Question-47407

Question Number 47407 by ajfour last updated on 09/Nov/18

Answered by ajfour last updated on 10/Nov/18

l e t B D = x, ∠ A = ϕ, ∠ C = θ

S_{A B C D} = \frac{a b \sin θ}{2} + \frac{c d \sin ϕ}{2}

x^{2} = a^{2} + b^{2} - 2 a b \cos θ

= c^{2} + d^{2} - 2 c d \cos ϕ

\Rightarrow (a b \sin θ) \frac{d θ}{d x} = (c d \sin ϕ) \frac{d ϕ}{d x} = 2 x

\frac{d S}{d x} = 0 \Rightarrow

(a b \cos θ) \frac{d θ}{d x} + (c d \cos ϕ) \frac{d ϕ}{d x} = 0

\Rightarrow \cos θ + (\cos ϕ) (\frac{\sin θ}{\sin ϕ}) = 0

\Rightarrow θ + ϕ = π

\Rightarrow a^{2} + b^{2} - 2 a b \cos θ

= c^{2} + d^{2} + 2 c d \cos θ

o r \cos θ = \frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b + c d)}

{(S_{A B C D})}_{m a x} = \frac{\sin θ}{2} (a b + c d)

= \frac{(a b + c d)}{2} \sqrt{1 - {[\frac{a^{2} + b^{2} - c^{2} - d^{2}}{2 (a b + c d)}]}^{2}}

S_{m a x} = \frac{1}{4} \sqrt{4 {(a b + c d)}^{2} - {[(a^{2} + b^{2}) - (c^{2} + d^{2})]}^{2}} .

Commented by mr W last updated on 10/Nov/18

t h a n k y o u s i r!

i t s h o w s t h e a r e a r e a c h e s m a x i m u m

w h e n t h e q u a d r i l a t e r a l i s i n s c r i b e d i n

a c i r c l e .

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