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Question Number 79731 by M±th+et£s last updated on 27/Jan/20

Commented by M±th+et£s last updated on 27/Jan/20

AH:HC=4:3 BG:GC=5:9 find AF:FB

$${AH}:{HC}=\mathrm{4}:\mathrm{3} \\ $$$${BG}:{GC}=\mathrm{5}:\mathrm{9} \\ $$$${find}\:{AF}:{FB} \\ $$

Commented by mr W last updated on 27/Jan/20

equilateral triangle?

$${equilateral}\:{triangle}? \\ $$

Commented by M±th+et£s last updated on 27/Jan/20

yes sir

$${yes}\:{sir} \\ $$

Commented by mr W last updated on 27/Jan/20

5:2

$$\mathrm{5}:\mathrm{2} \\ $$

Commented by M±th+et£s last updated on 27/Jan/20

its right but can you show your work and thank you sir

$${its}\:{right}\:{but}\:{can}\:{you}\:{show}\:{your}\:{work}\:{and}\:{thank}\:{you}\:{sir} \\ $$

Answered by mr W last updated on 27/Jan/20

Commented by M±th+et£s last updated on 27/Jan/20

thank you so much sir

$${thank}\:{you}\:{so}\:{much}\:{sir}\: \\ $$

Commented by mr W last updated on 27/Jan/20

let′s say the side length of equilateral triangle is 1. u^2 +((9/(14)))^2 =v^2 +((3/7))^2 (=QC^2 ) v^2 +((4/7))^2 =w^2 +(1−x)^2 (=QA^2 ) w^2 +x^2 =u^2 +((5/(14)))^2 (=QB^2 ) Σ: ((9/(14)))^2 +((4/7))^2 +x^2 =((3/7))^2 +(1−x)^2 +((5/(14)))^2 ((9/(14)))^2 −((5/(14)))^2 +((4/7))^2 −((3/7))^2 =(1−x)^2 −x^2 ((9−5)/(14))+((4−3)/7)=1−2x ⇒x=(2/7)=FB ⇒AF=1−(2/7)=(5/7) ⇒((AF)/(FB))=(5/2)

$${let}'{s}\:{say}\:{the}\:{side}\:{length}\:{of}\:{equilateral} \\ $$$${triangle}\:{is}\:\mathrm{1}. \\ $$$${u}^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} ={v}^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{7}}\right)^{\mathrm{2}} \:\:\:\:\left(={QC}^{\mathrm{2}} \right) \\ $$$${v}^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{7}}\right)^{\mathrm{2}} ={w}^{\mathrm{2}} +\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:\:\:\left(={QA}^{\mathrm{2}} \right) \\ $$$${w}^{\mathrm{2}} +{x}^{\mathrm{2}} ={u}^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{14}}\right)^{\mathrm{2}} \:\:\:\left(={QB}^{\mathrm{2}} \right) \\ $$$$\Sigma: \\ $$$$\left(\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{7}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{14}}\right)^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{9}}{\mathrm{14}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}}{\mathrm{14}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{7}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{7}}\right)^{\mathrm{2}} =\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$$\frac{\mathrm{9}−\mathrm{5}}{\mathrm{14}}+\frac{\mathrm{4}−\mathrm{3}}{\mathrm{7}}=\mathrm{1}−\mathrm{2}{x} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}}{\mathrm{7}}={FB} \\ $$$$\Rightarrow{AF}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{7}}=\frac{\mathrm{5}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{{AF}}{{FB}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

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