JEE MAIN - Mathematics (2022 - 26th June Morning Shift - No. 9)
Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $$\Delta$$PQR is :
$${{25} \over {4\sqrt 3 }}$$
$${{25\sqrt 3 } \over 2}$$
$${{25} \over {\sqrt 3 }}$$
$${{25} \over {2\sqrt 3 }}$$
Explanation
_26th_June_Morning_Shift_en_9_1.png)
Let, side of triangle = a.
$$h = {{|3 + 7 - 5|} \over {\sqrt {{1^2} + {1^2}} }}$$
$$ = {5 \over {\sqrt 2 }}$$
From figure, $$h = a\sin 60^\circ $$
$$ \Rightarrow a = {{2h} \over {\sqrt 3 }}$$
$$ = {2 \over {\sqrt 3 }} \times {5 \over {\sqrt 2 }}$$
$$ = {{10} \over {\sqrt 6 }}$$
$$\therefore$$ Area $$ = {3 \over 4}{\left( {{{10} \over {\sqrt 6 }}} \right)^2}$$
$$ = {{25} \over {2\sqrt 3 }}$$
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