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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=1:2:1whatfractionofΔABCisshaded?

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

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

say area of ΔABC is A.  ((CG)/(AC))=(1/4)   ⇒(A_1 /A)=(1/4)  ((ΔBGA)/A)=(3/4)  ((BE)/(BA))=(1/3)   ⇒(A_2 /(ΔBGA))=(1/3)   ⇒(A_2 /A)=(1/3)×(3/4)=(1/4)  ((ΔGEA)/(ΔBGA))=(2/3)   ⇒((ΔGEA)/A)=(2/3)×(3/4)=(1/2)  ((ED)/(EA))=(1/2)   ⇒((ΔGED)/(ΔGEA))=(1/2)   ⇒((ΔGED)/A)=(1/2)×(1/2)=(1/4)  ((KF)/(AF))=((DE)/(AE))=(1/2), ((AF)/(FG))=(1/2)   ⇒((KF)/(FG))=(1/2)×(1/2)=(1/4)  ((DH)/(HG))=((KF)/(FG))=(1/4)   ⇒((HG)/(DG))=(4/(1+4))=(4/5)   ⇒(A_3 /(ΔGED))=(4/5)   ⇒(A_3 /A)=(4/5)×(1/4)=(1/5)    ((shaded)/(ΔABC))=((A_1 +A_2 +A_3 )/A)=(1/4)+(1/4)+(1/5)=(7/(10))

sayareaofΔABCisA.CGAC=14A1A=14ΔBGAA=34BEBA=13A2ΔBGA=13A2A=13×34=14ΔGEAΔBGA=23ΔGEAA=23×34=12EDEA=12ΔGEDΔGEA=12ΔGEDA=12×12=14KFAF=DEAE=12,AFFG=12KFFG=12×12=14DHHG=KFFG=14HGDG=41+4=45A3ΔGED=45A3A=45×14=15shadedΔABC=A1+A2+A3A=14+14+15=710

Commented by Tawa11 last updated on 10/Oct/21

Great sir

Greatsir

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

Brilliant !

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=3bAC=4cs=ABC=6(b×c)AEF=(b×c)ADG=32(b×c)p=AEF+ADG=52(b×c)forHeq.ofDG:h=b+λ(2cb)eq.ofFE:h=c+μ(3bc)comparingcoefficientsofa,b1λ=3μ&1μ=2λ1μ=2(13μ)μ=15,λ=25h=b+25(2cb)=3b+4c5DAH=12b×(3b+4c5)HAF=12(3b+4c5)×cadding=12(bc)(3b+4c5)q=(DAH+HAF)=7(b×c)10shadedarea=ABC{AEF+AEG(DAH+HAF)}shadedarea=s(pq)shadedareaABC=sp+qsshadedareaABC=652+7106shadedareaABC=4260=710

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

great!

great!

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))

bylengthratiooftriangles(1)areaΔAEGareaΔABC=23×341=121(2)areashadedareaΔABC=12(1+21+2+2)1=310(3)fractionofarea=1310=710

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