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Question Number 25367 by ajfour last updated on 08/Dec/17

Commented by ajfour last updated on 08/Dec/17

$F i n d a r e a o f △ P Q R .$
	Answered by jota+ last updated on 09/Dec/17

	$l e t ∡ Q B O = ∡ A B Q = θ$ $p o i n t Q . Q (0, r \tan θ)$ $p o i n t P . M (- r / 2, q / 2)$ $e q u a t i o n o f M C$ $y = \frac{q / 2 - 0}{- r / 2 - p} (x - p) . . . (1)$ $P (0, \frac{p q}{r + 2 p})$ $p o i n t R . R = B Q ∤ M C$ $e q u a t i o n o f B Q$ $y = \tan θ (x + r) . . . (2)$ $r e s o l v i n g (1) a n d (2)$ $R (\frac{y_{P} - y_{Q}}{\tan θ + m}, \frac{y_{P} - y_{Q}}{1 + m / \tan θ} + y_{Q})$ $w h e r e m = \frac{q}{r + 2 p}$ $A r e a = \frac{a b s [x_{R} (y_{Q} - y_{P})]}{2}$ $= \frac{{(y_{Q} - y_{P})}^{2}}{2 (\tan θ + m)} .$
	Commented by ajfour last updated on 10/Dec/17

	$t h a n k s .$
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