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

Contents

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 $\triangle A B C$ ?

area of triangle - problem figure — Area of Triangle Problem

$4 \sqrt{3}$
$7 \sqrt{3}$
$14 \sqrt{3}$
$21$
$42$

Key Concepts

Triangle

Similarity

Geometry

Check the Answer

But try the problem first...

Answer: $7 \sqrt{3}$

Source

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