Question

A length of wire is bent into a semicircular arc. The length of EF is 21 chentimeters. Find the length of the wire. Use 22/7 as an approximation for п

Answer 1

L = πr = (22/7) * 6.727 = 21.14 centimeters.

Explanation:

Let's assume that the radius of the semicircular arc is 'r' centimeters.

The formula for the circumference of a circle is C = 2πr, where π is approximately equal to 22/7.

Since we are dealing with a semicircle, we need to divide the circumference by 2 to get the length of the wire. Therefore, the length of the wire is:

L = C/2 = (2πr)/2 = πr

Now, we need to find the value of 'r'. We know that the length of EF is 21 centimeters, which is half of the circumference of the semicircle.

So, we can write:

EF = (C/2) = (πr)

Substituting the value of π as 22/7, we get:

21 = (22/7) * r

Multiplying both sides by 7/22, we get:

r = 21 * (7/22) = 6.727 centimeters (approx.)

Therefore, the length of the wire is:

L = πr = (22/7) * 6.727 = 21.14 centimeters.



Answer 2

Answer: 65.94 cm

Explanation:

Since the wire is bent into a semicircular arc, we can use the formula for the circumference of a circle to find the length of the wire.

The formula for the circumference of a circle is C = 2πr, where C is the circumference and r is the radius.

In this case, the length of EF is equal to the diameter of the circle, so the radius is half of that, or 10.5 cm.

Plugging this into the formula, we get:

C = 2 x (22/7) x 10.5

C = 2 x 3.14 x 10.5

C = 65.94

Therefore, the length of the wire is approximately 65.94 cm.