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

Try this beautiful problem from Geometry: Area of triangle from AMC-10A, 2009, Problem-10. You may use sequential hints to solve the problem.

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