Power chords

Here’s our next Catriona Shearer puzzle.
Power Chords

What’s the area of the circle?

“I never learnt the intersecting chords theorem at school, so I love anything where I get to use it!” — CS

I never learned it either. Now I have.

There’s a vertical chord here, and a horizontal chord. The vertical chord is divided into a 10 unit piece and a 4 unit piece. The horizontal chord has a 5 unit piece and… what? By the intersecting chords theorem, 5\times x = 4 \times 10 so x=8. So the red square’s area is 64, but unfortunately it’s the circle whose area we need…

The perpendicular bisector of a chord contains the center of the circle, so the center is 6.5 left of the right end (or right of the left end) of the horizontal chord, or 1.5 left of the vertical chord; and the center is 7 below the top end (or above the bottom end) of the vertical chord, or 3 below the horizontal chord. Then we can draw a radius from the center to any of the chord end points. The square of the length of that radius we can calculate as r^2 = 1.5^2 + 7^2, or as r^2 = 6.5^2 + 3^2. Either way, r^2 = 51.25, and the area of the circle is 51.25\pi.

2 thoughts on “Power chords

  1. May I politely note that at the very end you give the area as 51.5 pi when it should be 51.25 pi – a typo I assume because all was well until then!

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