Enigmatic Code

Programming Enigma Puzzles

Puzzle #46: Pi-thagoras

From New Scientist #3269, 15th February 2020 [link] [link]

Pythagoras’s theorem says that for any right-angled triangle, the square of the hypotenuse (the longest side of the triangle) is equal to the sum of the squares of the other two sides.

There are some right-angled triangles whose sides are all whole number lengths. The simplest and best known is the “3-4-5” triangle (3² + 4² = 5²).

I have drawn a circle that fits precisely inside a 3-4-5 triangle. What is the area of the circle? Have a guess. And then see if you can prove that you are right.

[puzzle#46]

2 responses to “Puzzle #46: Pi-thagoras

  1. Jim Randell 15 February 2020 at 7:50 am

    The inradius, semiperimeter and area of a triangle are related as follows:

    area = inradius × semiperimeter

    In this case the area of the triangle is: (1/2) × 3 × 4 = 6

    And the semiperimeter is: (1/2) × (3 + 4 + 5) = 6.

    So the inradius is: 6 / 6 = 1, and the area of the incircle is π (pi) (as you might have guessed from the title of the puzzle).

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