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# Area of Triangle | AMC 10A, 2006 | Problem 21

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

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 $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=$16\sqrt2$