The length of the side of the square is twice the length of the radius.
B
b
If the radius doubles to 2r, then the length of the side of the square is 4r. Remember to square both the coefficient and r.
C
c
Use the ratios from parts A and B to start the table, look for patterns to complete it.
D
d
Notice the pattern of the ratios in the Part C table and consider why it is always the same.
A
a
Ï€/4
B
b
Ï€/4
C
c
Radius
Area of Circle
Area of Square
Ratio
r
Ï€ r^2
4r^2
Ï€/4
2r
4 π r^2
16r^2
Ï€/4
3r
9Ï€ r^2
36r^2
Ï€/4
4r
16Ï€ r^2
64r^2
Ï€/4
5r
25Ï€ r^2
100r^2
Ï€/4
6r
36Ï€ r^2
144r^2
Ï€/4
D
d
Given a circle inscribed in a square, the ratio of the area of the circle to the area of the square is always π4.
Practice makes perfect
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 2r and we can write an expression for the circle's area and the square's.
\begin{array}{cc}
A_\text{{\color{#0000FF}{circle}}}&=& {\color{#0000FF}{\pi r^2}}
\end{array}
We can use the formula for the square's area to write an expression in terms of r for the area of the square.
\begin{array}{cc}
A_\text{{\color{#009600}{square}}} &=& s^2 \\
&=& (2r)^2\\
&=&{\color{#009600}{ 4r^2}}
\end{array}
Now, we can write the ratio of the areas of the circle's to the square's and simplify.
π r^2/4r^2 = π/4
b
Let's do the same thing we did in Part A, but use 2r for the radius.
From this picture, we can see the side length of the square is 4r and we can write an expression for area of the circle with a radius of 2r.
\begin{array}{cc}
A_\text{{\color{#0000FF}{circle}}}&=& \pi (2r)^2
&=& 4 \pi r^2
\end{array}
We can use the formula for the square's area to write an expression in terms of r for the area of the square.
\begin{array}{cc}
A_\text{{\color{#009600}{square}}} &=& s^2 \\
&=& (4r)^2\\
&=&{\color{#009600}{ 16r^2}}
\end{array}
Now, we can write the ratio of the areas of the circle's to the square's and simplify.
4 π r^2/16r^2 = π/4
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
r
Ï€ r^2
2r
( 2r)^2=4r^2
π r^2/4r^2 = π/4
2r
π ( 2r)^2 = 4 π r^2
4r
( 4r)^2 = 16r^2
π r^2/16r^2 = π/4
3r
Ï€ ( 3r^)2= 9Ï€ r^2
6r
( 6r)^2 =36r^2
9 π r^2/36r^2=π/4
4r
Ï€ ( 4r)^2= 16Ï€ r^2
8r
( 8r)^2 = 64r^2
16 π r^2/64r^2=π/4
5r
Ï€ ( 5r)^2 = 25Ï€ r^2
10r
( 10r)^2= 100r^2
25 π r^2/100r^2=π/4
6r
Ï€ ( 6r)^2 = 36Ï€ r^2
12r
( 12r)^2 = 144r^2
36 π r^2/144r^2=π/4
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 π4.
Radius
Area of Circle
Area of Square
Ratio
r
Ï€ r^2
4r^2
Ï€/4
2r
4 π r^2
16r^2
Ï€/4
3r
9Ï€ r^2
36r^2
Ï€/4
4r
16Ï€ r^2
64r^2
Ï€/4
5r
25Ï€ r^2
100r^2
Ï€/4
6r
36Ï€ r^2
144r^2
Ï€/4
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 π4. This makes sense because the area of the circle and the square are increasing at the same rate.
