Area of Triangle Problem | 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 $\triangle ABC$?

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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