Persevering with Problems The figure on the right comprises a circle and a square with the circle touching the square at the midpoints of its sides. a. Determine the length of one side of the square. b. Use the formula A = πr^2 to calculate the area of the circle and A = s^2 for the area of the square. State the ratio of the area of the circle to that of the square in simplest terms. c. Fill in the table: | # | Area of Circle (units^2) | Side Length of Square | Area of Square | Ratio (Area of Circle : Area of Square) | |---|--------------------------|------------------------|----------------|------------------------------------------| | 1 | 12.57 | 2 | 4 | | | 2 | 50.27 | 4 | 16 | | | 3 | 113.10 | 6 | 36 | | d. What can be inferred about the relationship between the circle's area and the square's area? Lesson 3: Multiplication and Division of Monomials

a. To determine the length of one side of the square when the circle is touching the square at the midpoints of its sides, we need to know that the diameter of the circle will be equal to the length of the side of the square. Therefore, the radius of the circle is half the length of the side of the square. Let's denote the side length as "s". Hence, if "r" is the radius of the circle, r = s/2.

b. To calculate the area of the circle, we plug the radius into the area formula for a circle, A = πr^2. Since r = s/2, we get A_circle = π(s/2)^2 = (π/4)s^2. The area of the square is A_square = s^2. The ratio of the area of the circle to that of the square is A_circle / A_square = (π/4)s^2 / s^2 = π/4. Simplifying, we have the ratio in simplest terms: π:4.

c. To fill in the table, we use the given area of the circle and the side length of the square to calculate the missing ratio:

For #1: Ratio (Area of Circle : Area of Square) = 12.57 / 4 = 3.1425 which is approximately π to four decimal places. For #2: Ratio (Area of Circle : Area of Square) = 50.27 / 16 = 3.1419 which is approximately π to four decimal places. For #3: Ratio (Area of Circle : Area of Square) = 113.10 / 36 = 3.1411 which is approximately π to four decimal places.

d. From the above ratios, it can be inferred that regardless of the size of the circle and the square, as long as the circle touches the square at the midpoints of its sides, the ratio of the area of the circle to that of the square will approximate the value of π:4. This is due to the fact that the circle's diameter is equal to the side length of the square, and hence, its radius is half that length.

Extra: Understanding Area Formulas: The concepts involved are fundamental geometric properties. The area of a square is calculated by squaring the length of one of its sides (A = s^2), where "s" represents the side length. For a circle, the area is defined by the formula A = πr^2, where "r" is the radius of the circle, and π (Pi) is a mathematical constant approximately equal to 3.14159.

Circle within a Square: When a circle is perfectly inscribed in a square (touches the square at the midpoints of the sides), the diameter of the circle is the same as the side length of the square. Therefore, you can relate the dimensions of the circle directly to the dimensions of the square, which is why the areas of both shapes are related by a constant ratio (π:4), irrespective of their sizes.

Ratios and Proportionality: A ratio compares the relative sizes of two or more values. When the sizes of geometric shapes change, their areas do not always scale directly with the dimensions. In the case of the circle within the square, despite the change in size, the ratio of their areas remains constant because the shapes are scaling proportionally.

Multiplication and Division of Monomials: While the problem doesn’t directly address the multiplication and division of monomials, it is relevant to note that these are algebraic expressions composed of a single term. Operations with monomials involve using the properties of exponents, which are fundamental in manipulating and simplifying algebraic expressions. Monomial multiplication and division would come into play when dealing with the algebraic representations of areas and simplifying them, as we did with the ratio of the areas where exponents were involved (squared terms).