Notes
four squares and a quarter circle solution

Solution to the Four Squares and a Quarter Circle Puzzle

Four Squares and a Quarter Circle

The radius of the quarter circle is 55. What’s the total area of the four squares?

Solution by Pythagoras’ Theorem

FourSquaresandaQuarterCircleLabelled.png?

Consider the right-angled triangle in the above diagram formed by joining the centre of the circle to the point on the circumference of the circle where the two smaller squares touch it. Let this have height xx as in the diagram, then Pythagoras' Theorem says that:

5 2=x 2+(52+x2) 2=x 2+254+52x+x 24=254+52x+5x 24 5^2 = x^2 + \left( \frac{5}{2} + \frac{x}{2}\right)^2 = x^2 + \frac{25}{4} + \frac{5}{2} x + \frac{x^2}{4} = \frac{25}{4} + \frac{5}{2} x + \frac{5 x^2}{4}

Rearranging this yields

15=x 2+2x=x(x+2) 15 = x^2 + 2x = x(x + 2)

This has solutions x=3x = 3 and x=−5x = -5, but as xx is a length only x=3x = 3 is allows.

The area that we are looking for has area:

2×(52) 2+2×(x2) 2=5 22+x 22=252+92=17 2 \times \left(\frac{5}{2}\right)^2 + 2 \times \left(\frac{x}{2}\right)^2 = \frac{5^2}{2} + \frac{x^2}{2} = \frac{25}{2} + \frac{9}{2} = 17
Created on May 17, 2021 21:24:55 by Andrew Stacey?