McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
7. Solving ax^2+bx+c=0
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Exercise 39 Page 514

Practice makes perfect
a Let's start this exploration by drawing the two squares as described in the directions.

The area of the large square with sides length a is a^2. The area of the The area of the small square with sides length b is b^2.

b Let's cut the square out of the bottom corner.

Since the area of the larger square is a^2 and the smaller is b^2, by cutting the smaller one out, the remaining area is a^2-b^2.


c Let's first draw the diagonal line from the outside to the inside corner of the new figure.
Now, we can fold and cut along the diagonal line and we should have two polygons that look like the following.

If we flip one of the two polygons over, we can match along the magenta colored side that we just cut.

From the two dimension shown, we can see that the length of the newly formed rectangle is a+b and the width is a-b.


d From Part C, we determined that the length of the rectangle is a+b and its width is a-b. Since area is the product of the length and the width, let's multiply the two binomials to write an expression for the area.
A = (a+b)(a-b)
e First we can complete the statement with what we found in Part D.
a^2-b^2 = (a+b)(a-b) From Parts A through D, we can see the statement is true. First, we subtracted the area of the smaller square from the larger square to get a polygon with area a^2-b^2. Then, we cut the paper to rearrange the parts to make a rectangle with sides (a+b) and (a-b). Therefore, the original area did not change and so has the same area as the rectangle, (a+b)(a-b).