Prove that the area of the semicircle on the hypotenuse of a right triangle equals the sum of the areas of the semicircles on the other two sides.
Proved.
Step 1: Recall Pythagoras theorem
In a right triangle, if the two sides (legs) are \(a\) and \(b\), and the hypotenuse is \(c\), then:
\(c^2 = a^2 + b^2\)
Step 2: Formula for area of a semicircle
The area of a full circle with diameter \(d\) is:
\(A_{circle} = \dfrac{\pi d^2}{4}\)
So, the area of a semicircle (half of a circle) is:
\(A_{semicircle} = \dfrac{1}{2} \times \dfrac{\pi d^2}{4} = \dfrac{\pi d^2}{8}\)
Step 3: Areas of the three semicircles
Step 4: Substitute \(c^2\) from Pythagoras theorem
\(A_c = \dfrac{\pi c^2}{8} = \dfrac{\pi (a^2 + b^2)}{8}\)
Step 5: Separate into two terms
\(A_c = \dfrac{\pi a^2}{8} + \dfrac{\pi b^2}{8}\)
So, \(A_c = A_a + A_b\)
Conclusion: The area of the semicircle on the hypotenuse is exactly equal to the sum of the areas of the semicircles on the other two sides.