Question

The perimeter of a right triangle is $144cm$ and its hypotenuse measure $65cm$. Find the lengths of the other sides and calculate its area using heron's formula.

Answer 1

The perimeter of right angled triangle is $144cm$.

Hypotenuse is $65cm$ .

let the two sides are $x$ and $y$ .

Then from Pythagorean theorem: $x^2+y^2={65}^2=4225$ ..(i)

Since, perimeter of right angled triangle is 144 cm.

And $x+y+65=144$

$x+y=144-65=79$

squaring both sides, we get

$(x+y{)}^2={79}^2$

$x^2+y^2+2xy=6241$

$4225+2xy=6241$ [From (i)]

$2xy=6241-4225=2016$

$xy=\frac{2016}{2}$

$=1008$

$y=\frac{1008}{x}$

substituting $y=\frac{1008}{x}$ in $x+y=79$ , we get

$x+\frac{1008}{x}=79$

$x^2+1008=79x$

$x^2-79x+1008=0$

$x^2-63x-16x+1008=0$

$x(x-63)-16(x-63)=0$

$(x-63)(x-16)=0$

so, $x=63cm$ or $x=16cm$

the length of other sides are $63cm,16cm$.

Let $S=\frac{(16+63+65)}{2}=\frac{144}{2}=72$

Using Heron's Formula, Area, $A=\sqrt{\left(s\right(s-a\left)\right(s-b\left)\right(s-c\left)\right)}$

$=\sqrt{72(72-16)(72-63)(72-65)}$

$=\sqrt{(72\times 56\times 9\times 7)}$

$=\sqrt{9\times 8\times 8\times 7\times 9\times 7}$

$=9\times 8\times 7=504c{m}^2$