Question Number 212533 by emilagazade last updated on 16/Oct/24
$${given}\:{isoscele}\:{triangle}\:{with}\:{sides}\:\mathrm{10}\:{and}\:{inradius}\:\mathrm{3}.\:{how}\:{find}\:{base}? \\ $$
Answered by A5T last updated on 16/Oct/24
![Let base=b ;angle between non-equal sides=θ [△]=((10×10sin(180−2θ))/2)=((b×10×sinθ)/2) ⇒50×2cosθ=5b⇒cosθ=(b/(20))⇒sinθ=((√(400−b^2 ))/(20)) [△]=((3(10+10+b))/2)=((10bsinθ)/2)=((b(√(400−b^2 )))/4) ⇒120+6b=b(√(400−b^2 ))⇒b=12 or 4+2(√(19))](https://www.tinkutara.com/question/Q212535.png)
$${Let}\:{base}={b}\:;{angle}\:{between}\:{non}-{equal}\:{sides}=\theta \\ $$$$\left[\bigtriangleup\right]=\frac{\mathrm{10}×\mathrm{10}{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{\mathrm{2}}=\frac{{b}×\mathrm{10}×{sin}\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{50}×\mathrm{2}{cos}\theta=\mathrm{5}{b}\Rightarrow{cos}\theta=\frac{{b}}{\mathrm{20}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{400}−{b}^{\mathrm{2}} }}{\mathrm{20}} \\ $$$$\left[\bigtriangleup\right]=\frac{\mathrm{3}\left(\mathrm{10}+\mathrm{10}+{b}\right)}{\mathrm{2}}=\frac{\mathrm{10}{bsin}\theta}{\mathrm{2}}=\frac{{b}\sqrt{\mathrm{400}−{b}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{120}+\mathrm{6}{b}={b}\sqrt{\mathrm{400}−{b}^{\mathrm{2}} }\Rightarrow{b}=\mathrm{12}\:{or}\:\mathrm{4}+\mathrm{2}\sqrt{\mathrm{19}} \\ $$
Commented by emilagazade last updated on 16/Oct/24
$${thank}\:{you}\:{a}\:{lot} \\ $$
Answered by mr W last updated on 16/Oct/24
Commented by mr W last updated on 17/Oct/24
$${say}\:{x}=\frac{{b}}{\mathrm{2}} \\ $$$$\frac{{a}−{x}}{{r}}=\frac{\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{{x}} \\ $$$$\frac{{a}−{x}}{{r}^{\mathrm{2}} }=\frac{{a}+{x}}{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{3}} −{ax}^{\mathrm{2}} +{r}^{\mathrm{2}} {x}+{ar}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{90}=\mathrm{0} \\ $$$$\left({x}−\mathrm{6}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{15}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{6},\:\mathrm{2}\pm\sqrt{\mathrm{19}} \\ $$$$\Rightarrow{b}=\mathrm{12},\:\mathrm{4}+\mathrm{2}\sqrt{\mathrm{19}} \\ $$