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Question Number 211079 by Ari last updated on 27/Aug/24

	Answered by A5T last updated on 27/Aug/24

	$S u p p o s e D B ⊥ A C$ $A_{2} = \frac{D B \times A B}{2} = 63; A_{1} = \frac{E A \times A B}{2} = 144$ $\frac{A_{2}}{A_{1}} = \frac{63}{144} = \frac{D B}{E A} = \frac{7}{16} \Rightarrow \frac{A_{3}}{A_{1} + A_{2} + A_{3}} = {(\frac{7}{16})}^{2}$ $\Rightarrow A_{3} = 49 u^{2}$
	Commented by Ari last updated on 27/Aug/24

	$b u t D B i s n o t g i v e n p e r p e n d i c u l a r t o A C$
	Commented by A5T last updated on 27/Aug/24

	$T h e n i t c a n n o t b e u n i q u e o r i t {w o n}^{'} t b e$ $d e p e n d e n t o n i t b e i n g p e r p e n d i c u l a r$
	Commented by Ari last updated on 27/Aug/24

	$O k, y o u a r e r i g h t!$
	Commented by mm1342 last updated on 27/Aug/24

	$i f △_{1} \approx △_{2} \Rightarrow \frac{s_{1}}{s_{2}} = k^{2}$ $\Rightarrow \frac{A_{3}}{A_{1} + A_{2} + A_{3}} = {(\frac{D B}{E A})}^{2} = \frac{49}{256}$ $\Rightarrow A_{3} = 49 ✓$
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