note: figure not drawn to scale.
the circle shown has center o, circumference 144π, and diameters $\overline{pr}$ and $\overline{qs}$. the length of arc ps is twice the length of arc pq. what is the length of arc qr ?
a) 24π
b) 48π
c) 72π
d) 96π

Question

Accepted Answer

# Explanation:
## Step1: Find the total circumference and arc relationships
The circumference of the circle is $ 144\pi $. Since $ PR $ and $ QS $ are diameters, they intersect at the center $ O $, forming vertical angles. Let the measure of arc $ PQ $ be $ x $. Then the measure of arc $ PS $ is $ 2x $. Notice that arc $ PQ $ and arc $ SR $ are equal (vertical angles), and arc $ PS $ and arc $ QR $ are equal (vertical angles) or we can use the fact that the sum of arcs around a circle is the circumference. Wait, actually, the arcs $ PQ $, $ QR $, $ RS $, and $ SP $ make up the whole circle? Wait, no, since $ PR $ and $ QS $ are diameters, the circle is divided into four arcs: $ PQ $, $ QR $, $ RS $, $ SP $. But $ PQ $ and $ SR $ are vertical angles, so their arc lengths are equal. Similarly, $ PS $ and $ QR $ are vertical angles, so their arc lengths are equal? Wait, no, let's re-examine. The diameters $ PR $ and $ QS $ intersect at $ O $, so the central angles: $ \angle POQ $, $ \angle QOR $, $ \angle ROS $, $ \angle SOP $. Since $ PR $ and $ QS $ are diameters, $ \angle POQ + \angle QOR = 180^\circ $ (since $ PR $ is a straight line), and similarly for the other angles. Wait, maybe a better approach: the length of an arc is proportional to its central angle. Let the central angle for arc $ PQ $ be $ \theta $, then the central angle for arc $ PS $ is $ 2\theta $. Since $ PR $ and $ QS $ are diameters, the sum of arcs $ PQ $ and $ PS $ should be a semicircle? Wait, no, $ PQ $ and $ QR $ are along diameter $ PR $? Wait, no, $ PR $ is a diameter, so the arc from $ P $ to $ R $ is a semicircle, length $ \frac{144\pi}{2} = 72\pi $. Similarly, arc from $ Q $ to $ S $ is a semicircle, length $ 72\pi $. Wait, the problem says "the length of arc $ PS $ is twice the length of arc $ PQ $". Let's consider the arcs around the center. Let arc $ PQ = x $, then arc $ PS = 2x $. Now, since $ PR $ and $ QS $ are diameters, the sum of arc $ PQ $ and arc $ QR $ is a semicircle (since $ PR $ is a diameter), so $ x + \text{arc } QR = 72\pi $. Similarly, the sum of arc $ PS $ and arc $ SR $ is a semicircle, so $ 2x + \text{arc } SR = 72\pi $. But also, arc $ PQ $ and arc $ SR $ are vertical angles, so arc $ PQ = $ arc $ SR = x $. Wait, no, maybe not. Wait, the total circumference is $ 144\pi $, so a semicircle (half the circumference) is $ 72\pi $. Let's look at the arcs: arc $ PQ $ and arc $ PS $ are adjacent arcs forming a semicircle? Wait, $ QS $ is a diameter, so the arc from $ Q $ to $ S $ is a semicircle. So arc $ PQ + $ arc $ PS = $ semicircle? Wait, $ P $ to $ Q $ to $ S $: $ PQ $ and $ PS $ meet at $ P $ and $ Q $, $ S $. Wait, maybe the key is that arc $ PQ $ and arc $ QR $ are on diameter $ PR $, so their sum is a semicircle (length $ 72\pi $)? No, that's not right. Wait, let's start over.

