find areas of regular polygons. find areas of circles. bet ya didn’t see that coming!
TRANSCRIPT
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11.3Areas of Polygons and Circles.
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Objectives:
Find areas of regular polygons. Find areas of circles.
Bet ya didn’t see THAT coming!
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THINGS TO REMEMBER.
Regular polygons, have all sides CONGRUENT.
Properties of central, and inscribed angles.
The degree measure of one circle = 360°
30° – 60° – 90°
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Areas of POLYGONS.
When thinking of regular polygons, imagine them being inscribed in a circle. This will give you the basis to find initial angles for different problems. When a regular polygon is inscribed into a circle, from the center of the circle to a vertex of the polygon is a radius of the circle.
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Now for the new stuff.
APOTHEM: A segment, from the center of a circle, that is perpendicular to a side of the inscribed polygon.
N
B
A C
DSegment ND is an APOTHEM
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CN
B
A C
D
In ∆ ABC, Segments NB and NC are congruent, because they’re radii, making ∆CNB an isosceles triangle. When all of the other radii are drawn they break the big triangle into three congruent mini triangles.
Now to find the areas.
A= ½ bh, OR, A= ½ sa
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CN
B
A C
D
To calculate the area of the triangles, you use the normal polygon area formula, using the apothem for the height, and the side for the base.
Now to find the areas.
A= ½ bh, OR, A= ½ sa
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CN
B
A
D
Notice, the area of one small triangle is ½ sa square units. so the area of the BIG triangle is 3( ½ sa)
Now notice that the perimeter of the BIG triangle is 3s units. we can substitute P (perimeter) for 3s in the area formula.
C
So A= 3( ½ sa) becomes A= ½ Pa
This formula can be used for the area of any regular polygon.
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EXAMPLE #1( Remember the properties of a circle)
A
B
C
DE
Find area of a regular pentagon with a perimeter of 50 centimeters
The central angles of a regular pentagon are all congruent. So, the measure of each angle is 360/5 or 72
P
Q
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EXAMPLE #1( Remember the properties of a circle)
A
B
C
DE
Find area of a regular pentagon with a perimeter of 50 centimeters
P
Segment PQ is an apothem of pentagon ABCDE. It bisects angle CPB and is a perpendicular bisector of segment BC. Therefore, measure of angle BPQ = ½(72) or 36. Since the perimeter is 50 cm each side is 10cm, which makes segment BQ = 5 cm.
Q
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AC
DE
P
Q
B
Write a trig ratio to find the length of segment PQ.
Tan of angle BPQ = opposite/adjacent Tan 36° = 5/PQ (PQ) tan 36° = 5 PQ = 5/tan 36°
PQ ≈ 6.9.
Area = ½Pa A = ½(50)(6.9)
A = 172.5
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Area of a Circle
A = πr² (r = radius)
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4
Find the area of the shaded region. Assume that the triangle is equilateral.First find the are of the circle.
A = πr² = π(4)² ≈ 50.3
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4
To find the area of the triangle, use 30° - 60° - 90° rules.
First find the length of the base.
The hypotenuse is 4, so MP is 2.
MB is 2√3. AB is 4√3.
Next, find the height, MC.
MC = 2√3(√3) or 6
Use the formula to find the area of a triangle.
A = ½bh = ½(4√3)(6) ≈ 20.8
The area of the shaded region is 50.3 – 20.8 or 29.5
A B
C
M
P
60°
60°
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Homework
Assignment:Pg. 613 #8-22 evens, 26, 27, 30-33, 39 - 44