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 $$h=12$$ cm.

Get AI Help

Video Solution

Other Cracku’s Useful Resources:

JEE Questions JEE Question Bank JEE Mains Previous Papers JEE Advanced Previous Papers JEE Study Material PDF JEE Mains Mock Test JEE Advanced Mock Test JEE Daily Target

Free JEE Questions

JEE Physics Questions JEE Maths Questions JEE Chemistry Questions