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Question Number 156316 by cortano last updated on 10/Oct/21

Commented by cortano last updated on 10/Oct/21
$AD=DE=EB , AF : FG : GC=1:2:1 what fraction of ΔABC is shaded?$
$AD = DE = EB, AF : FG : GC = 1 : 2 : 1$ $what fraction of Δ ABC is shaded ?$
	Answered by mr W last updated on 10/Oct/21

	Commented by mr W last updated on 10/Oct/21

	$s a y a r e a o f Δ A B C i s A .$ $\frac{C G}{A C} = \frac{1}{4}$ $\Rightarrow \frac{A_{1}}{A} = \frac{1}{4}$ $\frac{Δ B G A}{A} = \frac{3}{4}$ $\frac{B E}{B A} = \frac{1}{3}$ $\Rightarrow \frac{A_{2}}{Δ B G A} = \frac{1}{3}$ $\Rightarrow \frac{A_{2}}{A} = \frac{1}{3} \times \frac{3}{4} = \frac{1}{4}$ $\frac{Δ G E A}{Δ B G A} = \frac{2}{3}$ $\Rightarrow \frac{Δ G E A}{A} = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$ $\frac{E D}{E A} = \frac{1}{2}$ $\Rightarrow \frac{Δ G E D}{Δ G E A} = \frac{1}{2}$ $\Rightarrow \frac{Δ G E D}{A} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ $\frac{K F}{A F} = \frac{D E}{A E} = \frac{1}{2}, \frac{A F}{F G} = \frac{1}{2}$ $\Rightarrow \frac{K F}{F G} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ $\frac{D H}{H G} = \frac{K F}{F G} = \frac{1}{4}$ $\Rightarrow \frac{H G}{D G} = \frac{4}{1 + 4} = \frac{4}{5}$ $\Rightarrow \frac{A_{3}}{Δ G E D} = \frac{4}{5}$ $\Rightarrow \frac{A_{3}}{A} = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}$ $\frac{s h a d e d}{Δ A B C} = \frac{A_{1} + A_{2} + A_{3}}{A} = \frac{1}{4} + \frac{1}{4} + \frac{1}{5} = \frac{7}{10}$
	Commented by Tawa11 last updated on 10/Oct/21

	$Great sir$
	Commented by otchereabdullai@gmail.com last updated on 10/Oct/21

	$Brilliant!$
	Answered by ajfour last updated on 10/Oct/21
	$AB=3b AC=4c s=△_(ABC) =6(b×c) △_(AEF) =(b×c) △_(ADG) =(3/2)(b×c) p=△_(AEF) +△_(ADG) =(5/2)(b×c) for H_(−) eq. of DG: h= b+λ(2c−b) eq. of FE: h= c+μ(3b−c) comparing coefficients of a,b 1−λ=3μ & 1−μ=2λ ⇒ 1−μ=2(1−3μ) ⇒ μ=(1/5) , λ=(2/5) h=b+(2/5)(2c−b)=((3b+4c)/5) △_(DAH) =(1/2)b×(((3b+4c)/5)) △_(HAF) =(1/2)(((3b+4c)/5))×c adding =(1/2)(b−c)(((3b+4c)/5)) q=(△_(DAH) +△_(HAF) ) =((7(b×c))/(10)) shaded area =△_(ABC) −{△_(AEF) +△_(AEG) −(△_(DAH) +△_(HAF) )} shaded area=s−(p−q) ((shaded area)/△_(ABC) )=((s−p+q)/s) ((shaded area)/△_(ABC) )=((6−(5/2)+(7/(10)))/6) ((shaded area)/△_(ABC) ) =((42)/(60))=(7/(10)) ✓$
	$AB = 3 b AC = 4 c$ $s = △_{ABC} = 6 (b \times c)$ $△_{AEF} = (b \times c)$ $△_{ADG} = \frac{3}{2} (b \times c)$ $p = △_{AEF} + △_{ADG} = \frac{5}{2} (b \times c)$ $\underline{for H}$ $eq . of DG : h = b + λ (2 c - b)$ $eq . of FE : h = c + μ (3 b - c)$ $comparing coefficients of a, b$ $1 - λ = 3 μ & 1 - μ = 2 λ$ $\Rightarrow 1 - μ = 2 (1 - 3 μ)$ $\Rightarrow μ = \frac{1}{5}, λ = \frac{2}{5}$ $h = b + \frac{2}{5} (2 c - b) = \frac{3 b + 4 c}{5}$ $△_{DAH} = \frac{1}{2} b \times (\frac{3 b + 4 c}{5})$ $△_{HAF} = \frac{1}{2} (\frac{3 b + 4 c}{5}) \times c$ $adding = \frac{1}{2} (b - c) (\frac{3 b + 4 c}{5})$ $q = (△_{DAH} + △_{HAF}) = \frac{7 (b \times c)}{10}$ $shaded area$ $= △_{ABC} - {△_{AEF} + △_{AEG} - (△_{DAH} + △_{HAF})}$ $shaded area = s - (p - q)$ $\frac{shaded area}{△_{ABC}} = \frac{s - p + q}{s}$ $\frac{shaded area}{△_{ABC}} = \frac{6 - \frac{5}{2} + \frac{7}{10}}{6}$ $\frac{shaded area}{△_{ABC}} = \frac{42}{60} = \frac{7}{10} ✓$
	Commented by mr W last updated on 10/Oct/21

	$g r e a t!$
	Answered by john_santu last updated on 11/Oct/21
	$by length ratio of triangles (1) ((area ΔAEG)/(area ΔABC)) = (((2/3)×(3/4))/1)=((1/2)/1) (2) ((area shaded)/(area ΔABC))=(((1/2)(((1+2)/(1+2+2))))/1)=(3/(10)) (3) fraction of area =1−(3/(10))=(7/(10))$
	$b y l e n g t h r a t i o o f t r i a n g l e s$ $(1) \frac{a r e a Δ A E G}{a r e a Δ A B C} = \frac{\frac{2}{3} \times \frac{3}{4}}{1} = \frac{\frac{1}{2}}{1}$ $(2) \frac{a r e a s h a d e d}{a r e a Δ A B C} = \frac{\frac{1}{2} (\frac{1 + 2}{1 + 2 + 2})}{1} = \frac{3}{10}$ $(3) f r a c t i o n o f a r e a = 1 - \frac{3}{10} = \frac{7}{10}$
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