Transcript Section 3

Transparency 11-3
5-Minute Check on Lesson 11-2
Find the area of each figure. Round to the nearest tenth if necessary.
1.
2.
12
A = 83.1 units²
A = 198 units²
3.
60°
4.
8
9
10
A = 234 units²
13
A = 39 units²
8
5. Trapezoid LMNO has an area of
55 square units. Find the height.
h
14
h = 5 units
6.
B
Rhombus ABCD has an
area of 144 square inches. Find AC if BD = 16.
C
Standardized Test Practice:
A
8 in
B
9 in
C
16 in
Click the mouse button or press the
Space Bar to display the answers.
D
A
18 in
D
Lesson 11-3
Areas of Regular Polygons and
Circles
Objectives
• Find areas of regular polygons
– A = ½ Pa
where P is the perimeter of the polygon
and a is the length of the apothem
• Find areas of circles
– A = πr²
Vocabulary
• Apothem – perpendicular bisector from center to
side of a regular polygon
Area of Regular Polygons
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
x
x
a
x
x
½x ½x
y
y
y
a
y
½y
½y
y
Hexagon Example Area
A=½*P*a
Hexagon: A = ½ * 6 * y * a
Area of Circles
S
Radius (r)
Center
Circle Area
A = π * r2 = π * (ST)2
r is a radius (ST)
S is the Center
T
Circle Example 1
Find the area of circle S
S
A = πr²
= π(8)² = 64π square units
8
T
Circle Example 2
R
Find the area of circle S, if RT = 20
S
A = πr²
but no r !
r=½d
A = π(d/2)² = π(20/2)²
= 100π square units
T
Regular Polygons Area Example 1
Find the area of the hexagon,
if the apothem is 4√3
A=½Pa
a
a = 4√3 and P = 6·8 = 48
A = ½ (48) (4√3)
= 96√3 square units
8
Regular Polygons Area Example 2
Find the area of the hexagon
A=½Pa
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° and
we have a 30-60-90 triangle problem.
a = ½ (10) √3 = 5√3
So A = ½ (60) (5√3)
= 150√3 square units
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.
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.
Find the area of a regular pentagon with a perimeter
of 120 inches.
Answer: about
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.
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:
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
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
.
Next, find the height of the triangle, XS.
Since m
3.5
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.
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:
– 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