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Question Number 201599 by mr W last updated on 09/Dec/23

Commented by deleteduser1 last updated on 09/Dec/23

$A r e t h o s e n u m b e r s a r e a s o r l e n g t h s ?$
Commented by mr W last updated on 09/Dec/23

$t h e y a r e l e n g t h s .$
	Answered by mr W last updated on 10/Dec/23

	Commented by mr W last updated on 10/Dec/23

	$3 x = y + z$ $3 x + \frac{20 - p}{p} y = 3 \times \frac{20 - p}{3 p} z$ $\Rightarrow y + z = \frac{(20 - p) (z - y)}{p}$ $x + \frac{20 - p}{3 p} z = \frac{20 - p}{p} y$ $\Rightarrow y + z = \frac{(20 - p) (3 y - z)}{p}$ $z - y = 3 y - z \Rightarrow z = 2 y \Rightarrow x = y$ $y + 2 y = \frac{(20 - p) (2 y - y)}{p}$ $\Rightarrow 3 = \frac{20 - p}{p} \Rightarrow p = 5$ $\frac{B F}{F A} = \frac{z}{y} = 2$ $\frac{{B F}^{2} + 5^{2} - 9^{2}}{2 \times 5 \times B F} = - \frac{{F A}^{2} + 5^{2} - 6^{2}}{2 \times 5 \times F A}$ ${B F}^{2} = - 2 \times {F A}^{2} + 78$ ${F A}^{2} = 13$ $\Rightarrow F A = \sqrt{13} \Rightarrow B A = 3 \sqrt{13}$ ${B A}^{2} = 9 \times 13 = 117 = 9^{2} + 6^{2}$ $\Rightarrow A D ⊥ B E$ $x = \frac{3 \times 6}{2} = 9$ $Δ_{A B C} = \frac{20}{p} (y + z) = 12 x = 12 \times 9 = 108 ✓$
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