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

Given:

A rhombus whose perimeter is $80\ m$ and one of whose diagonal is $24\ m$.

To do:

We have to find the area of the rhombus.

Solution:

Let the perimeter of the rhombus $ABCD$ be $80\ m$.

Each side $=\frac{80}{4}$

$=20 \mathrm{~m}$

One of the diagonals $=24 \mathrm{~m}$

Diagonals of a rhombus bisect each other at right angles.

In right angled triangle $\mathrm{AOB}$,

$\mathrm{AB}=20 \mathrm{~m}$

$\mathrm{AO}=\frac{1}{2} \mathrm{AC}$

$=\frac{1}{2} \times 24$

$=12 \mathrm{~m}$

$A B^{2}=A O^{2}+O B^{2}$

$(20)^{2}=(12)^{2}+O B^{2}$

$400=144+O B^{2}$

$O B^{2}=400-144$

$=256$

$=(16)^{2}$

$\Rightarrow OB=16 \mathrm{~m}$

$B D=2 \times O B$

$=2 \times 16$

$=32 \mathrm{~m}$

Area of the rhombus $=\frac{1}{2} \times d_{1} \times d_{2}$

$=\frac{1}{2} \times 24 \times 32$

$=384 \mathrm{~m}^{2}$.

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Updated on: 10-Oct-2022

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