Try this beautiful problem from Geometry based on Area of Triangle.

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

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

- $4 \sqrt{3}$
- $7 \sqrt{3}$
- $14 \sqrt{3}$
- \(21\)
- \(42\)

**Key Concepts**

Triangle

Similarity

Geometry

## Check the Answer

Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

We have to find out the area of the Triangle ABC where \(\angle B=90^{\circ}\) and \(BD \perp AC\)

Area of a Triangle = \(\frac{1}{2}\times \) Base \(\times\) Height.But we don know the value of \(AB\) & \(BC\). But we know \(AC=7\). So if we can find out the value of \(BD\) then we can find out the are of \(\triangle ABC\) by \(\frac{1}{2}\times AC \times BD\)

Can you now finish the problem ……….

Let \(\angle C=\theta\), then \(\angle A=(90-\theta)\) (as \(\angle B=90^{\circ}\), Sum of the angles in a triangle is \(180^{\circ}\))

In \(\triangle ABD\), \(\angle ABD=\theta\) \(\Rightarrow \angle A=(90-\theta\))

Again In \(\triangle DBC\), \(\angle DBC\)=(\(90-\theta\)) \(\Rightarrow \angle C=\theta\)

From the above condition we say that , \(\triangle ABD \sim \triangle BDC\)

Therefore , \(\frac{BD}{CD}=\frac{AD}{BD}\) \(\Rightarrow {BD}^2=AD.CD=4\times 3\)

\(\Rightarrow BD=\sqrt {12}\)

can you finish the problem……..

Therefore area of the \(\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}\)

## Other useful links

- https://www.cheenta.com/problem-from-probability-amc-8-2004problem-no-21/
- https://www.youtube.com/watch?v=VLyrlx2DWdA&t=3s