Really confused right now,
In triangle ABC, angle B = 90 degrees. Semicircles are constructed on sides AB, AC and BC as shown. Show that the total area of the shaded region is equal to the area of triangle ABC.
+118667 Let the sides be a, b and sqrt (a^2+b^2)
So the radii are \(\frac{a}{2},\quad \frac{b}{2},\quad \frac{\sqrt{a^2+b^2}}{2}\)
Area of triangle = ab/2
Area of big semicircle = \(0.5 \pi*\frac{(a^2+b^2)}{4}=\frac{(a^2+b^2)\pi}{8}\)
Sum of little white segments = \(\frac{(a^2+b^2)\pi}{8}-\frac{ab}{2}=\frac{(a^2+b^2)\pi-4ab}{8}\)
Sum of the 2 smaller semicircle = \(0.5*\pi((\frac{a}{2})^2+(\frac{b}{2})^2)= \pi(\frac{a^2+b^2}{8})\\ \)
You can finish it
