Area of Regular Polygons and Circles

continued …

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?

= 247.80 in2SOHCAHTOA

How do we find the apothem(a)?360/ 5 = 72o

72/ 2 = 36o

Tan (36o) = (6/a) a = (6/(tan(36o))) = 8.26 cm

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 Circle

Can we use the area of a

regular polygon?

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

If a circle has an area of A square units and a

radius of r units, then: A = 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

= (99)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 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

Area of a Circle – Area of a Square:(490.87 in2) – (312.5 in2)

178.37 in2

Areas of Irregular Figures

Objective: TLW find areas of irregular figures on and off the coordinate plane. SOL G.14b

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?

But how do we solve for the

area?

The area of an Irregular figure is the:

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?

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) bh = (1/2) (8 cm) (4 cm) = 16 cm2

Area of ☐ BFDE = (sides)2

= (8 cm)2

= 64 cm2

Area of Δ BEA = (1/2) bh = (1/2) (6 cm) (8 cm) = 24 cm2

Now add it all together to get your Area!

A = 104

cm2

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 = lw

= 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) bh =

(1/2) (6) (3√3) =

9√3 in2

Area of the figure= (114 in2 – 9√3 in2)

+ 4.5π in2

= 112.5 in2

YOUR TURN!!!!

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 Triangle:= (1/2) bh= (1/2)(6)(3)= 9 u2

Area of Irregular Figure:= 49 u2 + 9 u2

= 58 u2