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Question Number 79026 by TawaTawa last updated on 22/Jan/20

Commented by mathmax by abdo last updated on 22/Jan/20
$we consider the repere (D,DC^→ ,DA^→ ) ⇒D(0,0) ,C(1,0) A(0,1) (1=20cm) ⇒B(1,1) and F((1/2),1) coordonnate of I? M(x,y) ∈ (CF) ⇒det(CM^→ ,CF^→ )=0 ⇒ determinant (((x−1 −(1/2))),((y 1)))=0⇒ (x−1)+(1/2)y=0 ⇒2x−2+y =0 ⇒2x+y−2=0 equation of (BE) E(0,(1/2)) M(x,y) ∈(BE) ⇒det(BM^→ ,BE^→ )=0 ⇒ determinant (((x−1 −1)),((y−1 −(1/2))))=0 ⇒ −(1/2)(x−1)+y−1=0 ⇒(1/2)(x−1)−y+1=0 ⇒x−1−2y+2=0 ⇒ x−2y+1=0 let solve the systeme { ((2x+y−2=0)),((x−2y+1=0 )) :} ⇒ { ((y=−2x+2)),((x−2(−2x+2)+1=0 ⇒ { ((y=−2x+2)),((5x−3=0 ⇒)) :})) :} { ((x=(3/5) ⇒ { ((x=(3/5))),((y=(4/5))) :})),((y=−(6/5)+2 )) :} ⇒ I((3/5),(4/5)) A(BIF)=(1/2)∣det(IF^→ ,IB^→ ) we have IF^→ ((1/2)−(3/5),1−(4/5)) ⇒IF^→ (−(1/(10)),(1/5)) and IB^→ (1−(3/5),1−(4/5)) ⇒IB^→ ((2/5),(1/5)) ⇒ A(BIF) =(1/2)∣ determinant (((−(1/(10)) (2/5))),(((1/5) (1/5))))∣=(1/2)∣(−(1/(50))−(2/(25)))∣ =(1/2)×(5/(50)) =(1/(20)) we have 1 =20 cm ⇒1^2 =20^2 cm^2 ⇒ A(BIF) =(1/(20))×20^2 =20 cm^2 .$
$w e c o n s i d e r t h e r e p e r e (D, D \vec{C}, D \vec{A}) \Rightarrow D (0, 0), C (1, 0) A (0, 1)$ $(1 = 20 c m) \Rightarrow B (1, 1) a n d F (\frac{1}{2}, 1) c o o r d o n n a t e o f I ?$ $M (x, y) \in (C F) \Rightarrow d e t (C \vec{M}, C \vec{F}) = 0 \Rightarrow \| \begin{matrix} x - 1 - \frac{1}{2} \\ y 1 \end{matrix} \| = 0 \Rightarrow$ $(x - 1) + \frac{1}{2} y = 0 \Rightarrow 2 x - 2 + y = 0 \Rightarrow 2 x + y - 2 = 0$ $e q u a t i o n o f (B E) E (0, \frac{1}{2})$ $M (x, y) \in (B E) \Rightarrow d e t (B \vec{M}, B \vec{E}) = 0 \Rightarrow \| \begin{matrix} x - 1 - 1 \\ y - 1 - \frac{1}{2} \end{matrix} \| = 0 \Rightarrow$ $- \frac{1}{2} (x - 1) + y - 1 = 0 \Rightarrow \frac{1}{2} (x - 1) - y + 1 = 0 \Rightarrow x - 1 - 2 y + 2 = 0 \Rightarrow$ $x - 2 y + 1 = 0 l e t s o l v e t h e s y s t e m e {\begin{cases} 2 x + y - 2 = 0 \\ x - 2 y + 1 = 0 \end{cases}$ $\Rightarrow {\begin{cases} y = - 2 x + 2 \\ x - 2 (- 2 x + 2) + 1 = 0 \Rightarrow {\begin{cases} y = - 2 x + 2 \\ 5 x - 3 = 0 \Rightarrow \end{cases} \end{cases}$ ${\begin{cases} x = \frac{3}{5} \Rightarrow {\begin{cases} x = \frac{3}{5} \\ y = \frac{4}{5} \end{cases} \\ y = - \frac{6}{5} + 2 \end{cases}$ $\Rightarrow I (\frac{3}{5}, \frac{4}{5})$ $A (B I F) = \frac{1}{2} ∣ d e t (I \vec{F}, I \vec{B}) w e h a v e I \vec{F} (\frac{1}{2} - \frac{3}{5}, 1 - \frac{4}{5})$ $\Rightarrow I \vec{F} (- \frac{1}{10}, \frac{1}{5}) a n d I \vec{B} (1 - \frac{3}{5}, 1 - \frac{4}{5}) \Rightarrow I \vec{B} (\frac{2}{5}, \frac{1}{5}) \Rightarrow$ $A (B I F) = \frac{1}{2} ∣ \| \begin{matrix} - \frac{1}{10} \frac{2}{5} \\ \frac{1}{5} \frac{1}{5} \end{matrix} \| ∣= \frac{1}{2} ∣ (- \frac{1}{50} - \frac{2}{25}) ∣$ $= \frac{1}{2} \times \frac{5}{50} = \frac{1}{20} w e h a v e 1 = 20 c m \Rightarrow 1^{2} = 20^{2} {c m}^{2} \Rightarrow$ $A (B I F) = \frac{1}{20} \times 20^{2} = 20 {c m}^{2} .$
Commented by TawaTawa last updated on 22/Jan/20

