In the figure above, ABCD is a rectangle inscribed in a circle. Angle AOD = 60° and the radius of the circle is 1. What is the ratio of the length of minor arc AD to the length of segment AD?

A. 3/π

B. 1/1

C. π/3

D. 9/8

E. π/2

(C) If you are really confused with this question, you may be able to get to the correct answer without doing any math. Certainly, minor arc AD is longer than line segment AD, but not by very much. So we can expect the answer to be slightly larger than 1. We can also expect the answer to contain π, since circumferences usually (but not always) have a π in them. Remember that π is approximately 3.14, so only choices (C), (D), and (E) are greater than 1. Choice (D) does not contain π, so it is probably not correct, and choice (E) is greater than 1.5, so it is probably too big.

This only leaves choice (C), which is correct here.

Now let’s actually solve the problem. Let’s start by determining the circumference of circle O:

Circumference = 2π

*r* = 2π × 1 = 2π.

Because angle AOD measures 60°, we can deduce two things. First, the length of arc AD will be 60/360 = 1/6 of the total circumference of O. Second, if angle AOD measures 60° and the length of side AO is equal to the length of side OD, then all three angles of triangle AOD measure 60° and the triangle is an equilateral triangle with a side length of 1. Therefore, line segment AD has a length of 1.To determine the length of arc AD, multiply the circumference by 1/6:

ength of arc AD = 2π × 1/6= 2π/6 = π/3

The desired ratio is π/3 : 1, or π/3.

The correct answer is choice (C).

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**Perhaps an explanation for you to consider:**

Calculate line segment AD.

Triangle ADC is right angled because, the diameter forms 90 at the circumference. From that you can figure out that on line AOC angle AOD and angle DOC are 60 and 120 respectively. Triangle COD is isoceles. ODC is 30 and therefore AOD is 60. Making AOD equilateral. Therefore segment AD is the radius =1. Very little calculation even though the explanation is long.

Arc AD is as you discussed above.