If {c^2} = {a^2} + {b^2},then 4s(s - a)(s - b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the condition ${c^2} = {a^2} + {b^2}$,the triangle is a right-angled triangle with the hypotenuse $c$ and $\angle C = 90^{\circ}$.By Heron's

Vedclass · Accepted Answer

${a^2}{b^2}$

Vedclass · Answer

(D) Given the condition ${c^2} = {a^2} + {b^2}$,the triangle is a right-angled triangle with the hypotenuse $c$ and $\angle C = 90^{\circ}$.By Heron's formula,the area of the triangle $\Delta$ is given by $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.Squaring both sides,we get $\Delta^2 = s(s - a)(s - b)(s - c)$,which implies $4\Delta^2 = 4s(s - a)(s - b)(s - c)$.For a right-angled triangle with legs $a$ and $b$,the area is $\Delta = \frac{1}{2}ab$.Substituting this into the expression,we get $4\Delta^2 = 4(\frac{1}{2}ab)^2 = 4(\frac{1}{4}a^2b^2) = {a^2}{b^2}$.

If ${c^2} = {a^2} + {b^2}$,then $4s(s - a)(s - b)(s - c) = $

Similar Questions

The sum of solutions of the equation $\frac{\cos x}{1+\sin x}=|\tan 2 x|$,where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) - \left\{\frac{\pi}{4}, -\frac{\pi}{4}\right\}$,is:

In a triangle $ABC$,$\angle B < \angle C$ and the values of $B$ and $C$ satisfy the equation $2 \tan x - k (1 + \tan^2 x) = 0$,where $0 < k < 1$. Then the measure of angle $A$ is:

In a $\triangle PQR$,$m \angle R = \frac{\pi}{2}$. If $\tan \left(\frac{P}{2}\right)$ and $\tan \left(\frac{Q}{2}\right)$ are the roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

If in a triangle $ABC$,$a^2+2bc-(b^2+c^2)=ab \sin \frac{C}{2} \cos \frac{C}{2}$,then $\cot (B+C)=$

In a triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module