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A Circle inscribed in a right triangle" problem (need help!!)

Hello guys, I need help with the following problem:

A circle is inscribed in a right triangle with point P common to both the circle and hypotenuse AB. Find the area of the triangle if AP*BP=24 (hint: sketch a triangle!)

here's the drawing I made (see attached) and the work I have so far:

1. Basically, what I did was draw a point on the middle of the circle. From there, I connected the the segments AB, BC, and AC with a perpendicular line from point O. Then, I named AC "side b" and BC "side a". Next, I connected point O with the vertices A and B.

2. Now, triangles AOP and AOF are congruent as well as triangles BOP and BOE. Also, OF is parallel to BC and OE is parallel to AC. Thus, CF = CE = radius of circle, so AF = side b - radius (r) and BE = side a - r

3. By the congruent triangles, AP = b-r and BP = a - r

combining the results (AP = b-r and BP = a - r) I wrote a quadratic equation in r and I got: r = (a+b+c/2) where c is the hypotenuse (AB)

From here, I don't know what to do next. I'm confused and I have no idea about how to find the two legs (the legs are needed to find the area because one is the height and the other the base). Any form of help is appreciated. Please help me!!!

Thank you very much!

Re: A Circle inscribed in a right triangle" problem (need help!!)

I would use the Pythagorean theorem to state:

$\displaystyle (\bar{AP}+r)^2+(\bar{BP}+r)^2=(\bar{AP}+\bar{BP})^ 2$

$\displaystyle \bar{AP}^2+2r\bar{AP}+r^2+\bar{BP}^2+2r\bar{BP}+r^ 2=\bar{AP}^2+2\bar{AP}\bar{BP}+\bar{BP}^2$

$\displaystyle 2r\bar{AP}+2r\bar{BP}+2r^2=2\bar{AP}\bar{BP}$

$\displaystyle r\bar{AP}+r\bar{BP}+r^2=\bar{AP}\bar{BP}$

We are told $\displaystyle \bar{AP}\bar{BP}=24$ hence:

$\displaystyle r(\bar{AP}+\bar{BP}+r)=24$

Now, referring to your diagram, we may find the area $\displaystyle A$ of the triangle by adding together the square and 4 triangles making up the total triangle:

$\displaystyle A=r^2+r\bar{AP}+r\bar{BP}=r(\bar{AP}+\bar{BP}+r)$

Using our previous result, we find:

$\displaystyle A=24$

Re: A Circle inscribed in a right triangle" problem (need help!!)

Quote:

Originally Posted by

**MarkFL2** I would use the Pythagorean theorem to state:

$\displaystyle (\bar{AP}+r)^2+(\bar{BP}+r)^2=(\bar{AP}+\bar{BP})^ 2$

$\displaystyle \bar{AP}^2+2r\bar{AP}+r^2+\bar{BP}^2+2r\bar{BP}+r^ 2=\bar{AP}^2+2\bar{AP}\bar{BP}+\bar{BP}^2$

$\displaystyle 2r\bar{AP}+2r\bar{BP}+2r^2=2\bar{AP}\bar{BP}$

$\displaystyle r\bar{AP}+r\bar{BP}+r^2=\bar{AP}\bar{BP}$

We are told $\displaystyle \bar{AP}\bar{BP}=24$ hence:

$\displaystyle r(\bar{AP}+\bar{BP}+r)=24$

Now, referring to your diagram, we may find the area $\displaystyle A$ of the triangle by adding together the square and 4 triangles making up the total triangle:

$\displaystyle A=r^2+r\bar{AP}+r\bar{BP}=r(\bar{AP}+\bar{BP}+r)$

Using our previous result, we find:

$\displaystyle A=24$

Thank you so much!!!!!!!! :) So in other words, the area of the triangle equals 24?? (same as AP*PB=24??!!!!)