Question

The diagonals of a rhombus are 24 cm and 10 cm. Find its area and perimeter.​

Answer 1

Given that

Diagonals of the rhombus

Let d1 =10cm and d2= 24cm

Find out

We need to find the perimeter of the given rhombus

Solution

Diagonals meet at the centre and forms right-angled triangles.

So by using pythagoras theorem

Length of the base = 10/2 = 5cm
Length of the height = 24/2 = 12cm

Hypotenuse2 = side 2+ side2

Hypotenuse2= 52+ 122

Hypotenuse2 = 25 + 144

Hypotenuse2 = 169

On taking square root we get,

Hypotenuse = 13 { 13 X 13=169 => √169=13}

Hence the side of the rhombus is 13cm.

Perimeter of the rhombus = 4×side
= 4 × 13
= 52cm.

Answer

Therefore, the perimeter of the rhombus is 52cm.

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Answer 2

Answer:

120cm² and 52cm

Step-by-step explanation:

given; d1 = 24cm and d2 = 10cm

By using formula,

Area of rhombus= 1/2×d1×d2

= 1/2×24×10cm²

= 120cm²,ans..

and for perimeter firstly we have to find the side of rhombus so by using Pythagoras theorem,

H²= P²+B² (diagonal of rhombus bisect each other at 90°)

H²= (12)²+(5)²

H²= 144+25

H²= 169cm²

H= 13cm

so, side of rhombus= 13cm

perimeter of rhombus = 4×side

= 4× 13cm

= 52cm, ans...