If the area of a convex quadrilateral is 2k^2 and the sum of its diagonals is 4k^2 , then show that this quadrilateral is an orthodiagonal one.


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Question Number 151907 by mathdanisur last updated on 24/Aug/21

$If the area of a convex quadrilateral$ $is 2 k^{2} and the sum of its diagonals$ $is 4 k^{2}, then show that this quadrilateral$ $is an orthodiagonal one .$
Commented by mr W last updated on 24/Aug/21

$p l e a s e c h e c k t h e q u e s t i o n!$ $y o u c a n n o t c o m p a r e t h e a r e a w i t h$ $t h e s u m o f d i a g o n a l s! a s y o u c a n n o t$ $s a y 2 {c m}^{2} i s e q u a l t o 2 c m .$
Commented by mathdanisur last updated on 24/Aug/21

$Sorry S er, 4 k$
	Answered by mr W last updated on 24/Aug/21

	Commented by mr W last updated on 25/Aug/21

	$s a y A C = a, B D = b$ $a r e a = \frac{{a h}_{1}}{2} + \frac{{a h}_{2}}{2} = \frac{a b \sin θ}{2} = 2 k^{2}$ $\Rightarrow a b = \frac{4 k^{2}}{\sin θ}$ $a + b = 4 k$ $a, b a r e r o o t s o f :$ $x^{2} - 4 k x + \frac{4 k^{2}}{\sin θ} = 0$ $Δ = 16 k^{2} - \frac{16 k^{2}}{\sin^{2} θ} ⩾ 0$ $1 ⩾ \frac{1}{\sin^{2} θ}$ $\Rightarrow \sin^{2} θ ⩾ 1$ $s i n c e \sin θ ⩽ 1, i . e . \sin^{2} θ ⩽ 1$ $\Rightarrow t h e o n l y p o s s i b i l i t y i s \sin θ = 1,$ $i . e . θ = 90 °$
	Commented by mathdanisur last updated on 25/Aug/21

	$T h a n k y o u S e r$
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