar 90°.

Use Pythagoras therome to get sides.

In ∆OAB angle O is 90°.

Than,

AB^2 = OA^2 + OB ^2

AB = √ OA^2 + OB^2

Define AB as x

Than,

${x}^{2} = {(24)}^{2} + {(7)}^{2} \\ {x}^{2} = 576 + 49 \\ {x}^{2} = 625 \\ x = \sqrt{625 } \\ x = 25cm$

We get x = 25cm than,

•°• AB = 25cm

We know that in rhombus all sides are equal. then,

Perimeter = AB+BC+CD+ DA

$= 25 + 25 + 25 + 25 \\ = 100cm$

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Answer 2

Given : d1 = 14cm,

d2 = 48cm

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Side of a rhombus = √(d1/2)² + (d2/2)²

= √(14/2)² + (48/2)²

= √7² + 24²

= √49 + 576

= √625cm² = 25cm.

Perimeter of a rhombus = 4* side

= 4*25cm

= 100cm.