Question

A right triangle has a perimeter of $48cm$ and an area of $96c{m}^2$. Find its hypotenuse (in cm).

• A. $18$
• B. $20$
• C. $24$
• D. $27$

Answer 1

The correct option is B $20$

Given: Area $=96c{m}^2$ and Perimieter $=48cm$

Let $a,b,c$ be the sides of a right triangle, with $c$ as hypotenuse.

$Area=\frac{1}{2}ab=96$

$ab=192$

$b=\frac{192}{a}$

$Perimeter=a+b+c=48$

$c=48-a-b$

From Pythagoras Theorem,

$c^2=a^2+b^2$

$(48-a-b{)}^2=a^2+b^2$

$2304+(a+b{)}^2-96(a+b)=a^2+b^2$

$2304+a^2+b^2+2ab-96(a+b)=a^2+b^2$

$2304+2ab-96a-96b=0$

Divide by 2,

$1152+ab-48a-48b=0$

$1152+a\left(\frac{192}{a}\right)-48a-48\left(\frac{192}{a}\right)=0$

$1344-48a-\frac{9216}{a}=0$

$48a^2-1344a+9216=0$

$a^2-28a+192=0$

$a=16$ and $a=12$

There are 2 possibilities as we have 2 perpendicular side

If, $a=16$, $b=\frac{192}{a}=\frac{192}{16}=12$

If, $a=12$, $b=\frac{192}{a}=\frac{192}{12}=16$

If, $a=12$ then $b=16$ or if $b=12$ then $a=16$

It doesn't make a difference while computing $c$

$c^2=a^2+b^2$

$c^2=144+256=400$

$c=$ Hypotenuse $=20cm$