Area Problem #48

What fraction of the larger square is shaded?

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14 thoughts on “Area Problem #48

    • Also, I have assumed the similar triangles are isosceles, i don’t know if i can assume this, and i think this means the solution will be dependent on the angles in the similar triangles.

  1. Though I left a reply quite some time ago w/ an answer of 11/20, b/c of the varied answers provided other than 11/20 I thought it might be a good idea to provide some details for my answer.

    Note that the problem states “what fraction of the *larger square* is shaded – implying more than one square is shown. In fact, there are two squares in the problem – the one w/ solid line segments for its sides, and the other w/ a majority of its boundary shown as dashed line segments.

    Assume the far left dashed line segment has length 4x. Then, the top dashed line segment is split into two equal lengths of 2x. Lastly, the dashed line segment on the far right must have length x, by virtue of similar triangles (i.e., the small triangle w/ dashed line segment legs x and 2x is similar to the large triangle w/ dashed line segment legs 2x and 4x).

    The area of the smaller square is trivially (4x)^2, or 16x^2. The areas of the two right triangles w/ legs consisting of dashed line segments are 4x^2 [i.e., (1/2)(4x)(2x)] and x^2 [i.e., (1/2)(2x)(x)]. Therefore, the area of the shaded region is 16x^2 – 4x^2 – x^2 = 11x^2.

    But, the length of a side of the larger square is easily found from the Pythagorean Theorem to be sqrt[(4x)^2 + (2x)^2], or sqrt(20x^2). Therefore, the area of the larger square must be 20x^2.

    Therefore, the fraction of the larger square that is shaded is (11x^2)/(20x^2), or 11/20.

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