Question

prove that the are of a semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on other two sides

Answer 1

Answer: See attached image

The are of a semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on other two sides

Step-by-step explanation:

From the image:

By Pythagoras theorem:

y^2 + z^2 = x^2

Area of semicircle = 1/2π x Radius^2

Area of A = 1/2π(y/2)^2 = 1/8πy^2

Area of B = 1/2π(z/2)^2 = 1/8πz^2

Area of C = 1/2π(x/2)^2 = 1/8πx^2

Area A + Area B = 1/8πy^2 + 1/8πz^2

                           = 1/8π(y^2 + z^2)

But from the Pythagoras theorem: y^2 + z^2 = x^2

So, Area A + Area B = 1/8π(y^2 + z^2)= 1/8πx^2

Which is the same as area C = 1/8πx^2

Conclusion:

The are of a semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on other two sides

Answer 2

Step-by-step explanation:

hope it helps u...

follow me...

I'll be helpfull to... sure