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Question Number 203576 by sonukgindia last updated on 22/Jan/24

	Answered by mr W last updated on 22/Jan/24

	Commented by mr W last updated on 22/Jan/24

	$b o t h h a t c h e d t r i a n g l e s a r e s i m i l a r .$ $\frac{a}{0.5} = \frac{1}{b}$ $\Rightarrow a b = 0.5$ $a r e a o f r e c t a n g l e = a b = 0.5 ✓$
	Answered by ajfour last updated on 22/Jan/24

	Commented by ajfour last updated on 22/Jan/24
	$B^( 2) =(p^2 +q^2 ){1+(1−q)^2 } tan θ=t=1−q=(p/q)=(1/(2(1−p))) ⇒ p=1−(1/(2t)) , q=1−t , and 1−(1/(2t))=t(1−t) ⇒ t^3 −t^2 +t−(1/2)=0 ⇒ t^2 (t−1)+(t−1)=−(1/2) or (1−t)(t^2 +1)=(1/2) B^( 2) =q^2 (t^2 +1){1+t^2 ) B =(1−t)(1+t^2 ) B =(1/2)$
	$B^{2} = (p^{2} + q^{2}) {1 + {(1 - q)}^{2}}$ $\tan θ = t = 1 - q = \frac{p}{q} = \frac{1}{2 (1 - p)}$ $\Rightarrow p = 1 - \frac{1}{2 t}, q = 1 - t, a n d$ $1 - \frac{1}{2 t} = t (1 - t)$ $\Rightarrow t^{3} - t^{2} + t - \frac{1}{2} = 0$ $\Rightarrow t^{2} (t - 1) + (t - 1) = - \frac{1}{2}$ $o r (1 - t) (t^{2} + 1) = \frac{1}{2}$ $B^{2} = q^{2} (t^{2} + 1) {1 + t^{2})$ $B = (1 - t) (1 + t^{2})$ $B = \frac{1}{2}$
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