How to find the length of an arc - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

The clock at the town square has a minute hand eight feet long. How far has its tip traveled since noon if it is now 12:58 PM?

Answer

This question is asking for the length of an arc corresponding to $\frac{58}{60}$ of a circle with radius eight feet. The question can be answered by evaluating for :

$\frac{58}{60} \cdot 2 \pi r=\frac{58}{60} \cdot 2 \pi \cdot 8 \approx 48.6$

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Question

Note: Figure NOT drawn to scale

Refer to the above figure. $\angle B \cong \angle C$

Which is the greater quantity?

(a) $m \widehat{ABC}$

(b) $m \widehat{CAB}$

Answer

To compare $m \widehat{ABC}$ and $m \widehat{CAB}$ , we note that

$m \widehat{ABC} = m \widehat{AB} + m \widehat{BC}$

and

$m \widehat{CAB} = m \widehat{AB} + m \widehat{AC}$

We need to be able to compare $m \widehat{BC}$ and $m \widehat{AC}$ . If we know which of the intercepting angles is the greater, then we know which of the arcs is greater. The intercepting angles are $\angle A , \angle B$ , respectively. However, we are not given this relationship.

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Question

A giant clock has a minute hand that is six feet long. The time is now 3:50 PM. How far has the tip of the minute hand moved, in inches, between noon and now?

Answer

Every hour, the tip of the minute hand travels the circumference of a circle with radius six feet, which is

$C = 2 \pi r= 2 \pi \cdot 6 = 12 \pi$ feet.

Since it is now 3:50 PM, the minute hand made three complete revolutions since noon, plus $\frac{50}{60 } = \frac{5}{6}$ of a fourth, so its tip has traveled this circumference $3\frac{5}{6}$ times.

This is

$3\frac{5}{6} \times 12 \pi = \frac{23}{6} \times 12 \pi = 46 \pi$ feet. This is

$46 \pi \times 12 = 552 \pi$ inches.

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Question

A giant clock has a minute hand seven feet long. Which is the greater quantity?

(A) The distance traveled by the tip of the minute hand between 1:30 PM and 2:00 PM

(B) The circumference of a circle seven feet in diameter

Answer

The tip of a minute hand travels a circle whose radius is equal to the length of that minute hand, which, in this question, is seven feet long. The circumference of this circle is $2 \pi$ times the radius, or $2 \pi \times 7 = 14\pi$ feet; over the course of thrity minutes (or one-half of an hour) the tip of the minute hand covers half this distance, or $\frac{1}{2} \times 14 \pi = 7 \pi$ feet.

The circumference of a circle seven feet in diameter is $\pi$ times this diameter, or $7 \pi$ feet.

The quantities are equal.

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Question

A giant clock has a minute hand four and one-half feet in length. Since noon, the tip of the minute hand has traveled $110 \pi$ feet. Which of the following is true of the time right now?

Answer

Every hour, the tip of the minute hand travels the circumference of a circle, which here is

$C = 2 \pi r = 2 \pi \times 4 \frac{1}{2} = 9 \pi$ feet.

The minute hand has traveled $110 \pi$ feet since noon, so it has traveled the circumference of the circle

$\frac{110 \pi}{9 \pi} = \frac{110 }{9 } =12 \frac{2}{9}$ times.

Since $12 \leq 12 \frac{2}{9} \leq 12 \frac{1}{2}$ , between 12 and $12 \frac{1}{2}$ hours have elapsed since noon, and the time is between 12:00 midnight and 12:30 AM.

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Question

Acute triangle $\bigtriangleup ABC$ is inscribed in a circle. Which is the greater quantity?

(a) $m \overarc {ABC} - m \overarc {ACB}$

(b) $m \overarc {A B} - m \overarc {A C}$

Answer

Examine the figure below, which shows $\bigtriangleup ABC$ inscribed in a circle.

By the Arc Addition Principle,

$m \overarc {ABC} = m \overarc {AB}+ m \overarc {BC}$

and

$m \overarc {ACB} = m \overarc {AC}+ m \overarc {BC}$

Consequently,

$m \overarc {ABC} - m \overarc {ACB}$

$= (m \overarc {AB}+ m \overarc {BC} ) - ( m \overarc {AC}+ m \overarc {BC})$

$= m \overarc {AB}+ m \overarc {BC} - m \overarc {AC}- m \overarc {BC}$

$= m \overarc {A B} - m \overarc {A C}$

The two quantities are equal.

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