Trisected square

Geometry Level 2

A B C D ABCD is a square with area 720 720 . Points E , F , G , H E, F, G, H trisect the sides of the square, such that A E E B = B F F C = C G G D = D H H A = 2 \frac{ AE}{EB} = \frac{ BF}{FC} = \frac{ CG}{GD} = \frac{DH}{HA} = 2 . What is the area of square E F G H EFGH ?


The answer is 400.

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Arron Kau by Arron Kau

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2 solutions

Arron Kau Staff
May 13, 2014

Let the side length of square A B C D ABCD be 3 s 3s . Then, the side length of square E F G H EFGH is s 2 + ( 2 s ) 2 = 5 s \sqrt{ s^2 + (2s)^2 } = \sqrt{5} s . Hence, the ratio of their area is ( 5 s 3 s ) 2 = 5 9 \left( \frac{ \sqrt{5}s } {3s} \right) ^2 = \frac{5}{9} . Thus, the area of E F G H EFGH is 5 9 × 720 = 400 \frac{5}{9} \times 720 = 400 .

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Let the side of the square be 3s . And the area of the square is 720. Then the side of the square is 720^0.5 . Then find s . By using Pythagoras theorem find the length of the smaller square and then it's area the its area is found to be 400

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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