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Entry Task Find the unknown side lengths in each special right triangle. 1. a 30°-60°-90° triangle with hypotenuse 2 ft. 2. a 45°-45°-90° triangle with leg length 4 in. 3. a 30°-60°-90° triangle with longer leg length 3m. Areas of Regular Polygons 10.3. Learning Target. I can….
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Entry Task Find the unknown side lengths in each special right triangle. 1. a 30°-60°-90° triangle with hypotenuse 2 ft 2. a 45°-45°-90° triangle with leg length 4 in. 3. a 30°-60°-90° triangle with longer leg length 3m
Areas of Regular Polygons 10.3 Learning Target I can…...
Today’s Group Task • Find the area of a regular “n-gon” with a side length of x. If you need other side lengths, define a variable and use it. • In other words, create a formula that will work for ALL regular polygons. (a triangle, a hexagon, a square)
Vocabulary The center of a regular polygonis equidistant from the vertices. The apothemis the distance from the center to a side. A central angle of a regular polygonhas its vertex at the center, and its sides pass through consecutive vertices. Regular pentagon DEFGH has a center C, apothem BC, and central angle DCE.
radius apothem
10.3 Area of Regular Polygons m∠1= m∠2= m∠3= 360∕6 = 60° 60∕2 = 30° 180-90-30=60° 1 2 3 Example: Find the measure of each numbered angle
10.3 Area of Regular Polygons 6 4 5 m∠4= m∠5= m∠6= 360∕8 = 45° 45∕2 = 22.5° 180-90-22.5=67.5° Example: Find the measure of each numbered angle
Let’s see how this works… 10 A = 1/2ap A = ½(6.88)(50) A = 172 sq.units 6.88
10.3 Area of Regular Polygons 8 in 12.3 in Example: Find the area of the regular polygon
Example: Find the area of a regular pentagon with 7.2ft sides and a 6.1ft radius. p = (5)(7.2ft) = 36ft A = ½ ap = ½ (4.92ft)(36ft) = 88.6ft2 6.1ft a2 + b2 = c2 a2 + 3.62 = 6.12 a2 + 12.96 = 37.21 a2 = 24.25 a = 4.92 6.1ft a 3.6 ft
10.3 - Area of Regular Polygons Use Pythagorean Theorem! 18 ft 23.5 ft 21.7 ft 9 ft Example: Find the area of the regular polygon
Remember! Special Triangle shortcuts for 30-60-90 and 45-45-90 triangles. Things to Remember…. 2s 45 s 30 45 s s 60
Example 2: Find the apothem of a reg. hexagon with sides of 10mm. m1 = 360/6 = 60 m2 = ½ (60) = 30 m3 = 10mm 1 60 2 3 a = √3 • short side a = √3 • (5mm) a = 5√3 = 8.66 30 30-60-90 Δ shortcut a Could you find the area of it now?? 60 5mm
12 60° 60° 30° 60° Find the area of this one! ….. A circle has 360°… ….. How many degrees would the top angle of each Δ have? 60° ….. Since the Δs are isosceles, what are the measures of the base angles? 30° ….. If the apothem is an angle bisector, then what is the measure of the small top angle? The short side = 6 The apothem = 6√3 A = 1/2ap A = ½(6√3)(72) = 216√3 (exact) A = 374.12 (approx.)