Question Number 77420 by BK last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
$${the}\:{three}\:{triangles}\:{are}\:{similar}. \\ $$$${the}\:{radii}\:{of}\:{their}\:{incircles}\:{are}\:{in} \\ $$$${the}\:{same}\:{ratio}\:{as}\:{their}\:{side}\:{lengthes}, \\ $$$${therefore}\:{x}=\mathrm{5}. \\ $$
Commented by BK last updated on 06/Jan/20
$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is}\:? \\ $$$$\mathrm{sir}\:\mathrm{please}\:\mathrm{steb}\:\mathrm{by}\:\mathrm{step}\:\mathrm{solution}\: \\ $$
Commented by mr W last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
$$\alpha+\beta=\mathrm{90}° \\ $$$$\Delta{ACB}\sim\Delta{ADC}\sim\Delta{CEB} \\ $$$$\frac{{x}}{{BA}}=\frac{\mathrm{4}}{{AC}}=\frac{\mathrm{3}}{{CB}} \\ $$$${since}\:{BA}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{CB}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$
Commented by BK last updated on 06/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\boldsymbol{\mathrm{W}} \\ $$
Commented by BK last updated on 06/Jan/20
$$\mathrm{sir}\:\:\:\mathrm{answer}\:\mathrm{x}\:=\mathrm{4}\:,\:\:\mathrm{MN}=\mathrm{15}\:? \\ $$
Commented by BK last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
$${my}\:{solution}\:{is}\:{short}\:{and}\:{correct}. \\ $$$${but}\:{your}\:{book}\:{is}\:{very}\:{wrong}!\: \\ $$$${if}\:{the}\:{radii}\:\mathrm{3}\:{and}\:\mathrm{4}\:{are}\:{given},\:{then} \\ $$$${x}=\mathrm{5},\:{MN}=\mathrm{17},\:{AB}=\mathrm{25},\:{CB}=\mathrm{20},\:{BA}=\mathrm{15}. \\ $$$${you}\:{can}\:{check}\:{this}\:{by}\:{drawing}\:{the} \\ $$$${figure}.\:{but}\:{you}\:{can}\:{not}\:{draw}\:{the} \\ $$$${figure}\:{given}\:{in}\:{the}\:{book}'{s}\:{solution}. \\ $$
Commented by BK last updated on 06/Jan/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:! \\ $$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Great}\:\mathrm{sir} \\ $$