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

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

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

But try the problem first…

Answer: $7 \sqrt{3}$

Source

AMC-10A (2009) Problem 10

Pre College Mathematics

## Try with Hints

First hint

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 ……….

Second Hint

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……..

Final Step

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

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