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Learning Math Home
Measurement Session 1: Solutions
 
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A B C 
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Solutions for Session 1, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5


Problem C1

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Problem C2

All five polygons have the same area, since they are made up of the same three smaller polygons. Since there is no overlapping, the area of all five is the same as the sum of the area of the three triangles.

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Problem C3

First, you need to arbitrarily choose a unit with which to measure the perimeters. One way to do this is to choose the length of a leg of the smaller triangle as your unit. Whatever unit you choose, you will discover that the square has the smallest perimeter. The rectangle's perimeter is slightly larger. The parallelogram, triangle, and trapezoid are tied for the largest perimeters.

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Problem C4

Standard units provide a frame of reference that can always be relied on. Although we can say, "This one is three times longer than that one," standard units provide everyone with an equivalent value for "that one." With standard units, anyone measuring the same item could say, "This one is six times longer than 1 cm."

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Problem C5

Small triangle: Using the Pythagorean theorem to determine the length of the hypotenuse, we can write the following:

c2 = a2 + b2 = 12 + 12 = 2, or
c =

The length of the hypotenuse is , or approximately 1.414 units.

Medium triangle: The length of the legs is equal to the hypotenuse of the smaller triangle, or . So, to determine the hypotenuse, we can write the following:

c2 = a2 + b2 = 2 + 2 = 4
So the length c is equal to 2 units.

Remember, our unit is the length of one leg of the small triangle. So the perimeters are as follows:

Square:
4 • , or approximately 5.656 units

Rectangle that is not a square:
2 • 1 + 2 • 2 = 6 units

Parallelogram that is not a rectangle:
2 • + 2 • 2, or approximately 6.828 units

Triangle:
2 + 2 + (2 • ), or approximately 6.828 units

Trapezoid:
2 • + 1 + 3, or approximately 6.828 units

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