$God bless you sir$
Commented by msup trace by abdo last updated on 22/Jan/20

$y o u a r e w e l c o m e$
Commented by jagoll last updated on 22/Jan/20

$let area △ BIF = x, area AEB = \frac{1}{2} \times 200 = 100$ $EB = \sqrt{100 + 400} = 10 \sqrt{5}$ $now we have$ $\frac{area △ BIF}{area △ AEB} = {(\frac{10}{10 \sqrt{5}})}^{2}$ $are △ BIF = \frac{100}{5} = 20 {cm}^{2}$
Commented by TawaTawa last updated on 23/Jan/20

$God bless you sir$
	Answered by mind is power last updated on 22/Jan/20
	$first Idea let D(0,0),C(20,0),B(20,20),A(0,20) E(0,10),F(10,20) (EB):y=ax+b a=((10)/(20))=(1/2) y=(x/2)+b,⇒b=20−((20)/2)=10 y=(x/2)+10 CF:y=ax+b,a=−((20)/(10))=−2 y=−2x+b,b=40,y=−2x+40 I=(EB)∩(CF) { ((y=−2x+40)),((y=(x/2)+10)) :} ⇒−2x+40=(x/2)+10 (5/2)x=30⇒x=12,y=16 I(12,16) Area IFB=(1/2)∣Det(IF^→ ,IB^→ )∣ IF^→ =(−2,4),IB^→ (8,4) determinant (((−2 8)),(( 4 4)))=40 Area=((40)/2)=20$
	$f i r s t I d e a$ $l e t D (0, 0), C (20, 0), B (20, 20), A (0, 20)$ $E (0, 10), F (10, 20)$ $(E B) : y = a x + b a = \frac{10}{20} = \frac{1}{2}$ $y = \frac{x}{2} + b, \Rightarrow b = 20 - \frac{20}{2} = 10$ $y = \frac{x}{2} + 10$ $C F : y = a x + b, a = - \frac{20}{10} = - 2$ $y = - 2 x + b, b = 40, y = - 2 x + 40$ $I = (E B) \cap (C F) {\begin{cases} y = - 2 x + 40 \\ y = \frac{x}{2} + 10 \end{cases}$ $\Rightarrow - 2 x + 40 = \frac{x}{2} + 10$ $\frac{5}{2} x = 30 \Rightarrow x = 12, y = 16$ $I (12, 16)$ $A r e a I F B = \frac{1}{2} ∣ D e t (I \vec{F}, I \vec{B}) ∣$ $I \vec{F} = (- 2, 4), I \vec{B} (8, 4)$ $\| \begin{matrix} - 2 8 \\ 4 4 \end{matrix} \| = 40$ $A r e a = \frac{40}{2} = 20$
	Commented by TawaTawa last updated on 22/Jan/20

	$God bless you sir$
	Answered by mr W last updated on 22/Jan/20

	Commented by mr W last updated on 22/Jan/20

	$F G = \frac{A E}{2} = \frac{A D}{4} = \frac{B C}{4}$ $\frac{H B}{F B - H B} = \frac{H B}{F H} = \frac{B C}{F G} = 4$ $\Rightarrow H B = \frac{4}{5} F B = I K$ $\Rightarrow Δ_{I B C} = \frac{4}{5} Δ_{F B C}$ $\Rightarrow Δ_{B I F} = Δ_{F B C} - Δ_{I B C} = \frac{1}{5} Δ_{F B C} = \frac{1}{5} \times \frac{[A B C D]}{4}$ $= \frac{20 \times 20}{20} = 20 {c m}^{2}$
	Commented by TawaTawa last updated on 22/Jan/20

	$God bless you sir$
	Answered by john santu last updated on 22/Jan/20

	Commented by TawaTawa last updated on 22/Jan/20

	$God bless you sir$
	Answered by mr W last updated on 22/Jan/20

	$a n o t h e r w a y :$ ${i t}^{'} s c l e a r t h a t F C ⊥ E B .$ $E B = \sqrt{10^{2} + 20^{2}} = 10 \sqrt{5}$ $\frac{I B}{F B} = \frac{A B}{E B}$ $\Rightarrow I B = \frac{20}{10 \sqrt{5}} \times 10 = \frac{20}{\sqrt{5}}$ $\frac{F I}{F B} = \frac{A E}{E B}$ $\Rightarrow F I = \frac{10}{10 \sqrt{5}} \times 10 = \frac{10}{\sqrt{5}}$ $Δ_{B I F} = \frac{1}{2} \times I B \times F I = \frac{1}{2} \times \frac{20}{\sqrt{5}} \times \frac{10}{\sqrt{5}} = 20 {c m}^{2}$
	Commented by TawaTawa last updated on 22/Jan/20

	$God bless you sir . I appreciate$
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