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You are decorating the top of a table by covering it with small ceramic tiles. The table top is a regular octagon with 15 inch sides and a radius of about 19.6 inches . What is the area you are covering?. Find the area of a regular polygon. EXAMPLE 2. DECORATING. SOLUTION. STEP 1.

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slide1

You are decorating the top of a table by covering it with small ceramic tiles. The table top is a regular octagon with 15inch sides and a radius of about 19.6inches. What is the area you are covering?

Find the area of a regular polygon

EXAMPLE 2

DECORATING

SOLUTION

STEP 1

Find the perimeter Pof the table top. An

octagon has 8 sides, so P = 8(15) = 120inches.

slide2

Find the apothem a. The apothem is height

RSof ∆PQR. Because ∆PQRis isosceles,

altitude RSbisects QP.

1

1

So,QS = (QP) = (15) = 7.5 inches.

2

2

a = RS ≈ √19.62 – 7.52 = 327.91 ≈ 18.108

Find the area of a regular polygon

EXAMPLE 2

STEP 2

To find RS, use the Pythagorean

Theorem for ∆ RQS.

slide3

1

A = aP

2

1

≈ (18.108)(120)

2

ANSWER

So, the area you are covering with tiles is about

1086.5square inches.

Find the area of a regular polygon

EXAMPLE 2

STEP 3

Find the area Aof the table top.

Formula for area of regular polygon

Substitute.

≈ 1086.5

Simplify.

slide4

A regular nonagon is inscribed in a circle with radius 4 units. Find the perimeter and area of the nonagon.

360°

The measure of central JLKis , or 40°. Apothem LMbisects the central angle, so m KLMis 20°. To find the lengths of the legs, use trigonometric ratios for right ∆KLM.

9

Find the perimeter and area of a regular polygon

EXAMPLE 3

SOLUTION

slide5

LM

MK

sin 20° =

cos 20° =

LK

LK

MK

LM

cos 20° =

sin 20° =

4

4

4 sin 20° = MK

4 cos 20° = LM

The regular nonagon has side length

s = 2MK = 2(4 sin 20°) = 8(sin 20°)

and apothem a = LM = 4(cos20°).

Find the perimeter and area of a regular polygon

EXAMPLE 3

slide6

ANSWER

So, the perimeter is P = 9s = 9(8 sin 20°) = 72 sin 20° ≈ 24.6 units,

and the area is A = aP = (4 cos 20°)(72 sin20°)≈46.3 square units.

1

1

2

2

Find the perimeter and area of a regular polygon

EXAMPLE 3

slide7

3.

360

The measure of the central angle is = or 72°. Apothem abisects the central angle, so angleis 36°. To find the lengths of the legs, use trigonometric ratios for right angle.

5

for Examples 2 and 3

GUIDED PRACTICE

Find the perimeter and the area of the regular polygon.

SOLUTION

slide8

b

sin 36° =

1

1

and the area is A = aP = 6.5 46.6

hyp

2

2

b

sin 36° =

8

8 sin 36° = b

The regular pentagon has side length = 2b= 2

(8 sin 36°) = 16 sin 36°20°

So, the perimeter is P = 5s = 5(16 sin 36°)

= 80 sin 36°

for Examples 2 and 3

GUIDED PRACTICE

≈ 46.6 units,

≈151.5 units2.

slide9

4.

for Examples 2 and 3

GUIDED PRACTICE

Find the perimeter and the area of the regular polygon.

SOLUTION

The regular nonagon has side length = 7.

So, the perimeter is P = 10 · s = 10 · 7 = 70 units

slide10

3.5

tan 18° =

a

opp

tan 18° =

adj

3.5

a =

360

tan 18°

The measure of centralis = or 36°. Apothem abisects the central angle, so angleis 18°. To find the lengths of the legs, use trigonometric ratios for right angle.

10

1

1

and the area is A = aP = 10.8 70

2

2

for Examples 2 and 3

GUIDED PRACTICE

≈10.8

≈377 units2.

slide11

5.

360°

The measure of central angle is =120°. Apothem abisects the central angle, so is 60°. To find the lengths of the legs, use the trigonometric ratios.

3

for Examples 2 and 3

GUIDED PRACTICE

SOLUTION

slide12

a

b

cos 60° =

sin 60° =

x

10

10 sin 60° =

x cos 60° =

5

b

x 0.5 =

5

x =

10

The regular polygon has side length

s = 2 = 2 (10 sin 60°) = 20 sin 60° and apothem a = 5.

for Examples 2 and 3

GUIDED PRACTICE

slide13

So, the perimeter isP = 3 s = 3(20 sin 60°)

= 60 sin 60°

= 30 3 units

1

and the area is A = aP

2

1

= ×5 30 3

2

for Examples 2 and 3

GUIDED PRACTICE

= 129.9 units2

slide14

ANSWER

Exercise 5 can be solved using special right triangles.

The triangle is a 30-60-90 Right Triangle

for Examples 2 and 3

GUIDED PRACTICE

6. Which of Exercises 3–5 above can be solved

using special right triangles?

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