by Soumalya Pramanik
Last Updated May 16, 2019 04:20 AM

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).

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 $\frac 14 \cdot 12=3$

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?

$$\frac{R}{12}=\frac{2x}{48}=\frac{40-(24+x)}{40} \implies \frac{R}{12}=\frac{2x+2(16-x)}{48+2(40)}$$

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