Question-28574

Question Number 28574 by ajfour last updated on 27/Jan/18

Commented by ajfour last updated on 27/Jan/18

F i n d e n t i r e g r e e n a r e a i n t e r m s

o f r, θ . T h e t w o f r a m e s a r e

s q u a r e f r a m e s w i t h c o m m o n

c e n t r e . F i n d a l s o t h e m a x i m u m

o f t h e a r e a f o r c o n s t a n t r .

Commented by ajfour last updated on 29/Jan/18

m r W 2 S i r, w o u l d y o u s o l v e t h i s ?

Answered by mrW2 last updated on 29/Jan/18

Commented by ajfour last updated on 29/Jan/18

t h a n k y o u e n o r m o u s l y S i r!

e x t r e m e l y s a t i s f a c t o r y!

Commented by mrW2 last updated on 29/Jan/18

E q n . o f A B :

x = \frac{a}{2} w i t h a = \sqrt{2} r

E q n . o f C D (= A B r o t a t e d b y θ)

x \cos θ + y \sin θ = \frac{a}{2}

\frac{a}{2} \cos θ + y_{E} \sin θ = \frac{a}{2}

\Rightarrow y_{E} = \frac{a (1 - \cos θ)}{2 \sin θ}

x_{F} \cos θ + \frac{a}{2} \sin θ = \frac{a}{2}

\Rightarrow x_{F} = \frac{a (1 - \sin θ)}{2 \cos θ}

A_{Δ B E F} = \frac{1}{2} [\frac{a}{2} - \frac{a (1 - \sin θ)}{2 \cos θ}] [\frac{a}{2} - \frac{a (1 - \cos θ)}{2 \sin θ}]

A_{Δ B E F} = \frac{a^{2}}{8} [\frac{\cos θ - 1 + \sin θ}{\cos θ}] [\frac{\sin θ - 1 + \cos θ}{\sin θ}]

A_{Δ B E F} = \frac{a^{2}}{8} \times \frac{{(\cos θ + \sin θ - 1)}^{2}}{\sin θ \cos θ}

A_{Δ B E F} = \frac{a^{2}}{4} [1 - \frac{\cos θ + \sin θ - 1}{\sin θ \cos θ}]

A_{G r e e n} = 4 A_{Δ B E F}

A_{G r e e n} (θ) = a^{2} [1 - \frac{\cos θ + \sin θ - 1}{\sin θ \cos θ}]

A_{G r e e n} (θ) = 2 r^{2} [1 - \frac{\cos θ + \sin θ - 1}{\sin θ \cos θ}]

w i t h 0 ⩽ θ ⩽ \frac{π}{2}

A_{G r e e n} (\frac{π}{2} + θ) = A_{G r e e n} (θ)

f (θ) = 1 - \frac{\cos θ + \sin θ - 1}{\sin θ \cos θ}

f (θ) = 1 + \frac{2 [1 - \sqrt{2} \sin (θ + \frac{π}{4})]}{\sin 2 θ}

m a x . f (θ) = 1 + \frac{2 [1 - \sqrt{2}]}{1} = 3 - 2 \sqrt{2} \approx 0.17

a t θ = \frac{π}{4}

\Rightarrow m a x . A_{G r e e n} = (3 - 2 \sqrt{2}) a^{2} \approx 0.17 a^{2} \approx 0.34 r^{2}

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