Try this beautiful problem from Geometry based on Area of Triangle.
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 $\triangle A B C$ ?
Triangle
Similarity
Geometry
Answer: $7 \sqrt{3}$
AMC-10A (2009) Problem 10
Pre College Mathematics
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 ..........
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........
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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