Question Number 223836 by Tawa11 last updated on 06/Aug/25
Answered by som(math1967) last updated on 06/Aug/25
$$\bigtriangleup{ABG}\sim\bigtriangleup{HFG} \\ $$$${let}\:{AB}={x} \\ $$$$\:\:\therefore\frac{{x}}{{x}+\mathrm{105}}=\frac{{x}−\mathrm{112}}{{x}−\mathrm{105}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{112}{x}+\mathrm{105}{x}−\mathrm{112}×\mathrm{105} \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{2}} −\mathrm{105}{x} \\ $$$$\Rightarrow\mathrm{210}{x}−\mathrm{112}{x}=\mathrm{112}×\mathrm{105} \\ $$$$\Rightarrow\mathrm{98}{x}=\mathrm{112}×\mathrm{105} \\ $$$$\:{x}=\frac{\mathrm{112}×\mathrm{105}}{\mathrm{98}}=\mathrm{120}{cm} \\ $$
Commented by Tawa11 last updated on 06/Aug/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$