Question

In right triangle $PQR$, we have $\angle Q = \angle R$ and $PR = 6\sqrt{2}$. What is the area of $\triangle PQR$?

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

In right triangle PQR, ∠Q = ∠R.

It means, ∠P = 90°

We know, Side opposite to equal angles are equal

So, it means PQ = PR

As it is given that

$\rm \: PR = 6 \sqrt{2}$

$\rm\implies \:PQ = PR = 6 \sqrt{2}$

We know, Area of right angle triangle is given by

$\boxed{ \rm{ \:Area \: = \: \frac{1}{2} \times base \times height \: }}$

So,

$\rm \: Area_{(\triangle\:PQR)} \: = \: \dfrac{1}{2} \times PQ \times PR$

$\rm \: Area_{(\triangle\:PQR)} \: = \: \dfrac{1}{2} \times 6 \sqrt{2} \times 6 \sqrt{2}$

$\rm\implies \:Area_{(\triangle\:PQR)} = 36 \: square \: units$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$