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Transparency 11-3. 5-Minute Check on Lesson 11-2. Find the area of each figure. Round to the nearest tenth if necessary. 2. 3. 4.

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5-Minute Check on Lesson 11-2

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5 minute check on lesson 11 2

Transparency 11-3

5-Minute Check on Lesson 11-2

  • Find the area of each figure. Round to the nearest tenth if necessary.

  • 2.

  • 3. 4.

  • 5. Trapezoid LMNO has an area of 55 square units. Find the height.

  • 6. Rhombus ABCD has an area of 144 square inches. Find AC if BD = 16.

12

9

11

A = 83.1 units²

A = 198 units²

11

60°

9

8

9

A = 234 units²

3

13

6

A = 39 units²

10

8

h

14

h = 5 units

B

C

Standardized Test Practice:

A

D

8 in

9 in

16 in

18 in

D

A

B

C

D

Click the mouse button or press the Space Bar to display the answers.


Lesson 11 3

Lesson 11-3

Areas of Regular Polygons and Circles


Objectives

Objectives

  • Find areas of regular polygons

    • A = ½ Pawhere P is the perimeter of the polygonand a is the length of the apothem

  • Find areas of circles

    • A = πr²


Vocabulary

Vocabulary

  • Apothem – perpendicular bisector from center to side of a regular polygon


Area of regular polygons

Area of Regular Polygons

x

x

x

Polygons Area

A = 1/2 * P * a

P is the perimeter (# of sides * length)

a is apothem (perpendicular bisector)

Regular (all sides equal!)

Octagon example: A = ½ * 8 * x * a

x

x

a

x

x

½x

½x

y

y

y

Hexagon Example Area

A = ½ * P * a

Hexagon: A = ½ * 6 * y * a

a

y

y

½y

½y


Area of circles

Area of Circles

S

Radius (r)

T

Center

Circle Area

A = π * r2 = π * (ST)2

r is a radius (ST)

S is the Center


Circle example 1

Circle Example 1

Find the area of circle S

S

8

T

A = πr²

= π(8)² = 64πsquare units


Circle example 2

Circle Example 2

R

Find the area of circle S, if RT = 20

S

A = πr²

but no r !

T

r = ½ d

A = π(d/2)² = π(20/2)²

= 100πsquare units


Regular polygons area example 1

Regular Polygons Area Example 1

Find the area of the hexagon, if the apothem is 4√3

A = ½ P a

a

a = 4√3 and P = 6·8 = 48

A = ½ (48) (4√3)

= 96√3square units

8


Regular polygons area example 2

Regular Polygons Area Example 2

Find the area of the hexagon

A = ½ P a

10

P = 6·10 = 60, but no a!

The dotted line is the hypotenuse to the apothem !

10

Since the interior angle is 120°, then the ∆’s angle is 60° andwe have a 30-60-90 triangle problem.

a = ½ (10) √3 = 5√3

So A = ½ (60) (5√3)

= 150√3square units


Example 3 1a

Example 3-1a

Find the area of a regular pentagon with a perimeter of 90 meters.

Apothem: The central angles of a regular pentagon are all congruent. Therefore, the measure of each angle is (360/5) or 72. GF is an apothem of pentagon ABCDE. It bisects EGD and is a perpendicular bisector of ED. So, mDGF = ½(72) = 36°. Since the perimeter is 90 meters, each side is 18 meters and FD = 9 meters.

Write a trigonometric ratio to find the length of GF.


Example 3 1a1

Example 3-1a

Multiply each side by GF.

Divide each side by tan 36°.

Use a calculator.

Area:

Area of a regular polygon

Simplify.

Answer:The area of the pentagon is about 558 m².


Example 3 1b

Answer:about

Example 3-1b

Find the area of a regular pentagon with a perimeter of 120 inches.


Example 3 2a

Example 3-2a

An outdoor accessories company manufactures circular covers for outdoor umbrellas. If the cover is 8 inches longer than the umbrella on each side, find the area of the cover in square yards.

The diameter of the umbrella is 72 inches, and the cover must extend 8 inches in each direction. So the diameter of the cover is 8 + 72 + 8 or 88 inches. Divide by 2 to find that the radius is 44 inches.

Area of a circle

Convert 6082.1 square inches to square yards, by dividing by 1296.

Answer:The area of the cover is 4.7 square yards to the nearest tenth.


Example 3 2b

Example 3-2b

A swimming pool company manufactures circular covers for above ground pools. If the cover is 10 inches longer than the pool on each side, find the area of the cover in square yards.

Answer:


Example 3 3a

To find the area of the triangle, use properties of 30-60-90 triangles. First, find the length of the base. The hypotenuse of so RS is 3.5 and . Since .

Example 3-3a

Find the area of the shaded region. Assume that the triangle is equilateral. Round to the nearest tenth.

The area of the shaded region is the difference between the area of the circle and the area of the triangle. First, find the area of the circle.

Area of a circle


Example 3 3a1

Next, find the height of the triangle, XS. Since m 3.5

Example 3-3a

Area of a triangle

Use a calculator.

Answer:The area of the shaded region is 153.9 – 63.7 or 90.2 square centimeters to the nearest tenth.


Example 3 3b

Example 3-3b

Find the area of the shaded region. Assume that the triangle is equilateral. Round to the nearest tenth.

Answer: 46.0 in2


Summary homework

Summary & Homework

  • Summary:

    • A regular n-gon is made up of n congruent isosceles triangles

    • The area of a circle of radius r units is πr² square units

  • Homework:

    • pg 613-615; 14-22


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