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Unit 7 Triangles and Area

Unit 7 Triangles and Area. This unit begins to classify triangles. It addresses special right triangles and the Pythagorean Theorem (again). This unit covers the area equations required for quadrilaterals, and for circles. It also covers the area of any polygon using an apothem.

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Unit 7 Triangles and Area

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  1. Unit 7 Triangles and Area • This unit begins to classify triangles. • It addresses special right triangles and the Pythagorean Theorem (again). • This unit covers the area equations required for quadrilaterals, and for circles. • It also covers the area of any polygon using an apothem. • This unit differentiates between perimeter and area similarity ratios, and concludes with Geometric Probability.

  2. Standards • SPI’s taught in Unit 7: • SPI 3108.1.1 Give precise mathematical descriptions or definitions of geometric shapes in the plane and space. • SPI 3108.1.2 Determine areas of planar figures by decomposing them into simpler figures without a grid. • SPI 3108.4.3 Identify, describe and/or apply the relationships and theorems involving different types of triangles, quadrilaterals and other polygons. • SPI 3108.4.6 Use various area of triangle formulas to solve contextual problems (e.g., Heron’s formula, the area formula for an equilateral triangle and A = ½ ab sin C). • SPI 3108.4.7 Compute the area and/or perimeter of triangles, quadrilaterals and other polygons when one or more additional steps are required (e.g. find missing dimensions given area or perimeter of the figure, using trigonometry). • SPI 3108.4.11 Use basic theorems about similar and congruent triangles to solve problems. • SPI 3108.4.12 Solve problems involving congruence, similarity, proportional reasoning and/or scale factor of two similar figures or solids. • SPI 3108.5.1 Use area to solve problems involving geometric probability (e.g. dartboard problem, shaded sector of a circle, shaded region of a geometric figure). • CLE (Course Level Expectations) found in Unit 7: • CLE3108.2.3 Establish an ability to estimate, select appropriate units, evaluate accuracy of calculations and approximate error in measurement in geometric settings. • CLE 3108.3.1 Use analytic geometry tools to explore geometric problems involving parallel and perpendicular lines, circles, and special points of polygons. • CLE 3108.4.6 Generate formulas for perimeter, area, and volume, including their use, dimensional analysis, and applications. • CLE 3108.4.8 Establish processes for determining congruence and similarity of figures, especially as related to scale factor, contextual applications, and transformations. • CLE 3108.5.1 Analyze, interpret, employ and construct accurate statistical graphs. • CLE 3108.5.2 Develop the basic principles of geometric probability.

  3. Standards • CFU (Checks for Understanding) applied to Unit 7: • 3108.1.5 Use technology, hands-on activities, and manipulatives to develop the language and the concepts of geometry, including specialized vocabulary (e.g. graphing calculators, interactive geometry software such as Geometer’s Sketchpad and Cabri, algebra tiles, pattern blocks, tessellation tiles, MIRAs, mirrors, spinners, geoboards, conic section models, volume demonstration kits, Polyhedrons, measurement tools, compasses, PentaBlocks, pentominoes, cubes, tangrams). • 3108.4.9 Classify triangles, quadrilaterals, and polygons (regular, non-regular, convex and concave) using their properties. • 3108.4.10 Identify and apply properties and relationships of special figures (e.g., isosceles and equilateral triangles, family of quadrilaterals, polygons, and solids). • 3108.4.11 Use the triangle inequality theorems (e.g., Exterior Angle Inequality Theorem, Hinge Theorem, SSS Inequality Theorem, Triangle Inequality Theorem) to solve problems. • 3108.4.12 Apply the Angle Sum Theorem for polygons to find interior and exterior angle measures given the number of sides, to find the number of sides given angle measures, and to solve contextual problems. • 3108.4.20 Prove key basic theorems in geometry (i.e., Pythagorean Theorem, the sum of the angles of a triangle is 180 degrees, characteristics of quadrilaterals, and the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length). • 3108.4.28 Derive and use the formulas for the area and perimeter of a regular polygon. (A=1/2ap) • 3108.4.43 Apply the Pythagorean Theorem and its converse to triangles to solve mathematical and contextual problems in two- or three-dimensional situations. • 3108.4.44 Identify and use Pythagorean triples in right triangles to find lengths of an unknown side in two- or three-dimensional situations. • 3108.4.45 Use the converse of the Pythagorean Theorem to classify a triangle by its angles (right, acute, or obtuse). • 3108.4.46 Apply properties of 30° - 60° - 90° and 45° - 45° - 90° to determine side lengths of triangles. • 3108.5.2 Translate from one representation of data to another (e.g., bar graph to pie graph, pie graph to bar graph, table to pie graph, pie graph to chart) accurately using the area of a sector. • 3108.5.3 Estimate or calculate simple geometric probabilities (e.g., number line, area model, using length, circles).

