EXAMPLE 1

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EXAMPLE 1 Find angle measures in a regular polygon a. m AFB In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure. SOLUTION AFB is a central angle, so m AFB = , or 72 ° . 360 ° 5

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In the diagram, ABCDE is a regular pentagon inscribed in F . Find each angle measure. 360 °. AFB is a central angle, so m AFB = , or 72 °. 5. a. m AFB. Find angle measures in a regular polygon. EXAMPLE 1. SOLUTION. - PowerPoint PPT Presentation

Transcript of EXAMPLE 1

Page 1: EXAMPLE 1

EXAMPLE 1 Find angle measures in a regular polygon

a. m AFB

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

AFB is a central angle, so m AFB = , or 72°.

360°

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EXAMPLE 1 Find angle measures in a regular polygon

b. m AFG

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

FG is an apothem, which makes it an altitude of isosceles ∆AFB. So, FG bisects AFB and m AFG = m AFB = 36°.1

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EXAMPLE 1 Find angle measures in a regular polygon

c. m GAF

In the diagram, ABCDE is a regular pentagon inscribed in F. Find each angle measure.

SOLUTION

The sum of the measures of right ∆GAF is 180°.So, 90° + 36° + m GAF = 180°, and m GAF = 54°.

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GUIDED PRACTICE for Example 1

In the diagram, WXYZ is a square inscribed in P.

1. Identify the center, a radius, an apothem, and a central

angle of the polygon.

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GUIDED PRACTICE for Example 1

SOLUTION

center – P

radius – PY or XP

an apothem – PQ

central angle – XPY

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GUIDED PRACTICE for Example 1

2. Find m XPY, m XPQ, and m PXQ.

m XPY is a central angle so m XPY = 4

360

= XPQ = 45°

SOLUTION

m XPY = 90°

QP is an apothem, which make it an altitude of isosceles ∆ XPQ so QP bisects XPY and m XPQ = m 1

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GUIDED PRACTICE for Example 1

90° + 45 + m PXQ = 180 and PXQ = 45°

The sum of measures of right ∆ PXQ is 180° so