How to find the radius of this smaller circle?

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).

Tags : geometry


Answers 3


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 $\frac 14 \cdot 12=3$

enter image description here

Ross Millikan
Ross Millikan
May 16, 2019 04:05 AM

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?

Michael Rozenberg
Michael Rozenberg
May 16, 2019 04:09 AM

enter image description here

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

CY Aries
CY Aries
May 16, 2019 04:13 AM

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