# Find the perimeter of Triangle ABC

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)?

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 $$\Delta=rs$$.

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:

$$P=2\times \left(\frac {7\cdot 35}{2}+50\right)=7\cdot 35+100=\boxed {345}$$

• Ohh sorry there's typo error the circle is inscribed and only tangent to line AB of Triangle ABC – Denver Feb 4 at 12:57
• Ohh sorry there's typo error the circle is inscribed and only tangent to line AB of Triangle ABC – Denver Feb 4 at 12:58
• Can anyone again provide a solution or I just consider SIROUS's answer?? – Denver Feb 4 at 14:07
• @Denver, did you want a method for solution or just a number to pass the four option test? – sirous Feb 6 at 10:42
• @DR SK MOBINUl HAQUE, Thanks for correcting my error – sirous Feb 7 at 12:39