Question

If a pipe is \[2{\text{ inches}}\] diameter what is the circumference?

Answer 1

Hint: We are given the diameter of the circle and we have to find the circumference of this circle. Now, diameter is the distance between two points on a circle passing through the center. Hence, the diameter of the circle is 2 times its radius. Using this formula, we can find the radius of the given circle. Now, the circumference of circle is given by the formula
\[ \Rightarrow {\text{Circumference}} = 2\pi r\]

Complete step-by-step solution:
In this question, we are given a circle and its diameter. We have to find its circumference.
Diameter \[ \Rightarrow D = 2{\text{ inches}}\]
Then we need to find radius,
\[ \Rightarrow {\text{r}} = ?\]
Now, first of all let us understand what circle, diameter and radius are.
So, a circle is the collection of all the points lying in a plane at a given distance from the center point.
The distance between the center and all the points on the circle is known as radius.
The distance between two points on the circle passing from the center is known as the diameter of the circle.
Now, the diameter is double in length as compared to the radius.
So, we can say that the diameter of a circle is equal to 2 times its radius.
\[ \Rightarrow D = 2r\]
Substituting value of diameter \[ \Rightarrow 2r = 2\]
Divide both sides by 2, we get
Radius of circle \[ \Rightarrow r = 1{\text{ inch}}\]
Hence, the radius of the circle having diameter \[2{\text{ inches}}\] is \[1{\text{ inch}}\].
Now, circumference of a circle is given by the formula
\[ \Rightarrow \text{circumference} = ?\]
\[ \Rightarrow \text{circumference} = 2\pi r = 2 \times 3.14 \times 1 = 6.28{\text{ inches}}\]
Hence, the circumference of the circle whose diameter is \[2{\text{ inches}}\] is \[6.28{\text{ inches}}\].
So, the correct answer is “\[6.28{\text{ inches}}\]”.

Note: Here the circumference of a circle formula can be written as \[\text{circumference} = D\pi \] , where \[D\] is the diameter and we also know that the radius is the half of the diameter and diameter is the twice of the radius and we directly substitute the value of \[D\] and we get the solution and don’t forget to mention the unit.