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-

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