Question 328030
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A tangram consists of the following pieces, with side measurements in proportion to the total square with side measurement of 1 unit.


5 right triangles
*[tex \LARGE \ \ \cdot\ ]2 small (hypotenuse of *[tex \Large \frac{1}{2}] and sides of *[tex \Large \frac{\sqrt{2}}{4}])
*[tex \LARGE \ \ \cdot\ ]1 medium (hypotenuse of *[tex \Large \frac{\sqrt{2}}{2}] and sides of *[tex \Large \frac{1}{2}])
*[tex \LARGE \ \ \cdot\ ]2 large (hypotenuse of *[tex \Large 1] and sides of *[tex \Large \frac{\sqrt{2}}{2}])
1 square (side of *[tex \Large \frac{\sqrt{2}}{4}])
1 parallelogram (sides of *[tex \Large \frac{1}{2}] and *[tex \Large \frac{\sqrt{2}}{4}])


So the area of an isosceles right triangle is side squared divided by 2.  For example, the area of the small triangle is *[tex \Large \frac{\frac{\sqrt{2}}{4}\,\cdot\,\frac{\sqrt{2}}{4}}{2}\ =\ \frac{\frac{2}{16}}{2}\ =\ \frac{1}{16}]


Use the dimensions above to calculate the area of the other two size triangles.


The area of the square is simply the measure of the side (given above) squared.


The parallelogram area is a little trickier.  The short dimension is *[tex \Large \frac{\sqrt{2}}{4}].  If you draw in a height line perpendicular to the long side and passing through a vertex, then you form an isosceles right triangle with hypotenuse *[tex \Large \frac{\sqrt{2}}{4}].  Multiplied by *[tex \Large \frac{\sqrt{2}}{2}] is *[tex \Large \frac{1}{4}], the measure of the altitude.  Then the area of the parallelogram is just the altitude times the long side, which arithmetic you can do for yourself.


To check your work, add up all of the areas of all of the shapes.  Since the large square has sides of 1 unit, the sum  of the areas of the shapes must be 1 square unit.


As for the second part of the problem, you didn't happen to mention which of the shapes was shape (a).  I'll just guess that it is the square.  The square has an area of *[tex \Large \frac{1}{8}] of the overall square.  Hence if you assign the value 1 to the area of the square, then the area of the overall square must be 8.  Take each of the fractional areas calculated in the previous step and multiply by 8 to get the rational area of each of the other pieces in relation to the unit area assigned to the small square.  If shape (a) is some other piece, use the denominator of its fractional area from the earlier part of the problem as the multiplier.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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