McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
2. Division Properties of Exponents
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Exercise 60 Page 404

a We are comparing the area of a circle inscribed in a square to the area of the square. The length of the side of the square is dependent on the area of the circle. Let's look at the relationship.
From the diagram, we can see the side length of the square is and we can write an expression for the circle's area and the square's.
We can use the formula for the square's area to write an expression in terms of for the area of the square.
Now, we can write the ratio of the areas of the circle's to the square's and simplify.


b Let's do the same thing we did in Part A, but use for the radius.
From this picture, we can see the side length of the square is and we can write an expression for area of the circle with a radius of
We can use the formula for the square's area to write an expression in terms of for the area of the square.
Now, we can write the ratio of the areas of the circle's to the square's and simplify.


c For this part of the exercise, we need to complete the table. Keep in mind, from Parts A and B, that the length of the side of the square is twice that of the radius.
Radius Area of Circle Side of Square Area of Square Ratio
d For this part, we are asked to draw a conclusion. This means we need to look for patterns. From the table in Part C, we can see that the ratio is always
Radius Area of Circle Area of Square Ratio

One conclusion we can make is that the ratio of the area of circle inscribed in a square to the area of the square is This makes sense because the area of the circle and the square are increasing at the same rate.