Try this beautiful problem from Geometry: Area of a triangle

## Triangle – AMC-10A, 2006- Problem 21

A circle of radius 1 is tangent to a circle of radius 2 . The sides of $\triangle A B C$ are tangent to the circles as shown, and the sides $\overline{A B}$ and $\overline{A C}$ are congruent. What is the area of $\triangle A B C ?$

,

i

- $15 \sqrt{2}$
- $\frac{35}{2} $
- $\frac{64}{3}$
- $16 \sqrt{2}$
- \(24\)

**Key Concepts**

Geometry

Circle

Triangle

## Check the Answer

Answer: $16 \sqrt{2}$

AMC-10A (2006) Problem 21

Pre College Mathematics

## Try with Hints

Given that there are two circle of radius 1 is tangent to a circle of radius 2.we have to find out the area of the \(\triangle ABC\).Now draw a perpendicular line \(AF\) on \(BC\).Clearly it will pass through two centers \(O_1\) and \(O_2\). and $\overline{A B}$ and $\overline{A C}$ are congruent i.e \(\triangle ABC\) is an Isosceles triangle. Therefore \(BF=FC\)

So if we can find out \(AF\) and \(BC\) then we can find out the area of the \(\triangle ABC\).can you find out \(AF\) and \(BC\)?

Can you now finish the problem ……….

Now clearly $\triangle A D O_{1} \sim \triangle A E O_{2} \sim \triangle A F C$ ( as \(O_1D\) and \(O_2E\) are perpendicular on \(AC\) , R-H-S law )

From Similarity we can say that , $\frac{A O_{1}}{A O_{2}}=\frac{D O_{1}}{E O_{2}} \Rightarrow \frac{A O_{1}}{A O_{1}+3}=\frac{1}{2} \Longrightarrow A O_{1}=3$

By the Pythagorean Theorem we have that $A D=\sqrt{3^{2}-1^{2}}=\sqrt{8}$

Again from $\triangle A D O_{1} \sim \triangle A F C$

$\frac{A D}{A F}=\frac{D O_{1}}{C F} \Longrightarrow \frac{2 \sqrt{2}}{8}=\frac{1}{C F} \Rightarrow C F=2 \sqrt{2}$

can you finish the problem……..

The area of the triangle is $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=\(16\sqrt2\)

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=gGT15ls_brU