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Question Number 188177 by cortano12 last updated on 26/Feb/23

Commented by cortano12 last updated on 26/Feb/23

$[BCDH] = {6 m}^{2}$ $[ABI] = {4 m}^{2}$ $[DGE] = {2 m}^{2}$ $Find [AIF]$
Commented by cortano12 last updated on 26/Feb/23

$the problem possible or impossible ?$
	Answered by mr W last updated on 26/Feb/23

	Commented by cortano12 last updated on 26/Feb/23

	$(1) 10 + C = 8 + D \Rightarrow D - C = 2$ $(2) 4 + C + 2 + D = 6 \Rightarrow C + D = 0 ?$
	Commented by mr W last updated on 26/Feb/23

	$x + 4 = A + 2$ $\Rightarrow A = x + 2$ $x + 4 = 2 + D + 6$ $\Rightarrow D = x - 4$ $x + B + D = 2 (x + 4)$ $\Rightarrow B = x + 8 - D = x + 8 - (x - 4) = 12$ $x + 4 = 4 + C + 6$ $\Rightarrow C = x - 6$ $\frac{B_{1}}{A} = \frac{D}{2} \Rightarrow B_{1} = \frac{A \times D}{2} = \frac{(x + 2) (x - 4)}{2}$ $\frac{B_{2}}{x} = \frac{C}{4} \Rightarrow B_{2} = \frac{C \times x}{4} = \frac{(x - 6) x}{4}$ $B_{1} + B_{2} = B = 12$ $\frac{(x + 2) (x - 4)}{2} + \frac{(x - 6) x}{4} = 12$ $3 x^{2} - 10 x - 64 = 0$ $x = \frac{5 + \sqrt{25 + 3 \times 64}}{3} = \frac{5 + \sqrt{217}}{3} \approx 6.577$
	Commented by cortano12 last updated on 26/Feb/23

	$what wrong my solution ?$ $D - C = 2$ $D + C = 0 \Rightarrow D = 1 and C = - 1 ?$
	Commented by cortano12 last updated on 26/Feb/23

	$the answer x = 7$
	Commented by cortano12 last updated on 26/Feb/23

	$Let [BCH] = u = 4 + C$ $let [CDH] = v = 2 + D$ $then u + v = 6 \Rightarrow D + C = ?$ $i think impossible$
	Commented by mr W last updated on 26/Feb/23

	$w e s h o u l d s e l e c t [A B I] + [D G E] \neq [B C D H]$
	Commented by cortano12 last updated on 26/Feb/23

	$why ?$
	Commented by cortano12 last updated on 26/Feb/23

	$[ABH] = [BCH], since AB = BC$ $[CDH] = [DHE], since CD = DE$ $so [BCDH] = [BCH] + [CHD]$ $6 = 4 + C + 2 + D$
	Commented by mr W last updated on 26/Feb/23

	Commented by mr W last updated on 26/Feb/23

	$e q n . o f A D :$ $y = \frac{x b}{2 a}$ $e q n . o f B F :$ $y = - \frac{2 b}{a} (x - \frac{a}{2})$ $e q n . o f B E :$ $y = \frac{2 b}{a} (x - \frac{a}{2})$ $e q n . o f D F :$ $y = b - \frac{x b}{2 a}$ $i n t e r s e c t i o n p o i n t I :$ $y_{I} = \frac{x_{I} b}{2 a} = - \frac{2 b}{a} (x_{I} - \frac{a}{2})$ $\Rightarrow x_{I} = \frac{2 a}{5}$ $\Rightarrow y_{I} = \frac{b}{5}$ $a r e a X = \frac{a}{4} \times \frac{b}{5} = \frac{a b}{20}$ $i n t e r s e c t i o n p o i n t G :$ $y_{G} = \frac{2 b}{a} (x_{G} - \frac{a}{2}) = b - \frac{x_{G} b}{2 a}$ $\Rightarrow x_{G} = \frac{4 a}{5}$ $\Rightarrow y_{G} = \frac{3 b}{5}$ $a r e a Y = \frac{b}{4} \times (a - \frac{4 a}{5}) = \frac{a b}{20} = X$ $i n t e r s e c t i o n p o i n t H :$ $y_{H} = \frac{x_{H} b}{2 a} = \frac{2 b}{a} (x_{H} - \frac{a}{2})$ $\Rightarrow x_{H} = \frac{2 a}{3}$ $\Rightarrow y_{H} = \frac{b}{3}$ $a r e a Z = \frac{a}{4} \times \frac{b}{3} + \frac{b}{4} \times (a - \frac{2 a}{3}) = \frac{a b}{6} > X + Y$ $w e s e e :$ $a r e a X = a r e a Y$ $a r e a Z \neq a r e a X + a r e a Y$
	Commented by mr W last updated on 26/Feb/23

	$g i v e n i n q u e s t i o n :$ $X = 4, Y = 2 \Rightarrow X \neq Y \Rightarrow w r o n g!$ $Z = 6 \Rightarrow Z = X + Y \Rightarrow w r o n g!$
	Commented by mr W last updated on 26/Feb/23

	$f o l l o w i n g i s e . g . s o l v a b l e :$
	Commented by mr W last updated on 26/Feb/23

	Answered by horsebrand11 last updated on 26/Feb/23

	$i t s h o u l d b e [B C H D] = 10$ $\Rightarrow 4 + C + 2 + D = 10; C + D = 4$ $\Rightarrow 14 + C = 12 + D \Rightarrow D - C = 2$ $t h e n {\begin{cases} D = 3 \\ C = 1 \end{cases}, i t c l e a r [A F I] = 11$
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