Question

15 B!!! VERY EASY QUESTION!!!PLS HELP ME!!!
In right triangle $PQR$, we have $\angle Q = \angle R$ and $QR = 6\sqrt{2}$. What is the area of $\triangle PQR$?

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Answer 1

$\textbf{Given:}$

$\text{In right angled \triangle PQR, \angle{Q}=\angle{R} and QR=6 \sqrt{2} \;cm}$

$\textbf{To find:}$

$\text{Area of triangle PQR}$

$\textbf{Solution:}$

$\textbf{We know that, the sides opposite to equal angles are equal}$

$\implies\,PQ=PR=x(say)$

$\text{In right triangle PQR, by pythagoras theorem}$

$PQ^2+PR^2=QR^2$

$x^2+x^2=(6\sqrt{2})^2$

$2x^2=72$

$\implies\bf\,x^2=36$

$\text{Now,}$

$\textbf{Area of triangle PQR}$

$=\dfrac{1}{2}{\times}\text{Base}{\times}\text{Height}$

$=\dfrac{1}{2}{\times}\text{PR}{\times}\text{PQ}$

$=\dfrac{1}{2}{\times}x{\times}x$

$=\dfrac{1}{2}{\times}x^2$

$=\dfrac{1}{2}{\times}36$

$=18\;\text{Square cm.}$

$\therefore\textbf{Area of triangle PQR is 18 square cm.}$

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