Question

The difference of two line adjoining at right angle in right angle triangle is 14 and thr area of triangle is 120cm^2.Hence prove that this area is satisfied by heron's formula ​

Answer 1

Consider Heron's formula.

$A=\sqrt{s(s-a)(s-b)(s-c)}\quad\text{where}\quad s=\dfrac{a+b+c}{2}.$

To find the area of any triangle using Heron's Formula we should have the length of the three sides of the triangle.

So, to prove whether the area 120 cm² is obtained by Heron's Formula, we have only to check whether the length of the three sides can be found out.

We're only given that the area of a right triangle is 120 cm² and the difference between the perpendicular sides of the triangle is 14 cm.

Let the perpendicular sides be 'a' and 'b', so that,

$\text{ a-b=14 cm}$

The area can be written as,

$\text{ \dfrac{1}{2}ab=120 cm \quad\implies\quad ab=240 cm}$

Let the hypotenuse be 'c'.

So,

$\begin{aligned}&(a-b)^2=14^2\\\\\implies\ \ &a^2+b^2-2ab=196\\\\\implies\ \ &a^2+b^2-2ab+4ab=196+4\times240\quad[\ \because\ ab=240\ ]\\\\\implies\ \ &a^2+b^2+2ab=196+960\\\\\implies\ \ &(a+b)^2=1156\\\\\implies\ \ &a+b=34\ \text{cm}\end{aligned}$

Now,

$\begin{aligned}&\dfrac{(a+b)+(a-b)}{2}=\dfrac{34+14}{2}\\\\\implies\ \ &\dfrac{2a}{2}=\dfrac{48}{2}\\\\\implies\ \ &a=24\ \text{cm}\end{aligned}$

And,

$\begin{aligned}&\dfrac{(a+b)-(a-b)}{2}=\dfrac{34-14}{2}\\\\\implies\ \ &\dfrac{2b}{2}=\dfrac{20}{2}\\\\\implies\ \ &b=10\ \text{cm}\end{aligned}$

And by Pythagoras' Theorem,

$c=\sqrt{10^2+24^2}=\sqrt{100+576}=\sqrt{676}=26\ \text{cm}$

So we could find out the lengths of the three sides. So we can find out the area of the triangle by using Heron's formula.