Question 19
In triangle $$ABC$$, $$AB=15\ \text{cm}$$, $$BC=20\ \text{cm}$$ and $$CA=25\ \text{cm}$$. Then the length of the shortest altitude of the triangle in $$\text{cm}$$ is ______.
Correct Answer: 12
Solution
Since $$15^2+20^2=25^2$$, the triangle is right-angled and its area is $$\frac{1}{2}\times 15\times 20=150\ \text{cm}^2$$. The shortest altitude is drawn to the longest side, which is $$25\ \text{cm}$$. Thus $$150=\frac{1}{2}\times 25\times h$$, giving $$h=12$$.
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