Question

In $\Delta ABC$, $c^2=a^2+b^2$, then $4s(s-a)(s-b)(s-c)=?$

• A. $a^2{b}^2$
• B. $c^2{a}^2$
• C. $b^2{c}^2$
• D. $s^4$

Answer 1

The correct option is A $a^2{b}^2$

Since, it follows Pythagoras Theorem

given triangle is a Right Angled Triangle with

$Base=bHeight=aHypotenuse=c$

Therefore, its Area will be $△=\frac{1}{2}ab................\left(1\right)$

Also, we know that ${△}^2=s(s-a)(s-b)(s-c)$

Multiplying by 4 to both sides

$4{△}^2=4s(s-a)(s-b)(s-c)$

So, from $\left(1\right)$

$4{△}^2=4\ast \frac{1}{4}\ast a^2\ast b^2=a^2{b}^2$

Therefore Answer is $A$