Proof Problem #2

proof2

The orange square is constructed using lines from midpoints to opposite vertices.

Prove that the orange square is 1/5 the area of the blue square.

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5 thoughts on “Proof Problem #2

  1. I approximated the 1/5 area by assuming each side of the original square has length of two.
    Area of original square = 2^2 = 4 sq units.
    Hypotenuse of right triangle made by one full side and half of adjacent side = sq rt 5
    Smallest angle of that right triangle = 26.565 degrees
    Side length of orange square = sq rt 5 – cos 26.565 – sin 26.565 = 0.894427191
    Area of orange square = 0.894427191^2 = 0.8
    Ratio of areas = 0.8/4 = 1/5

  2. We have four right angle triangles, visible.
    Each is a part of a larger triangle, leaning on the big square’s edge, demonstrating each trapezoid is actually same triangle + two other forming a rectangle.
    Hence, on each edge of the big square we have 1+1+2=4 same triangles.
    Around the orange square we have, then, 4*4=16 triangles.
    The orange square itself is equal in area to four triangles.
    All in all—the big square is 20 triangles in areas, four are with the orange one, which makes its area 4/20=1/5.
    QED.

  3. Assume that the side of the blue square is “a”; assume that the side of the orange square is “z”
    Area of blue square = a*a
    Area of triangle (where the line joining the midpoint to the vertex is the hypotenuse) is = 1/2 * base * height
    =1/2*a*a/2= (a*a)/4

    Hypotenuse in this triangle (Pythagorus theorem) = √(a*a + a/2*a/2) = √5a*a/4 = √5*(a/2)

    Area of parallelogram with orange square = base * height = hypotenuse * side of orange square = (√5*a*z)/2…….(1)

    Area of parallelogram with orange square = area of big blue square – area of two triangles
    = a*a – 2*a*a/4
    =a*a/2…….(2)

    Thus, using (1) & (2)

    √5*a*z/2 = a*a/2
    √5*z=a
    z=a/√5

    Area of orange square = z*z = a*a/5 =1/5 of area of square

  4. The white area of the large square consists of four (congruent) trapeziums and four right-angled triangles. If we put together one trapezium and one right-angled triangle they form a square exactly congruent with the orange square. Hence from the four trapeziums and the four right-angled triangles of the white area we can form four squares each congruent with the orange square. Hence the white area equals 4 times the area of the orange square and so the total area of the large square equals 5 times the area of the orange square.

  5. Take the square as 2 x 2. The large triangles are each 1 x 2 on the perpendicular sides with hypotenuses of (5)^1/2. Consider the small triangles formed in each corner. They are similar to the large triangles, but have hypotenuses of 1 unit, a ratio of 1/(5)^1/2 compared to the larger triangles, so each measurement is reduced by that factor and the area of each small triangle is then 1/5. There are 4 small triangles with a total area of 4/5 which is exactly 1/5 of the total area of the large square. These four small triangles equal the square in question. This can be seen as follows: The length of the hypotenuse of a large triangle is (5)^1/2 but consists of 1 long side of a small triangle + 1 short side of a small triangle and the length of a side of the square in the middle. This means the length of the side of the square in the middle must be 2/(5)^1/2, the length of 1 long side of a small triangle and twice the length of 1 short side of a small triangle. Thus 4 small triangles will fit in the square exactly.

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