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June 22, 2020

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

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


Answer: $7 \sqrt{3}$

AMC-10A (2009) Problem 10

Pre College Mathematics

Try with Hints


area of triangle - problem

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 ..........

area of triangle - 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........

area of triangle

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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