The circumference $ C = 144\pi $. Let the length of arc $ PQ = x $, then arc $ PS = 2x $. Since $ PR $ and $ QS $ are diameters, the arcs $ PQ $ and $ SR $ are equal (vertical angles), and arcs $ PS $ and $ QR $ are equal (vertical angles). Wait, no, vertical angles: $ \angle POQ $ and $ \angle ROS $ are vertical angles, so their arcs $ PQ $ and $ SR $ have the same length. $ \angle QOR $ and $ \angle SOP $ are vertical angles, so their arcs $ QR $ and $ PS $ have the same length. Ah, that's the key! So arc $ PQ = $ arc $ SR = x $, and arc $ QR = $ arc $ PS = 2x $. Now, the total circumference is the sum of all four arcs: $ PQ + QR + RS + SP = x + 2x + x + 2x = 6x $. Wait, but the circumference is $ 144\pi $, so $ 6x = 144\pi $? No, that can't be, because a diameter divides the circle into two semicircles, each of length $ 72\pi $. Wait, maybe my mistake is in the number of arcs. Let's see: the circle has center $ O $, diameters $ PR $ and $ QS $. So the circle is divided into four arcs: $ PQ $, $ QR $, $ RS $, $ SP $. But $ PR $ is a diameter, so the arc from $ P $ to $ R $ is a semicircle, which is $ PQ + QR $, so $ PQ + QR = 72\pi $. Similarly, $ QS $ is a diameter, so $ QR + RS = 72\pi $, $ RS + SP = 72\pi $, $ SP + PQ = 72\pi $. Now, the problem states that arc $ PS $ (which is $ SP $) is twice arc $ PQ $, so $ SP = 2 \times PQ $. Let $ PQ = x $, then $ SP = 2x $. From $ SP + PQ = 72\pi $ (since $ QS $ is a diameter, arc $ SQ $ is a semicircle? Wait, no, $ QS $ is a diameter, so the arc from $ Q $ to $ S $ is a semicircle, which is $ QR + RS $, or $ QP + PS $? Wait, $ Q $ to $ S $ through $ P $: $ QP + PS $, which is a semicircle, so $ QP + PS = 72\pi $. So $ x + 2x = 72\pi $, so $ 3x = 72\pi $, so $ x = 24\pi $. Then, since $ PQ + QR = 72\pi $ (because $ PR $ is a diameter, arc $ PR $ is a semicircle), so $ x + QR = 72\pi $, so $ 24\pi + QR = 72\pi $, so $ QR = 48\pi $? Wait, no, that contradicts. Wait, maybe I mixed up the arcs. Let's define the arcs properly:

- Arc $ PQ $: from $ P $ to $ Q $
- Arc $ QR $: from $ Q $ to $ R $
- Arc $ RS $: from $ R $ to $ S $
- Arc $ SP $: from $ S $ to $ P $

Since $ PR $ is a diameter, the sum of arcs $ PQ $ and $ QR $ is a semicircle (length $ \frac{144\pi}{2} = 72\pi $), so $ PQ + QR = 72\pi $.

Since $ QS $ is a diameter, the sum of arcs $ QR $ and $ RS $ is a semicircle, so $ QR + RS = 72\pi $.

Similarly, $ RS + SP = 72\pi $ (since $ PR $ is a diameter, arc $ RS + SP $ is the other semicircle? Wait, no, $ PR $ is a straight line, so the arc from $ R $ to $ P $ through $ S $ is a semicircle, so $ RS + SP = 72\pi $.

And $ SP + PQ = 72\pi $ (arc from $ S $ to $ Q $ through $ P $ is a semicircle, since $ QS $ is a diameter), so $ SP + PQ = 72\pi $.

Given that $ SP = 2 \times PQ $ (length of arc $ PS $ is twice arc $ PQ $; arc $ PS $ is $ SP $). Let $ PQ = x $, then $ SP = 2x $.

From $ SP + PQ = 72\pi $: $ 2x + x = 72\pi $ → $ 3x = 72\pi $ → $ x = 24\pi $. So $ PQ = 24\pi $, $ SP = 48\pi $.

Now, from $ PQ + QR = 72\pi $: $ 24\pi + QR = 72\pi $ → $ QR = 72\pi - 24\pi = 48\pi $. Wait, but let's check with another equation. From $ RS + SP = 72\pi $: $ RS + 48\pi = 72\pi $ → $ RS = 24\pi $. Then from $ QR + RS = 72\pi $: $ 48\pi + 24\pi = 72\pi $, which works. So arc $ QR $ is $ 48\pi $? Wait, but the options include $ 48\pi $ as option B. Wait, but let's re-examine the problem statement: "the length of arc $ PS $ is twice the length of arc $ PQ $". Arc $ PS $ is from $ P $ to $ S $, which is $ SP $, so that's correct. So arc $ PS = SP = 2x $, arc $ PQ = x $. Then using the semicircle $ PQ + PS = x + 2x = 3x = 72\pi $ (since $ QS $ is a diameter, the arc from $ Q $ to $ S $ through $ P $ is a semicircle), so $ x = 24\pi $, then arc $ QR = 72\pi - x = 72\pi - 24\pi = 48\pi $. So the answer should be $ 48\pi $, which is option B.

Wait, but let's check the total circumference. The four arcs: $ PQ = 24\pi $, $ QR = 48\pi $, $ RS = 24\pi $, $ SP = 48\pi $. Sum: $ 24\pi + 48\pi + 24\pi + 48\pi = 144\pi $, which matches the circumference. So that's correct.

## Step2: Verify the arc lengths
We found that arc $ PQ = 24\pi $, arc $ PS = 48\pi $, arc $ QR = 48\pi $, arc $ RS = 24\pi $. The sum is $ 24\pi + 48\pi + 24\pi + 48\pi = 144\pi $, which is the given circumference. So the length of arc $ QR $ is $ 48\pi $.

# Answer:
B) $ 48\pi $

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Explanation:

Step1: Find the total circumference and arc relationships

Answer: