Question-153901

Question Number 153901 by mr W last updated on 11/Sep/21

Commented by mr W last updated on 12/Sep/21

1) f i n d {(a b)}_{m a x} = ?

2) p r o v e t h a t S_{1} = S_{2} .

Answered by EDWIN88 last updated on 12/Sep/21

Commented by mr W last updated on 12/Sep/21

t h a n k s a l o t s i r!

Commented by Tawa11 last updated on 12/Sep/21

Weldone sir .

Answered by mr W last updated on 12/Sep/21

Commented by mr W last updated on 12/Sep/21

(1)

α + β = 45

\frac{\tan α + \tan α}{1 - \tan α \tan β} = 1

\frac{1}{\tan α} + \frac{1}{\tan β} = \frac{1}{\tan α} \times \frac{1}{\tan β} - 1

\Rightarrow \frac{1}{\tan β} = \frac{\frac{1}{\tan α} + 1}{\frac{1}{\tan α} - 1}

\frac{\sin (α + 45)}{\sin α} = \frac{l}{a}

\frac{1}{\tan α} + 1 = \frac{\sqrt{2} l}{a} \dots (i)

s i m i l a r l y

\frac{1}{\tan β} + 1 = \frac{\sqrt{2} l}{b} \dots (i i)

\frac{2 l^{2}}{a b} = (\frac{1}{\tan α} + 1) (\frac{1}{\tan β} + 1)

= 2 (\frac{1}{\tan α} + 1) (\frac{\frac{1}{\tan α}}{\frac{1}{\tan α} - 1})

l e t t = \frac{1}{\tan α}

\Rightarrow P = \frac{l^{2}}{a b} = \frac{t (t + 1)}{t - 1}

\frac{d P}{d t} = \frac{2 t + 1}{t - 1} - \frac{t (t + 1)}{{(t - 1)}^{2}} = 0

t^{2} - 2 t - 1 = 0

\Rightarrow t = \frac{1}{\tan α} = 1 + \sqrt{2}

\Rightarrow \tan α = \sqrt{2} - 1 \Rightarrow α = 22.5 ° = β

\Rightarrow \tan β = \sqrt{2} - 1

\frac{2 l^{2}}{{(a b)}_{m a x}} = {(2 + \sqrt{2})}^{2}

\Rightarrow {(a b)}_{m a x} = {(\sqrt{2} - 1)}^{2} l^{2} \approx 0.171 l^{2}

Commented by mr W last updated on 12/Sep/21

(2)

\frac{L}{c} = \frac{\sin (45 + α)}{\sin 45} = \frac{\sin (45 + 45 - β)}{\sin 45} = \sqrt{2} \cos β

\frac{L}{d} = \frac{\sin (45 + β)}{\sin 45} = \frac{\sin (45 + 45 - α)}{\sin 45} = \sqrt{2} \cos α

\frac{L^{2}}{c d} = 2 \cos α \cos β

\Rightarrow c d = \frac{L^{2}}{2 \cos α \cos β}

e = \frac{L}{\cos α}

f = \frac{L}{\cos β}

\Rightarrow e f = \frac{L^{2}}{\cos α \cos β} = 2 c d

S_{1} = \frac{c d \sin 45 °}{2}

S_{1} + S_{2} = \frac{e f \sin 45 °}{2} = 2 S_{1}

\Rightarrow S_{1} = S_{2}

Commented by liberty last updated on 12/Sep/21

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