Buzz
The area of a circle is the size of its surface that we can observe, denoted as A, in square units (such as cm2, m2...). Calculating the area of a circle is quite simple, we just need to apply the formula: A = π .r2 (where: A is the area of the circle; π is a constant (π = 3.14); r is the radius). However, for the problem of calculating the area of a circle with a circumference of 12.56 cm, what should we do to get it right? Students, follow our article to learn how to solve this exercise.
Calculate the area of a circle with a circumference of 12.56 cm
How to solve a problem calculating the area of a circle with a circumference of 12.56 cm
Method: In this exercise, you need to remember two extremely important pieces of knowledge:
- The formula for calculating the area of a circle (as mentioned above).
- The formula for calculating the circumference of a circle: C = d.π or C = r x 2π.
Where:
- C is the circumference of the circle.
- d is the diameter of the circle.
- r is the radius of the circle.
- π = 3.14
Solution:
- Given circumference => Find radius: r = C : 2 : π.
- After finding the radius => Calculate the area of the circle.
Detailed Solution:
The radius of the circle is:
12.56 : 2 : 3.14 = 2 (cm).
The area of the circle is:
3.14 x 22 = 12.56 (cm2).
Answer: 12.56 (cm2).
Similar Exercise (Students do it themselves to reinforce knowledge)
1. Calculate the area of the circle, given the circumference:
a) C = 37.68 cm
b) C = 15.7 m
c) C = 56.52 dm
2. A bicycle wheel has a circumference of 43.96 cm, what is the area of the wheel?
By learning how to solve exercises calculating the area of a circle given a circumference of 12.56 cm, students must have found it easier to solve similar exercises, right? Hopefully, our concise instructions will be helpful for students to learn Geometry more easily.
In addition, students can also refer to other methods of calculating the area of a circle such as calculating the area of a circle with diameter d, which is a quite common type of exercise. Students should try to understand it well to apply it easily to specific problems.
