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

Try this beautiful problem from Geometry: Circle from AMC-10A (2006) You may use sequential hints to solve the problem.

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

But try the problem first…

Answer: $16 \sqrt{2}$

Source

AMC-10A (2006) Problem 21

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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……..

Final Step

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

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