The
The Pythagoragram Problems
Given: The figure on the left above is the familiar outline of the diagram showing
the squares drawn on the sides of a right triangle. We will call this figure a Pythagoragram.
The figure below is a Pythagoragram with three additional line segments drawn in
that connect the outlying corners of the three squares in the diagram.
The triangles created numbered I to IV (including the original right triangle.)
1)Show: the areas of the four triangles are all equal.
[Click here for a Hint: ®
]
2)Show: algebraically that the area of the entire hexagon is 2(a2 + b2) + 2ab = a2 + b2 + (a + b)2,
assuming the lengths of the two legs of the right triangle are equal to a and b.
3)Show: that the triangles II, III, and IV can be pushed together to form triangle.
4) A Pythagoragram can be categorized by the central right triangle. Pictured below is an isosceles right Pythagoragram.
Next to it are two isosceles right Pythagoragrams partially overlapping each other.
Show: that in the overlapping Pythagoragrams, the sum of the areas of the blue squares plus the areas of the two
green isosceles right triangles is equal to the sum of the areas of the two red squares.
5) If the measure of the shortest side of the central triangle in a Pythagoragram is a and the measure the other leg of the central
triangle is b, then with a proper coordinate system, the centroid of the central triangle is given by (b + (2/3)a, a + (1/3)b).
Show: that the average of the nine x-coordinates of the vertices of the Pythagoragram and the average of the nine y-coordinates
of the vertices of the Pythagoragram, are also the coordinates of the centroid of the central triangle of the Pythagoragram.
[Click here for a Hint: ®
]
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© Thomas M. Green, Contra Costa College