Question

In triangle ABC, if c square=a square bsquare, 2s=abc. Then 4s(s-a)(s-b)(s-c) is?

Answer 1

2s=a+b+c
s=(a+b+c)/2
By Heron's formula
Area=(s(s-a)(s-b)(s-c))1/2

a2+b2=c2
Area=1/2 * a * b

(s(s-a)(s-b)(s-c))1/2=1/2 * a * b
s(s-a)(s-b)(s-c)=a2b2/4
4s(s-a)(s-b)(s-c)=a2b2

Answer 2

∴ the values of $4{s(s-a)(s-b)(s-c)}$ is $2s$

Step-by-step explanation:

Given;

$c^{2}=a^{2}b^{2}$ and $2s=abc$ then find $4s(s-a)(s-b)(s-c)=?$

Now,

$2s=abc$

⇒$4s^{2} =a^{2} b^{2} c^{2}$ (By squaring both sides)

⇒$4s^{2} =c^{2} c^{2}$  (∵$c^{2}=a^{2}b^{2}$)

⇒$c^{2} =2s$ (By square root on both sides)

By Heron's formula;

Area of a triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

Also,

Area of a triangle$=\frac{1}{2} ab$

Then easily write;

$\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2} ab$

⇒${s(s-a)(s-b)(s-c)}=\frac{1}{4}a^{2} b^{2}$

⇒$4{s(s-a)(s-b)(s-c)}=a^{2} b^{2}$

⇒$4{s(s-a)(s-b)(s-c)}=c^{2}$

Plug the value $c^{2} =2s$

∴$4{s(s-a)(s-b)(s-c)}=2s$

So the values of $4{s(s-a)(s-b)(s-c)}$ is $2s$