Question Number 185938 by Mingma last updated on 30/Jan/23
Answered by HeferH last updated on 30/Jan/23
$${Let}\:{S}\:{be}\:{the}\:{pink}\:{area}\:{and}\:{let}\:\mathrm{5}{k}\:{be}\:{the}\:{area} \\ $$$$\:{of}\:{ABC}. \\ $$$$\:{S}\left(\frac{\mathrm{3}{k}}{\mathrm{2}}\:+\:{S}\right)\:=\:\left({k}\:−\:{S}\right)\left(\frac{\mathrm{3}{k}}{\mathrm{2}}−{S}\right) \\ $$$$\:{S}\:\:=\:\frac{\mathrm{3}{k}}{\mathrm{8}} \\ $$$$\:\frac{{Pink}}{{Area}\:{of}\:{ABC}}\:=\:\frac{\frac{\mathrm{3}{k}}{\mathrm{8}}}{\mathrm{5}{k}}\:=\:\frac{\mathrm{3}}{\mathrm{40}}\:\: \\ $$
Commented by HeferH last updated on 30/Jan/23
Answered by mr W last updated on 30/Jan/23
Commented by mr W last updated on 30/Jan/23
![DF=(3/5)×DC=(3/5)×BD GE=(3/5)×BG=(3/8)×BE [GDE]=(3/8)×[BED] =(3/8)×(1/2)×[BEC] =(3/8)×(1/2)×(2/5)×[ABC] =(3/(40))×[ABC] ✓](https://www.tinkutara.com/question/Q186003.png)
$${DF}=\frac{\mathrm{3}}{\mathrm{5}}×{DC}=\frac{\mathrm{3}}{\mathrm{5}}×{BD} \\ $$$${GE}=\frac{\mathrm{3}}{\mathrm{5}}×{BG}=\frac{\mathrm{3}}{\mathrm{8}}×{BE} \\ $$$$\left[{GDE}\right]=\frac{\mathrm{3}}{\mathrm{8}}×\left[{BED}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{2}}×\left[{BEC}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{8}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{5}}×\left[{ABC}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{40}}×\left[{ABC}\right]\:\checkmark \\ $$