LMN is a triangle right angled at L. Semi-circles have been drawn on its sides. If MN = x, then sum of the areas of these semi-circles is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi x^2}{4}\)
A semicircle drawn on a side of length a has that side as its diameter, so its radius is \(\frac{a}{2}\) and its area is \(\displaystyle\frac{1}{2}\pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{8}\). Since the triangle is right-angled at L, MN is the hypotenuse, and Pythagoras’ theorem gives \(LM^2+LN^2=MN^2=x^2\).
Sum of the three areas: \(\displaystyle\frac{\pi}{8}\left(LM^2+LN^2+MN^2\right)\)
Putting \(LM^2+LN^2=x^2\) and \(MN^2=x^2\): \(\displaystyle\frac{\pi}{8}\left(x^2+x^2\right)=\frac{\pi x^2}{4}\)
