The difference between the sides at right angles in a right-angled triangle is $14 \mathrm{~cm}$. The area of the triangle is $120 \mathrm{~cm}^{2}$. Calculate the perimeter of the triangle.

Given:

The difference between the sides at right angles in a right-angled triangle is $14 \mathrm{~cm}$. The area of the triangle is $120 \mathrm{~cm}^{2}$.

To do:

We have to calculate the perimeter of the triangle.

Solution:

Let $ABC$ the right-angled triangle with right angle at $B$.

This implies,
$AB-AC=14\ cm$......(i)
Area $=120\ cm^2$

$\frac{1}{2}\times AB \times AC=120$

$AB \times AC=120\times2=240$.....(ii)

We know that,

$(a+b)^2=(a-b)^2+4ab$

Therefore,

$(AB+AC)^2=(AB-AC)^2+4AB.AC$

$(AB+AC)^2=(14)^2+4\times240$

$(AB+AC)^2=196+960$

$(AB+AC)^2=1156$

$(AB+AC)=\sqrt{1156}$

$(AB+AC)=34$.....(iii)

Adding (i) and (ii), we get,

$AB+AC+AB-AC=34+14$

$2AB=48$

$AB=24$

$\Rightarrow AC=AB-14$

$=24-14$

$=10$

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

$AC^2=(24)^2+(10)^2$

$=576+100$

$=676$

$\Rightarrow AC=\sqrt{676}$

$AC=26$

Perimeter of the triangle $=AB+BC+AC$

$=24+10+26$

$=60\ cm$

The perimeter of the triangle is $60\ cm$.

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Updated on: 10-Oct-2022

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