Do NOW 04-15-2014
 Take out your notes from yesterday and work on the
two area of a polygon problems using the A = (1/2)Pa
formula.
 Homework: Irregular Polygons Worksheet
TEST FRIDAY!!!!
Area of a Regular Polygon:
Let’s begin with an example using a regular pentagon:
What is the Perimeter of MNOPQ when QP = 12 inches?
P = # of sides  side length
= 5 (12 in.)
= 60 inches
What is the Area?
How do we find the apothem(a)?
360/ 5 = 72o
72/ 2 = 36o
Tan (36o) = (6/a)
a = (6/(tan(36o))) = 8.26 cm
= 247.80 in2
How can we use all of this information
to find the area of a circle? Can we get
to the area of a circle from the equation
for Perimeter?
Area of a
If a circle has an area of A square units and a radius of r




units, then: A =
Yes!
A = (1/2) Pa
PC = 2πr
Plug in!
 A = (1/2)(2πr)a
 What’s our “a”? It’s our radius!!
 A = (1/2)(2πr)(r)
= πr2
πr2
Let’s Try One!
 Let the circle shown below have a radius of 9 centimeters, what
is the perimeter and the area of the circle?
P = 2πr
= 2π(9)
= 18π cm
≈ 56.55 cm
A = πr2
= π(9)2
= (99)π cm2
= 81π cm2
≈ 254.47 cm2
Your Turn!!
What if we want to know the area of the
shaded region around an inscribed polygon?
 Let r = 12.5 inches
 What do we need to calculate??
Area of a Circle:
= πr2
= π(12.5)2
= 156. 25π in2
≈ 490.87 in2
Area of a Circle – Area of a Square:
(490.87 in2) – (312.5 in2)
178.37 in2
Area of the Square:
= s2
s2 + s2 = (2r)2, where r = 12.5
2s2 = (25)2 = 625
s2 = 312.5 in2
Square root both sides!
s = 17.68 inches
So we’ve already discussed
regular polygons … what if we
have an irregular figure? What
does irregular mean?
An Irregular Figure is a figure that:
 Is not comprised of equal side and angle
measurements
 Can it be classified like the other
polygons we’ve studied?
 NO
 Do irregular figures exist in our everyday
lives? What are some examples?
The area of an
 Sum of all it’s distinct parts
 What do I mean by distinct parts??
 Non-overlapping figures that can be combined
to create the irregular figure
 These parts can be made up of rectangles,
squares, triangles, circles, and other polygons!
 Let’s think about our original figure, how could
we break it up?
How would breaking our image up help us to find the
area?
is the:
Let’s evaluate that same figure!
 Do we have all of the
measurements we need?
 What is missing?
 The base of our bottom Δ

How could we find this?

The Pythagorean Theorem!!
 AE = 6 cm
 Now what can we do?
Let’s evaluate that same figure!
Area of Δ CFB = (1/2) bh
= (1/2) (8 cm) (4 cm)
= 16 cm2
Area of ☐ BFDE = (sides)2
= (8 cm)2
= 64 cm2
Area of Δ BEA = (1/2) bh
= (1/2) (6 cm) (8 cm)
= 24 cm2
Now add it all together to get your Area!
What if we have a figure that is
missing a portion of it’s area?
Consider the image from the previous slide …
 Can this figure be
separated into other
figures?
 Yes
 What are they?
Consider the image from the previous slide …
I’m going to set up the problem for the
area of a rectangle + a semi-circle,
minus the triangle. Also, assume the
triangle is an equilateral.
Area of the rectangle = lw
= 19 in 6 in
= 114 in2
Area of the semi-circle = (1/2)πr2
= (1/2)π (3 in)2
= 4.5π in2
Area of the Δ = (1/2) bh
= (1/2) (6) (3√3)
= 9√3 in2
Area of the figure
= (114 in2 – 9√3 in2) + 4.5π in2
= 112.5 in2
What if we were on the coordinate plane?
What do we need to know about a figure to evaluate it?
 Consider the image to your left!
 How can we break this image into two
portions that we can evaluate with our
prior knowledge?
Area of Trapezoid:
= (1/2) h(b1 + b2)
= (1/2)(7) (8+6)
= 49 u2
Area of Irregular Figure:
= 49 u2 + 9 u2
= 58 u2
Area of Triangle:
= (1/2) bh
= (1/2)(6)(3)
= 9 u2