Find areas of regular polygons. Find areas of circles.

Apothem- A segment drawn from the center of a regular polygon perpendicular to a side of the polygon

Apothem=a or BD

Triangle ABC is an isosceles triangle, since the radii are congruent. If all of the radii were drawn, they would separate the hexagon into 6 nonoverlapping congruent isosceles triangles

BA

C

D

Area of a triangle is A=1/2(base)(height)

The height of the triangle is equal to a and the base is s then ½ bh= ½ sa

The total area of the hexagon is 6( ½ sa)

Perimeter is 6s so A=6( ½ sa) becomes A= ½ Pa

BA

D

C

a

s

Find the area of a regular hexagon with a perimeter of 60

Central angles of a regular pentagon are all congruent. Therefore, the measure of each angle is 360/6 or 60. BD is an apothem of the hexagon. It bisects angle ABC and is a perpendicular bisector to AC. So measure of angle CBD= ½ (60) or 30. Since the perimeter of hexagon is 60 each is 10 units long and DC=5 units.

Tan CBD= DC/BD Tan 30 = 5/BD(BD) tan 30 =5BD=5/tan 30 BD 8.7

A= ½ Pa

= ½ (60)(8.7)A=261

Area of a circle=πr

Find the area of the shaded region

2

4

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.

A= πr

= π(4)

50.3

4

2

2

To find the area of the triangle, use properties of 30-60-90 triangles. First, find the length of the base. The hypotenuse of ABC is 4, so BC is 2 √3. Since EC= 2(BC), EC=4 √3.

4

60

A

BC E

D

Next, find the height of the triangle, DB. Since m DCB is 60, DB= 2 √3(√3) or 6.

A= ½ bh = ½ (4 √3)(6) 20.8The area of the shaded region is

50.3-20.8 or 29.5 square units to the nearest tenth.

60

A

BC E

D

2 √3

Pg. 613 #8 – 22, 26 – 32 evens

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