Question 1071670: A square ABCD, there is a diagonal line from A to c. M is the midpoint of side AD and MN is perpendicular to AC, What is the ratio of the area of the shaded triangle MNC to the area of the square?
this is grade 8 kangaroo math question. Kangaroo 2012.Cadet level question 22
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! It is 3:16. The way I looked at it is that the small triangle MNA is a 45-45-90 triangle with hypotenuse (1/2)x, so that the line MN, the altitude of the shaded triangle, is (1/2)x/sqrt(2).
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The line MC is the hypotenuse of the triangle MNC, and its length squared is (0.5x^2)+x^2 or (5/4)x^2. Its length is sqrt(5)*x/2
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We need NC, which is the base. MN squared + MC squared is NC squared.
MN squared is (1/8)x^2, and MC squared is (10/8)x^2, just rewriting it with denominator 8.
The difference between those two is (9/8)x^2, and its length is the square root of that or 3x/2sqrt(2)
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the area of MNC is half the base and altitude
that is (1/2)(x/2sqrt(2))*3x/2sqrt(2)
That is 3x^2/16.
The area of the square is x^2.
The ratio is that product with the x^2 cancelling, and is 3/16.
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