A presentation by the Group 5 Period 2 Society We are proud to bring to you

Gonzaga vs. Area and Sektor Problems by Tyler Golian, Evan Connolly, Aidan Kenney and Elijah H. Davis By Elijah Davis

A= ½ ap A= ½ (6 √ 3)(72) A= 216 √ 3 6√36√ Find the area of a regular polygon with perimeter 72 Find side 72/6 =12 Use angle formula (6-2)180/6 = 120 Make Triangle Plug numbers into Area equation

26. Theorem 11-6 The area of a regular polygon is equal to half theproduct of the apothem and the perimeter. A= ½ ap A= ½ ( ½ )(3 √ 3) A= ¼ (3 √ 3) A= ¾ √ 3 Find the Area the Equilateral Triangle Make Triangle Plug numbers into the area formula

27.Find The Area of the Regular Polygon 6 sides means a Hexagon Make a Area = half the apothem X the perimeter A= ½ AP A= X 6 A= or

28. Area of Triangle = ½ B × H A = 2 × 4 =8 90 / 360 × π 4 2 = = is the area of the striped section Find the area of the striped section

29. Find the area of the shaded region 6 is radii Find area of semi-circle 2.Find area of Triangle 3.Subtract triangle from circle Area of semi circle Area of Triangle A= ½ Pi R*R A= ½ BH A=1/2 Pi 6*6 A= ½ 6*6 A= ½ PI 36 A= ½ 36 A= 18Pi A= 18 Answer: 6 6

30. Find Area of shaded region is the distance between // lines and also the radius 1.Divide into 4 parts: 2 sectors and 2 triangles 2.Find Area of sector, add to area of triangle 2.5 make sure denominators are equal 3 Multiply by 2 Sector A= x/360 r*r A= 60/360 2*2 A= 1/6 *4 A= 4 /6 = 2 /3 Triangle A= ½ Bh A= ½ 2 *1 A= ½ 2 A= *2 Answer

33. What is the chance of cutting a 6inch wire and both piece be at least 1 inch long 6 -2 = 4 The probability of cutting two pieces of wire and both pieces are at least 1 in long is 4/6 = 2/3 4

31. Find the ratio of the regions I and II. (x=x marks congruent angles) Ratio of perimeters 3:7 Ratio of areas 9:49 So, final ratio: 9:49-9 9:40 *Theorem 11-7: If the scale factor of two similar figures is a:b 1)the ratio of perimeters is a:b 2) the ratio of areas is

32.The area of the parallelogram ABCD is 48 square inches, and DE= 2EC Find the area of: 16 3 Find Sides divide 48 by 3x= 16 Divide 48 by 16 = 3 Height of ABE is the same as a side.

The Armory (Area of regular P-gon) Triangles Triangles A=(n-2)180/360 (Angle of a regular P-gon) A=½ Pi R*R (Area of semi-circle Area of arc = x/360* 2(pi*r) A= ½ Bh (Area of a Triangle If the scale factor of two similar figures is a:b A Triangle inscribed in a semi-circle is a right triangle Probability= want/total 1)the ratio of perimeters is a:b 2) the ratio of areas is

Move List Inscribed= Within the object Circumscribed= the object is within it. Regular= angles are the same Apothem= line perpendicular from side to center Sector= region bound by 2 radii and an arc of the circle

All right class, lesson’s over. Now get out there and…

You Don’t Mess with Pikachu