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 Two Crescents (Posted on 2004-08-11)
Given a right triangle, draw the three following semicircles:
1. The semicircle with diameter formed by one of the legs and extending away from the triangle.
2. The semicircle with diameter formed by the other leg and extending away from the triangle.
3. The semicircle with diameter formed by the hypotenuse and extending towards the triangle.

Prove that the area of the two crescents (shown in RED and BLUE) formed by the three semicircles equals the area of the triangle.

 No Solution Yet Submitted by ThoughtProvoker Rating: 3.2000 (10 votes)

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 2x | Comment 7 of 15 |

Be S de desired area value. Be 2a the smaller side, 2c the hypotenusa and 2b the other side of the triangle.
So the area of the triangle is (4ab)/2.
The area of the first semicircle is (pi*(a^2))/2.
The area of the second semicircle is (pi*(b^2))/2.
The area of the third semicircle is (pi*(c^2))/2.
The desired area is
(pi*(a^2))/2 + (pi*(b^2))/2 - ((pi*(c^2))/2 - (4ab)/2).
But 4(c^2) = 4(a^2) + 4(b^2) (right triangle), so we have
(c^2) = (a^2) + (b^2)
The desired area is (4ab)/2, the same as the triangle area.

 Posted by Elisabeth on 2004-08-13 21:04:56

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