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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