Question 1

Let $$ABC$$ be a right-angled triangle with $$\angle ABC = 90^\circ$$, $$BD = 2a + 2$$, $$DC = 3a - 3$$, $$BE = 2a + 1$$ and $$EA = 4a + 1$$, where $$D$$ is the mid-point of side $$BC$$ and $$E$$ is a point on side $$AB$$. The area of the semicircle with diameter $$AC$$ is ______.

Solution

Since $$D$$ is the mid-point of $$BC$$ we get $$2a + 2 = 3a - 3$$, so $$a = 5$$ and $$BD = DC = 12$$, giving $$BC = 24$$. Also $$AB = BE + EA = 11 + 21 = 32$$, so $$AC = \sqrt{24^2 + 32^2} = \sqrt{1600} = 40$$. The semicircle on $$AC$$ has radius 20, so its area is $$\frac{1}{2}\pi (20)^2 = 200\pi$$.

Get AI Help

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

Let $$ABC$$ be a right-angled triangle with $$\angle ABC = 90^\circ$$, $$BD = 2a + 2$$, $$DC = 3a - 3$$, $$BE = 2a + 1$$ and $$EA = 4a + 1$$, where $$D$$ is the mid-point of side $$BC$$ and $$E$$ is a point on side $$AB$$. The area of the semicircle with diameter $$AC$$ is ______.

Solution

Get AI Help

Other Cracku’s Useful Resources:

Free JEE Questions

50,000+ JEE Students Trusted Our Score Calculator

Download resources