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 :

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 $$\frac 14 \cdot 12=3$$

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
May 16, 2019 04:09 AM

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

CY Aries
May 16, 2019 04:13 AM

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