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McGraw Hill Integrated II, 2012

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McGraw Hill Integrated II, 2012 View details

3. Arcs and Chords

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Exercise 39 Page 739

Consider an arc with a measure of 60^(∘). Find the length of the corresponding chord. Then, triple the arc measure.

No, see solution.

Practice makes perfect

Let's consider a circle with a radius of 6 cm and an arc with a measure of 60^(∘).

Since △ ABC is isosceles, by the Isosceles Triangle Theorem we get that m∠ A = m∠ B. Also, we have that m∠ C=60^(∘). Let's apply the Angle Sum Theorem.

m∠ A + m∠ B + m∠ C = 180^(∘)

m∠ B= m∠ A, m∠ C= 60^(∘)

m∠ A + m∠ A + 60^(∘) = 180^(∘)

▼

Solve for m∠ A

LHS-60^(∘)=RHS-60^(∘)

m∠ A + m∠ A = 120^(∘)

Add terms

2m∠ A = 120^(∘)

.LHS /2.=.RHS /2.

m∠ A = 60^(∘)

Since the three angles of △ ABC have the same measure, it is an equilateral triangle. In consequence, AB=6 cm. Next, if we triple the arc measure we will get an arc with a measure of 180^(∘), which implies the arc is a semicircle. Consequently, the corresponding chord is a diameter.

Then the length of AB is twice the radius — that is, AB=12 cm. As we can see, the length of the chord was not tripled. Thus, the answer to the question is no.

Subchapter links

Exercises p.736-740