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Question Number 40637 by Tawa1 last updated on 25/Jul/18

	Answered by MJS last updated on 25/Jul/18

	$l_{1} : y = \frac{x}{3} + \frac{a}{3}$ $l_{2} : y = 2 x$ $l_{3} : y = a$ $l_{1} \cap l_{2}$ $\frac{x}{3} + \frac{a}{3} = 2 x \Rightarrow x = \frac{a}{5}; y = \frac{2 a}{5}$ $pink area$ $\underset{0}{\int^{\frac{a}{5}}} l_{2} d x + \underset{\frac{a}{5}}{\int^{a}} l_{1} d x = {[x^{2}]}_{0}^{\frac{a}{5}} + {[\frac{x^{2}}{6} + \frac{a x}{3}]}_{\frac{a}{5}}^{a} = \frac{7 a^{2}}{15}$ $grey area$ $\underset{0}{\int^{\frac{a}{5}}} (l_{3} - l_{1}) d x + \underset{\frac{a}{5}}{\int^{\frac{a}{2}}} (l_{3} - l_{2}) d x = {[\frac{2 a x}{3} - \frac{x^{2}}{6}]}_{0}^{\frac{a}{5}} + {[a x - x^{2}]}_{\frac{a}{5}}^{\frac{a}{2}} = \frac{13 a^{2}}{60}$ $\frac{13 a^{2}}{60} = 13 \Rightarrow a = 2 \sqrt{15} \Rightarrow \frac{7 a^{2}}{15} = 28$
	Commented by Tawa1 last updated on 25/Jul/18

	$God bless you sir .$
	Answered by MrW3 last updated on 26/Jul/18

	$l e t$ $S = a r e a o f s q u a r e$ $G = a r e a o f g r e y r e g i o n = 13$ $P = a r e a o f p i n k r e g i o n$ $T = a r e a o f t h e s m a l l t r i a n g l e b e t w e e n G a n d P$ $G + T = \frac{S}{4} . . . (i)$ $P + T = \frac{S}{2} . . . (i i)$ $2 \times (i) - (i i) :$ $\Rightarrow P = 2 G + T$ $\frac{T}{G + T} = \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}$ $\Rightarrow T = \frac{2}{13} G$ $\Rightarrow P = 2 G + T = \frac{28}{13} G = 28$
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