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KCC's Quizzes AQQ258 about a partial eclipse Sun-Moon

1. Quote of the week: "People say nothing is impossible; but I do nothing every day!" - Anonymous

                     

2. New quiz: AQQ258 about a partial eclipse Sun-Moon

     

Some of us (in Dallas) had probably the chance to have observed a total eclipse of the Sun last week.

To celebrate that event in our quizzes, we propose the following related challenge: 

The above pictures describe a partial eclipse between the moon (circle B) and the sun (circle R).

At a certain instant, the eclipse is such that points MNO form a quarter area of circle R; meaning arc MN is a quarter circle).

At that moment, point O is the center of R and the segment OM measures 4 cm.

Question:

What is the area of the portion of blue circle B that is situated outside circle R?

Good luck! And try to be among the firsts!

Kuo-Chang

P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

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Parents
  • I think this might be a crude solution, feel free to help me improve my logic:

    1. We can form a square on circle B from the fact that OM and ON are same valued sides, then from this subtract it to the area of B, which circumscribes this square. This leaves us with 4 leftover pieces of the circle, of which we need only 2 : 

    1.a. radius sqaured of B can be acquired from this formed square : 4^2 = 2r^2 -> r^2 = 4^2 / 2

    1.b. multiply 1.a. by pi to get area of B : pi*(r^2) = 8pi.

    1.c. Subtract 1.b. to 4^2 and divide by 2 to get 2 leftover circle pieces : 9.133.

    2.Subtract the area of the square to the quarter Circle made by circle R: 

    2.a 4^2 - (pi*(4^2)/4) = 3.434

    3. Add 2.a and 1.c to get the final result:

    A = 3.434 + 9.133

    A = 4pi.

Reply
  • I think this might be a crude solution, feel free to help me improve my logic:

    1. We can form a square on circle B from the fact that OM and ON are same valued sides, then from this subtract it to the area of B, which circumscribes this square. This leaves us with 4 leftover pieces of the circle, of which we need only 2 : 

    1.a. radius sqaured of B can be acquired from this formed square : 4^2 = 2r^2 -> r^2 = 4^2 / 2

    1.b. multiply 1.a. by pi to get area of B : pi*(r^2) = 8pi.

    1.c. Subtract 1.b. to 4^2 and divide by 2 to get 2 leftover circle pieces : 9.133.

    2.Subtract the area of the square to the quarter Circle made by circle R: 

    2.a 4^2 - (pi*(4^2)/4) = 3.434

    3. Add 2.a and 1.c to get the final result:

    A = 3.434 + 9.133

    A = 4pi.

Children