Find Area of Isoscels Trianlge with 2 Inscribed Circles

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april19

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I am helping my daughter with a geometry problem and I haven't touched geometry for many years so it's very rusty.
Sorry for not including an image because I haven't figured out how to do it.

Here's the problem:
Triangle ABC is an isosceles triangle with two inscribed circles. The large circle has radius 2r, and the smaller circle with radius r is tangent to the large circle and the two equal sides. What is the area of triangle ABC?

Although the original problem did not state it, I have to assume the answer has to be in terms r (the radius).

Description of the picture:
C is the vertex where the 2 equal sides meet and A & B forms the base of the isosceles triangle. So for this triangle, C is on top and A & B are at the bottom. The larger circle touches the base AB and the 2 equal sides and the smaller circle sits on top of the larger circle touches the 2 equal sides.

What I know so far:
* I draw one line between the 2 circles parellel to the base and another line on top of the smaller circle also parallel to the base. Now I have 2 smaller triangles and they are similar to triangle ABC.
* According to the Power of Point Theorem, I can prove there are a lot of congruent triangles.
* Using the same theorem, the perimeter of the triangle on top of the smaller circle is 2x(length of vertex C to the point of tangency on one side of the smaller circle)
* Since ABC is an isosceles triangle, the altitude is also the angle bisector and the median.
* If I know the length of all the sides, I can easily find the area of ABC but unfortunately that is not given.

Can anyone give me some pointers on what to do to find the area of ABC?

Please help and thank you.
 
I'll take a stab at this.
From the center of each circle draw a radius perpendicular to one of the equal sides of the triangle. In the small circle that radius is r. In the large circle that radius is 2r. From the center of the small circle, draw a line parallel to the side that will intersect the radius of the large circle that you just drew. This will form a rectangle whose small sides are r in length. It also forms a right triangle with short leg of r and hypotenuse of 3r (2r+r). Using the Pythagorean Thm. you can determine that the long leg of this triangle (the length of the rectangle) is \(\displaystyle 2r\sqrt{2}\).
Now extend the line joining the centers of the two circles up to apex of triangle ABC. This will form a right triangle (from ptC to center of small circle to right angle where the BC is tangent to the small circle). These two triangles are congruent. They are right triangles having equal short legs and equal base angles since the base angles are corresponding angles of parallels cut by a transversal. This will enable you to calculate the altitude of ABC. Also each of these triangles is similar to the larger right triangle (1/2 of ABC) which will enable you to calculate the length of its short side which is 1/2 the base of ABC.
Good luck.
 
Thank your for replying my post.

I am having trouble visualizing the rectangle in the first paragraph. The line that I drew starts from the center of the small circle is parallel to one side of the isosceles triangle and intersect the radius of the large circle. The hypotenuse is drawn from the center of the small circle to point where the large circle radius touches the isoscels side. The short leg is r and the long leg should be 3r and the hypotenuse is r x sqrt(10).

Please help and thank you.
 
I made a rough drawing in Paint. Sorry, I could not get it to work any better. It was a pain trying to get the circles situated correctly.

As a consequence, they are ellipses. But, hopefully it imparts the idea.

Let h be the height of triangle ABC. Let b be the length of half the base.

Then, by similar triangles CBH and CEF: \(\displaystyle \frac{b}{h}=\frac{2r}{\sqrt{(h-2r)^{2}-4r^{2}}}\)

\(\displaystyle b=\frac{2rh}{\sqrt{(h-2r)^{2}-4r^{2}}}\)

Since the isosceles is two right triangles with area \(\displaystyle \frac{bh}{2}\), the area of the whole isosceles triangle is A=bh.

so the area is then

\(\displaystyle A=\frac{2rh^{2}}{\sqrt{(h-2r)^{2}-4r^{2}}}\)

By similar triangles CDG and CEF, \(\displaystyle \frac{r}{h-5r}=\frac{2r}{h-2r}\)

\(\displaystyle h=8r\)

Sub into A and get \(\displaystyle \boxed{A=16\sqrt{2}r^{2}}\)

Perhaps Soroban or someone else will be along to confirm or refute :) Seems good, though.

Easy to make a mistake in all those similar triangles.
 
Thank you for replying.

I finally figured out the height from the first reply which is 8r.

I have trouble seeing the similar triangles suggested by both replies.
The second reply said triangles ABH and AEF are similar but ABH is a line not a trianlge.
How did you get the square root value in the denominator for b/h?

Thank you.
 
Oops. Need to label midpoint of DG as H and midpoint of AB as J.

EF=r
DG=2r
DE=3r
(EH)^2 = (DE)^2 - (DH)^2

Triangle DEH congruent to triangle ECF.
Triangle DEH similar to triangle BCJ.
 
Thank you, Loren.
After staring at the picture long enough, I finally figured out what you meant. As long as I stared at the picture before posting the question, I never saw that triangle. Excellent!

I finally figured out how the triangles are similar. Now I just need to figure the base.
Thank you so much.
 
You should end up getting base = 8r / sqrt(2),
height being 8r, you'll then end up with area being "as per Galactus" (we don't need Soroban!).
 
Yes! I got the answer!!! Thank you all for your help. This is so exciting.
 
Sorry about the mislabeling. I fixed the similar triangles I used. I inadvertently used A instead of C.

I mentioned Soroban because he enjoys these types of problems and normally responds.

There is always more than one way to go about it. I just used the similar triangles as when tackling

the famous 'sphere inside a cone' optimization problem.

Loren, you're better at Paint then me. Everytime I try to draw circles tangent to one another they seem to have a mind of their own.

Good to see everything worked out. :D
 
The height = 8r can be determined right off the bat:
4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + .......for ever and ever, Amen...... = 8
Right?
 
Yes, Denis, I saw that as well .

This problem reminded me of the "infinite circles in the triangle" problem, except it stopped at two. :D
 
Denis said:
The height = 8r can be determined right off the bat:
4 + 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + .......for ever and ever, Amen...... = 8
Right?
Click to expand...

Randy Travis - whatever happened to him....
 
galactus said:
Everytime I try to draw circles [using MS Paint] … they seem to have a mind of their own.
Click to expand...

I find, in Paint, that circles are easier to deal with when you draw them off-to-the-side and then slide them into position using the transparent-overlay feature (versus opaque-overwrite).

Of course, we draw a "true" circle by dragging the mouse from one corner of an imaginary circumscribed-square to the diagonally-opposite corner.

That's my terminology, above, regarding the features. If you're not sure what I'm talking about, let me know, and I'll upload something more specific.

These comments relate to MS Paint bundled with WindowsXP. With Microsoft's propensity for making things "slower and more difficult", I don't want to think what MS Paint has become in Windows Vista or 7.
 
I am using Vista and it does not appear to have changed much. Thanks for the Paint tips. I will give it a go next time.

Anyway, I created the graph with a graphing utility that looks better than I could do in Paint.

Loren done a fine job, but I suppose I need more practice.
 
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