Question

Prove that area of triangle
Rut ovr. S(s-a)(s-b)(s-c)​

Answer 1

To Prove :-

• Heron's formula .

Some formula used here are :-

• (a+b)² = a²+b²+2ab

• (a-b)² = a²+b²-2ab

• Pythagoras - h² = b²+P²

• a²-b² = [a+b][a-b]

Now we know the value of h

(refer to the attachment)

→ Area of triangle = $\sf{\frac{1}{2} \times height \times base }$

→ Putting the value of h here

$\sf{Area \: = \: \frac{1}{2} \times a \: \times \frac{2}{a} \sqrt{s(s - a)(s - b)(s - c)}}$

$\boxed{\sf{Area = \: \sqrt{s(s-a)(s-b)(s-c)} }}$

Answer 2

From the fig., we get the semiperimeter,

$\longrightarrow s=\dfrac{a+b+c}{2}=x+y+z$

Also we get,

• $x+y=b$

• $y+z=c$

• $z+x=a$

So that,

• $s-a=y\quad\quad\dots(1)$

• $s-b=z\quad\quad\dots(2)$

• $s-c=x\quad\quad\dots(3)$

In the fig., OA and O'A are angle bisectors drawn from the same point. Thus,

$\longrightarrow\angle OAO'=90^o$

Hence triangles OAD and O'AG are similar. So we get,

$\longrightarrow \dfrac{r'}{x}=\dfrac{y}{r}$

$\longrightarrow r'=\dfrac{xy}{r}\quad\quad\dots(4)$

From the fig. it is clear that the triangles ODB and O'GB are similar. So,

$\longrightarrow \dfrac{r'}{x+y+z}=\dfrac{r}{z}$

From (4),

$\longrightarrow \dfrac{xy}{r(x+y+z)}=\dfrac{r}{z}$

$\longrightarrow r^2=\dfrac{xyz}{x+y+z}$

$\longrightarrow r=\sqrt{\dfrac{xyz}{x+y+z}}$

Since $s=x+y+z,$

$\longrightarrow r=\sqrt{\dfrac{xyz}{s}}$

From (1), (2) and (3),

$\longrightarrow r=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}\quad\quad\dots(5)$

But, if A is the area of the triangle ABC, then,

$\longrightarrow A=Area(\triangle AOB)+Area(\triangle BOC)+Area(\triangle COA)$

$\longrightarrow A=\dfrac{1}{2}\,ra+\dfrac{1}{2}\,rb+\dfrac{1}{2}\,rc$

$\longrightarrow A=\dfrac{r(a+b+c)}{2}$

Since $\dfrac{a+b+c}{2}=s,$

$\longrightarrow A=rs$

$\longrightarrow r=\dfrac{A}{s}$

Hence (5) becomes,

$\longrightarrow \dfrac{A}{s}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$

$\longrightarrow A=s\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}}$

$\longrightarrow\underline{\underline{A=\sqrt{s(s-a)(s-b)(s-c)}}}$

Hence Proved!