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# EXAMPLE 1 - PowerPoint PPT Presentation

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.

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## PowerPoint Slideshow about ' EXAMPLE 1' - sandra-neal

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Presentation Transcript

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

360°

AFB is a central angle,somAFB = , or 72°.

5

a. m AFB

Find angle measures in a regular polygon

EXAMPLE 1

SOLUTION

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

b. m AFG

FG is an apothem, which makes it an altitude ofisosceles ∆AFB. So, FGbisectsAFB andmAFG = mAFB = 36°.

1

2

Find angle measures in a regular polygon

EXAMPLE 1

SOLUTION

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

c. m GAF

The sum of the measures of right ∆GAF is 180°.

So, 90° + 36° + m GAF = 180°, andm GAF = 54°.

Find angle measures in a regular polygon

EXAMPLE 1

SOLUTION

In the diagram, WXYZis a square inscribed in P.

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

for Example 1

GUIDED PRACTICE

an apothem – PQ

central angle – XPY

for Example 1

GUIDED PRACTICE

SOLUTION

center –P

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

360

m XPY is a central angle so

m XPY =

4

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

1

m XPY= 90°

2

= XPQ = 45°

for Example 1

GUIDED PRACTICE

SOLUTION

90° + 45 + m PXQ = 180 and PXQ = 45°

for Example 1

GUIDED PRACTICE

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