Question

If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm${}^2$; find its perimeter ( in $cm.$)

Answer 1

Let ABC be the right angled triangle with right angle at A.
AB is the base, AC is the height and BC is the hypotenuse.
As the difference between the base and height is $3cm$, let AB $=xcm$ and AC $=(x+3)cm$
Area of triangle $=\frac{1}{2}\times base\times height=54$ sq cm
$\frac{1}{2}\times x\times (x+3)=54$
$=>x^2+3x=108$

$x^2+12x-9x-(12\times 9)=0$
Solving it $x=9;-12$
Since side cannot be negative, AB $=x=9cm$
AC $=x+3=12cm$
Since the third side BC is the hypotenuse , ${BC}^2=9^2+{12}^2=225$
Hence, BC $=15cm$
Perimeter of the triangle $=$ Sum of all sides $=15+9+12=36cm$