To find the radius of the circle that could represent the base of the vase with the area constraints given, we need to determine the radii corresponding to the lower and upper bounds of the area.
We know that the area [tex]\( A \)[/tex] of a circle is given by the formula:
[tex]\[ A = \pi r^2 \][/tex]
where [tex]\( r \)[/tex] is the radius. We need to find the radius that will produce an area within the bounds of 135 cm² and 155 cm².
1. Determine the radius for the lower bound area:
Given:
[tex]\[ A = 135 \, \text{cm}^2 \][/tex]
[tex]\[ \pi = 3.14 \][/tex]
Using the area formula:
[tex]\[ 135 = 3.14 \times r^2 \][/tex]
To isolate [tex]\( r \)[/tex], divide both sides by 3.14:
[tex]\[ r^2 = \frac{135}{3.14} \][/tex]
Now compute [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 \approx 43.0 \][/tex]
Taking the square root of both sides to find [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{43.0} \][/tex]
[tex]\[ r \approx 6.56 \][/tex]
So, the radius for the lower bound area is approximately 6.56 cm.
2. Determine the radius for the upper bound area:
Given:
[tex]\[ A = 155 \, \text{cm}^2 \][/tex]
[tex]\[ \pi = 3.14 \][/tex]
Using the same area formula:
[tex]\[ 155 = 3.14 \times r^2 \][/tex]
To isolate [tex]\( r \)[/tex], divide both sides by 3.14:
[tex]\[ r^2 = \frac{155}{3.14} \][/tex]
Now compute [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 \approx 49.36 \][/tex]
Taking the square root of both sides to find [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{49.36} \][/tex]
[tex]\[ r \approx 7.03 \][/tex]
So, the radius for the upper bound area is approximately 7.03 cm.
Therefore, any circle with a radius between approximately 6.56 cm and 7.03 cm can represent the base of Liliana's vase.
