How to find the radius of this smaller circle? The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."
I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$\frac{1}{2} * r * (40 + 40 + 48) = \sqrt{\left(\frac{40 + 40 + 48}{2}\right) \left(\frac{40 + 40 + 48}{2}-40\right)\left(\frac{40 + 40 + 48}{2}-40\right)\left(\frac{40 + 40 + 48}{2}-48\right)}$$
Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.
But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).
The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - \pi*(12)^2$ cm].
And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).
 A: See the figure below.  One unit on the paper is six units in your problem.  $AB=48,AC=40,BC=40$.  Circle $D$ has radius $12$ as you say.  $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$.  $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $\frac 14$ the size of the large one.  That says the radius of the small circle is $\frac 14 \cdot 12=3$

A: 
$$\frac{R}{12}=\frac{2x}{48}=\frac{40-(24+x)}{40} \implies \frac{R}{12}=\frac{2x+2(16-x)}{48+2(40)}$$
A: Let in $\Delta ABC$ we have $AB=AC=40$ and $BC=48.$
Also, let $(I,12)$  be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively,  $(O,x)$ be the little circle, which touched to $AC$ in the point $F$. 
Thus,  $$AE=\frac{40+40-48}{2}=16$$ and since $\Delta AIE\sim \Delta AOF,$ we obtain:
$$\frac{AF}{AE}=\frac{OF}{IE}$$ or
$$\frac{AF}{16}=\frac{x}{12},$$ which gives
$$AF=\frac{4}{3}x,$$
$$FE=16-\frac{4}{3}x$$ and by the Pythagoras's theorem we obtain:
$$FE=2\sqrt{IE\cdot OF}$$ or
$$16-\frac{4}{3}x=2\sqrt{12x}.$$
Can you end it now?
