1. Unit 10 Measuring Length and Area Arun Kammanadiminti and Jenny Oh

2. In this unit, you learn about… ●Area of common shapes like rectangles, parallelograms, trapezoids, rhombuses, kites, and triangles ●Heron’s Formula and the formula to find the area of an equilateral triangle ●How to find the perimeter and area of similar figures ●The area of regular polygons like hexagons, octagons, etc., which will come in handy for Unit 11 ●The area of shapes like circles and sectors ●Using geometric probability Overview

3. Area of Shapes ●Triangle - 1/2×base×height ●Rectangles - base×height ●Parallelogram - base×height ●Trapezoid - 1/2×height×(b 1 +b 2 ) ●Rhombus - 1/2×d 1 ×d 2 ●Kite - 1/2×d 1 ×d 2 b h b h b2b2 b1b1 h h b d1d1 d2d2 d1d1 d2d2 CONNECTION: Some of these will show up in Unit 11 to figure out surface areas and volumes of pyramids and prisms. COMMON MISTAKES: For the area of the rhombus and kite, don’t forget to multiply by ½ !

4. Heron’s Formula & Equilateral Triangles Heron’s Formula- A=√(s×(s-a)×(s-b)×(s-c)) ●A stands for the area of the triangle ●s is the semiperimeter ●a, b, and c are the side lengths of the triangle ●To find the semiperimeter (s), you do a+b+c Equilateral Triangles- A= s 2 ×√(3) ●To find this you have to start with the triangle area formula ●The height would then be s√(3) ●A=1/2 × s√(3) × s= s 2 ×√(3) 2 4 s s√(3) 2 2 24 s2s2 For the semiperimeter, DIVIDE the sum of the sides by 2, not 3.

5. Area of Shapes and Equilateral Triangles Examples: A.In a game of pool, there is an equilateral triangle item called a rack. If 15 balls fit in one rack, and the radius of each ball is 2.25 inches, what is the approximate length of the rack? B.There is a structure in a garden. It is an odd structure, but your contracter still wants to know what the area is if you look at it from the top. Figure it out.(There is a square, a triangle, and a trapezoid.) 2 ft 1 ft 1.5 ft 1 ft

6. Explanation A.15πr 2 = s 2 ×√(3) 15π(2.25) 2 = s 2 ×√(3) 75.94π= s 2 ×√(3) 303.75π= s 2 ×√(3) 175.37π= s 2 550.94= s 2 23.4≈ s Answer: ~23 inches 4 4 4 B.s 2 +(1bh)+(b 1 +b 2 )h 2 2 +(1(1)(2))+(1.5+2)1 4+1+1.75 = 6.75 ∴ 6.75 Answer: 6.75 feet 2 2 2 2

7. Perimeters & Area of Similar figures If the two figures are similar, then….. -the ratio of the side lengths = A:B The ratio of perimeters = A:B The ratio of areas = A 2 :B 2 ∴ Ratio of the perimeters = Ratio of the side length -the ratio of areas = E:F The ratio of side length = √E:√F The ratio of perimeters is √E:√F ∴ Ratio of the side length = Ratio of perimeters ஃ = SO. IN CONCLUSION Remember- The area is squared

8. Area of Regular Polygons Area for all regular polygons: A= 1 ×a×P 2 ●a is apothem ●P is perimeter To find the apothem, you have to find the central angle(=360÷ # of sides). In a hexagon each angle measure is 60°. ( hexagon = 6 sides 360÷ 6 = 60°) The apothem always bisects the side and the angle. If the base is 6, then the apothem would be 3√(3). (Use 30, 60, 90 triangle) And the perimeter would be 36. So, the area is 36×3√(3)×1 which is 54√(3). 2 x x (x√3) 2 COMMON MISTAKES: These will actually come in handy so remember how to figure out the area for regular polygons. 2

9. Area of Regular Polygons and Perimeters and Areas of Similar Figures Examples: A.In the movie Star Wars, the bad guys’ fighter jets have regular hexagons for wings. Say they are regular hexagons and each side is 5 feet. What is the total area of both wings? Answer: 130 ft 2 B.A soccer ball has a diameter of 22 centimeters. A golf ball has an area of 18 square centimeters. Find the golf ball's radius. (This is if you are looking at it from the side.) Answer: 4.25 cm

10. Explanation apothem 30 60 x√3 = 2.5 √3 x = 2.5 ft 2x = 5 ft 5 a= apothem p= perimeter B.You just have to do √(18) to figure it out. A.Can solve the question by using 30, 60, 90 triangle A = ½ ・ a ・ p A = ½ x 2.5√3 x (5x6) A = 75√3 A = 75√3 x 2 (Trying to figure out 2 wings, not 1 ) A = 130 feet 2

11. Circles Arc length central angle = arc length 360 2πr IF you shorten it, THEN it will be N° = 2πr 360 Area of sector area of sector central angle area of circle = 360 sector= shaded part of a circle sector arc You will be working with this in Unit 11 for spheres, cylinders, and cones, so remember it.

12. Probability is represented by a number between 0 and 1. Geometric probability = favorable outcomes total outcomes To figure out geometric probability, you have to find the total area of the figure. That value goes on the bottom of your fraction. The shaded part is the part that goes on the top. For example, in the example below, the square’s area would go on the bottom and the triangles area would go on top. What is the probability of landing something in the triangle? Answer: 50% Using Geometric Probability 2 ft

13. Circles and Geometric Probability Examples: A.The following figure is an easy darts board. The bigger circle has a radius of 5 and the small one has a radius of 2. Find the probability of throwing a dart in the shaded area. Answer: 84 %

14. Recap Just in case you forgot in the last few minutes, this unit is about… ●Area’s of shapes like rectangles, parallelograms, trapezoids, rhombuses, kites, and triangles ●Heron’s Formula and the formula to find the area of an equilateral triangle ●How to find the perimeter and area of similar figures ●The area of regular polygons, which will come in handy for Unit 11 ●The area of circles and sectors ●Using geometric probability