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# Area of Triangle Problem | AMC-10A, 2009 | Problem 10

Try this beautiful problem from Geometry: Area of triangle

## Area of the Triangle- AMC-10A, 2009- Problem 10

Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?

• $$8$$
• $$7\sqrt 3$$
• $$8\sqrt 3$$

### Key Concepts

Geometry

Triangle

similarity

Answer: $$7\sqrt 3$$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

We have to find out the area of $$\triangle ABC$$.now the given that $$BD$$ perpendicular on $$AC$$.now area of $$\triangle ABC$$ =$$\frac{1}{2} \times base \times height$$. but we don't know the value of $$AB$$ & $$BC$$.

Given $$AC=AD+DC=3+4=7$$ and $$BD$$ is perpendicular on $$AC$$.So if you find out the value of $$BD$$ then you can find out the area .can you find out the length of $$BD$$?

Can you now finish the problem ..........

If we proof that $$\triangle ABD \sim \triangle BDC$$, then we can find out the value of $$BD$$

Let $$\angle C =x$$ $$\Rightarrow DBA=(90-X)$$ and $$\angle BAD=(90-x)$$,so $$\angle ABD=x$$ (as sum of the angles of a triangle is 180)

In Triangle $$\triangle ABD$$ & $$\triangle BDC$$ we have...

$$\angle BDA=\angle BDC=90$$

$$\angle ABD=\angle BCD=x$$

$$\angle BAD=\angle DBC=(90-x)$$

So we can say that $$\triangle ABD \sim \triangle BDC$$

Therefore $$\frac{BD}{AD}=\frac{CD}{BD}$$ $$\Rightarrow (BD)^2=AD .CD \Rightarrow BD=\sqrt{3.4}=2\sqrt 3$$

can you finish the problem........

Therefore area of the $$\triangle ABC =\frac {1}{2} \times AC \times BD=\frac {1}{2} \times 7 \times 2\sqrt 3=7 \sqrt 3$$ sq.unit

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