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Question Number 50640 by ajfour last updated on 18/Dec/18

Commented by mr W last updated on 18/Dec/18

$w h e n t h e a n g l e b e t w e e n k a n d h i s$ $f i x e d, t h e n t h e s h a p e a n d t h e r e f o r e$ $a l s o t h e a r e a o f t h e q u a d r i l a t e r a l i s$ $f i x e d . t h e a r e a i s a c o n s t a n t .$ $c = \sqrt{h^{2} + k^{2}}$ $A = \frac{h k}{2} + \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}$
Commented by ajfour last updated on 18/Dec/18

$y e s m r W S i r, s o r r y f o r t h e$ $u n n e c e s a r y, t h a n k s y e t .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18
	$sinθ=(l/a) area shaded =(1/2)(k+l)×(h+acosθ)−(1/2)(acosθ)×(asinθ) =(1/2)(k+asinθ)(h+acosθ)−(a^2 /4)sin2θ =(1/2)(kh+akcosθ+ahsinθ+(a^2 /2)sin2θ)−(a^2 /4)sin2θ =((kh)/2)+(a/2)×(√(k^2 +h^2 )) {(k/(√(k^2 +h^2 )))cosθ+(h/(√(k^2 +h^2 )))sinθ} =((kh)/2)+((a(√(k^2 +h^2 )))/2)sin(θ+α) so max area =((kh)/2)+((a(√(k^2 +h^2 )))/2) pls check...$
	$s i n θ = \frac{l}{a}$ $a r e a s h a d e d = \frac{1}{2} (k + l) \times (h + a c o s θ) - \frac{1}{2} (a c o s θ) \times (a s i n θ)$ $= \frac{1}{2} (k + a s i n θ) (h + a c o s θ) - \frac{a^{2}}{4} s i n 2 θ$ $= \frac{1}{2} (k h + a k c o s θ + a h s i n θ + \frac{a^{2}}{2} s i n 2 θ) - \frac{a^{2}}{4} s i n 2 θ$ $= \frac{k h}{2} + \frac{a}{2} \times \sqrt{k^{2} + h^{2}} {\frac{k}{\sqrt{k^{2} + h^{2}}} c o s θ + \frac{h}{\sqrt{k^{2} + h^{2}}} s i n θ}$ $= \frac{k h}{2} + \frac{a \sqrt{k^{2} + h^{2}}}{2} s i n (θ + α)$ $s o m a x a r e a$ $= \frac{k h}{2} + \frac{a \sqrt{k^{2} + h^{2}}}{2}$ $p l s c h e c k . . .$
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