Question

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Answer 1

Given : Perimeter of a rhombus is 80 m and one of whose diagonal is 24 m.

Let ABCD be the rhombus and is divided by the diagonal BD = 24 m into two equal parts.

Therefore, Area of ∆BCD = Area of ∆ABD

Perimeter of rhombus ABCD = 4 × side  

80 = 4 × side  

Side = 80/4

Side = 20 m

Side (AB = BC = CD = DA) = 20 m  

Now,

Semi perimeter of Δ (s) = (a + b + c)/2

Semi perimeter of ΔBCD(s) = (20 + 20 + 24)/2

s = 64/2 m  

s = 32 m

 

Using heron’s formula :   Area of ΔBCD, A  = √s (s - a) (s - b) (s - c)

A = √32(32 - 20) × (32 - 20) × (32 - 24)

A = √32 × 12 × 12 × 8

A = √(32 × 8) × (12 × 12)

A = √256 × (12 × 12)

A = √(16 × 16) × (12 × 12)

A = 16 × 12

A = 192 m²

Area of ΔBCD = 192 m²

Area of ΔBCD = Area of ΔABD = 192 m²

Area of rhombus ABCD = 2 × Area of ∆BCD = 2 × 192 = 384 m²

Hence, the Area of rhombus is 384 m².

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Answer 2

Given:

Perimeter = 80 m

Diagonal = 24 m

To find:

Area of rhombus

Solution:

Side = Perimeter/4

=> Side = 80/4

=> Side = 20 m

On drawing the diagonal, we get 2 triangles with sides 20 m, 20 m, 24 m.

Formula to be used:

Heron's Formula

Heron's Formula: $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

Where:

s = semi perimeter

a, b, c = sides of triangle

a, b, c = 20 cm, 20 cm, 24 cm respectively

s = (a + b + c)/2 = (20 + 20 + 24)/2 = 64/2 = 32

Area of triangle = $\tt{\sqrt{s(s-a)(s-b)(s-c)}}$

=> $\tt{\sqrt{32(32-20)(32-20)(32-24)}}$

=> $\tt{\sqrt{32(12)(12)(8)}}$

=> 12 × 8 × 2

=> 192 sq cm

Area of both the triangles is equal. Thus, area of the rhombus:

192 × 2 sq cm

=> 396 sq cm