This practice engineering board exam which I've tried seems very tricky (question no.2). The inscribed circle of $\triangle ABC$ is tangent to $AB$ at $P$ . If the radius is $21$, $AP=23$ and $PB=27$, find the perimeter of $\triangle ABC$ .I've attempted to use the Pythagorean Theorem and Heron's Formula but the problem is that I've missed the values of the two sides $BC$ and $CA$ .Can you please provide solutions by two methods(geometry and trigonometry)?
2 Answers
$\begingroup$Let the circle touch side $BC$ and $CA$ at points $Q$ and $R$ respectively. Let $CQ=CR=m$.
$AR=AP=23$ and $BQ=BP=27$.
$AB=50, BC=27+m, CA=23+m$
Now, apply Heron's Formula and then use $\Delta=rs$.
$\endgroup$ $\begingroup$Another approach:
$\tan(\frac{\alpha}2)=\frac{21}{23}$
$\tan(\frac{\beta}2)=\frac{21}{27}$
$\frac{\alpha}2+\frac{\beta}2+\frac{\gamma}2=90^o$
$\tan(\frac{\alpha}2+\frac{\beta}2)=\tan(90-\frac{\gamma}2)=cotan(\frac{\gamma}2)$
Puting values we get $\tan(\frac{\gamma}2)=\frac6{35}$
Now we use this formula:
$\tan(\frac{\gamma}2)=\frac r{p-c}$
Where r=21, p is half perimeter, and $c=AB=23+27=50$
Plugging these values we get:
$p=\frac r{tan(\frac{\gamma}2)}+c=\frac {21}{\frac 6{35}}+50=\frac {7\cdot 35}{2}+50$
So perimeter P is:
$P=2\times \left(\frac {7\cdot 35}{2}+50\right)=7\cdot 35+100=\boxed {345}$
$\endgroup$ 7
