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

[megadeth95]

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!

 

[megadeth95]

Your r = (a+b+c/2) should be r = (a + b - c) / 2.
Btw, that's a standard formula: no need to calculate it.

I can't get anywhere with your problem (I may have missed something obvious).
Are you sure that it is the original IN FULL, plus no typoes?
Could the "AP*AB=24" be AP*PB=24 ?

Assuming it is YOUR posted version: let AP = p; then:
cp = 24 ; p = 24/c; so b - r = 24 / c ; and:
BP = (c^2 - 24)/c; so a - r = (c^2 - 24)/c

Any reasons why there is no such attempt in what you did?
Not that it helps much!

Sometimes it helps if you make one up; take a 3-4-5 right triangle;
the radius will be 1, AP = 2 ; so AP*AB = 2 * 5 = 10.

Sorry, I made a mistake. It's suppose to be AP*PB = 24

Thanks for helping me!

 

[megadeth95]

That wasted some 2 hours of my time...not that I'm over-busy :sad:

I may give it another shot tomorrow...

I'm sorry, I did not mean to waste your time. Thanks once again.

 

[megadeth95]

Finally got the darn thing...t'was staring me in the face!

Let k = AP * PB (that's the "given").

AP = a - r and PB = b - r

So (a - r) (b - r) = k

Substitute r = (a + b - c) / 2 :
[a - (a + b - c) / 2] * [b - (a + b - c) / 2] = k

Expand and simplify:
c^2 - a^2 - b^2 + 2ab = 4k
Since c^2 - a^2 - b^2 = 0, then:
2ab = 4k
ab = 2k
area ab/2 = k

I'll be charging 5 bucks for autographs

Thank you sooo much Denis!!!!!!!!! You are the man!!!!!!!!! Yes, I owe you a favor

So in other words, the area of the triangle is 24?????!!!!!!!

Another person was helping me with this problem and he came out with this:

I would use the Pythagorean theorem to state:

We are told

hence:

Now, referring to your diagram, we may find the area

of the triangle by adding together the square and 4 triangles making up the total triangle:

Using our previous result, we find:

 

[megadeth95]

Yes. Area of any right triangle is ALWAYS equal to (a - r)*(b - r)

Thank you soo much for helping me Denis, I really appreciate it!

Best Regards

Diego