AMC 10 Geometry Math Olympiad USA Math Olympiad

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 $\triangle ABC$?

Problem to find the area of triangle
  • \(8\)
  • \(7\sqrt 3\)
  • \(8\sqrt 3\)

Key Concepts




Check the Answer

Answer: \(7\sqrt 3\)

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints

Right- angled triangle

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 ……….

 area of triangle 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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