  4. Pythagorean Theorem • In a RIGHT triangle (a triangle with one 90 degree angle), the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse (the longest side). • a2 + b2 = c2 a c b

  5. Pythagorean Triple • A set of non-zero whole numbers a, b, and c that satisfy the equation a2 + b2 = c2 • Some triples include: • 3,4,5…5,12,13…8,15,17…7,24,25 • Recognize 3,4,5 and 5,12,13they are the most commonly used triples on standardized tests! • If you multiply each number in a Pythagorean triple by the same whole number, the 3 numbers that result also form a Pythagorean triple. • For example 3,4,5 x the whole number 2 equals 6 (3 x 2), 8 (4 x 2), and 10 (5 x 2), a new triple

  6. Converse of the Pythagorean Theorem • If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other 2 sides, then the triangle is a right triangle. • In other words, if you calculate a2 + b2 and it does in fact = c2, then in fact you can conclude that it is a right triangle

  7. Proofs of the Pythagorean Theorem • The Bride’s Chair •  The area of a square is Side x Side (or side squared) • Here, the side is a + b. So the area of the square is (a+b)2 • The area of a triangle is ½ b x h • This makes sense because the area of a rectangle is base times height, so the area of a triangle would be half of that • Continuing on then, the area of the triangle in the square then is ½ a x b • There are 4 of these triangles, so the area would be 4 x ½ x a x b • Adding the area of the center square (c2), we get (a+b)2 = 2ab + c2, which simplifies to a2 + b2 = c2

  8. A Proof Discovered by a High School Student • There are easily over 400 proofs of the Pythagorean Theorem • This one was discovered by a high school student (Jamie deLemos) in 1995. • We will learn that the area of a trapezoid is the top base plus the bottom base divided by 2 (in other words we average the bases) x the height. • Here, that would be (2a + 2b)/2 x ab • The area of a triangle is ½ b x h • This would be 2a·b/2 + 2b·a/2 + 2·c²/2 for all of the triangles • If you set them equal to each other, and reduce, you get a2 + b2 = c2 • Remember, all of this is used to prove the LENGTHS of the sides, not area

  9. Area of a Triangle • The area of a triangle is half the product (multiply) of a base and the corresponding height • So, A = 1/2 x b x h • The base of a triangle can be any of its sides. The corresponding height is the length of the altitude to the line containing that base. • Right triangles are unique, in that you can pick a base that automatically has a corresponding height h b

  10. Find the area of an Isosceles Triangle • Here, the height is not so easily seen as it is in a right triangle. To find the height, we draw a line perpendicularly from the base to the highest point of the triangle • To calculate the height, we can project a Right Triangle, and determine the length of the 3rd side. • 102+h2 = 122 • h2= 122-102 • h= 6.6 • Therefore, the area of this triangle is ½ x 6.6 x 20 • Or, area = 66 m2 12 m h 20 m

  11. Obtuse Triangles • If the square of the length of the longest side of a triangle is greater than the sum of the squares of the length of the other 2 sides, the triangle is obtuse • If C2>A2+B2 then the triangle is obtuse A B C

  12. Acute Triangles • If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other 2 sides, the triangle is acute. • If C2<A2+B2 then the triangle is acute B A C

  13. Assignment • Page 495-96 7-22,24-32 • Page 497 36-42 • Worksheet Practice 8-1 • Worksheet 7-1

  14. Unit 7 Quiz 1 Find the missing variable: • A = 6, B= 9, C = ? • A = 4, B = ? C = 11 • A = ? B = 4, C = 13 • A = 2, B = 13, C = ? • A =7, B = ? C = 14 • D = 12, L = 18, H = ? • E = 11, D = 11, L = 20, H = ? • H = 14, D = 20, L = ? • L = 6, H = 4, D = ? • D = 22, L = 30, H = ? A C B D E H L

  15. Area of a Parallelogram • The area of a parallelogram is the base times the height (b x h) • Here is why  • Remember the area of a rectangle is base times height also Basically you can cut a parallelogram in half, and put the two halves next to each other to make a rectangle Height Base

  16. Perimeter • The perimeter of a polygon is the sum of the lengths of the sides –it is the measure of how far around it is –like the perimeter of your back yard, you would measure how far around your back yard it is.

  17. Perimeter • For any “regular” polygon, having “n-sides”, the perimeter is n x the length of one side (length = s) • So a square would be 4s, a pentagon would be 5s and so on • Otherwise, just add the lengths of the sides for a given irregular polygon

  18. Area • For simple polygons which have 90 degree angles –like squares and rectangles, the area is equal to side x side • For a square, we write this: S x S –or S2 • For a rectangle we write this: L x W S W S L

  19. Example • What if you have an irregular shape like below? • Pick out different shapes and add them up • 12 cm + 8 cm + 4 cm = 24 cm2 6 cm this is 6 cm x 2 cm, for a total area of 12 cm This is (2+2) cm x 2 cm, for a total of 8cm This is 2 x 2 for 4 cm 2 cm

  20. Area of a Trapezoid • The area of a trapezoid is ½ times the height times the top base (b1) + the bottom base (b2) • In other words, you average the top and bottom (add them and divide by 2) and multiply by the height

  21. Area of a Trapezoid • Here is how the area of a trapezoid (1/2h(b1+b2) works:  • Take a trapezoid, and make a mirror image • Rotate the mirror image • Now we have a parallelogram again, with the same height, but the base is b1+b2 So the area of our new parallelogram is base times height, or (b1+b2)xh. Since we only need one of them (we only need 1 trapezoid, not 2), we use 1/2h(b1+b2) B1 B1 B2 B2 B1+B2

  22. Example • To find the area of the trapezoid calculate 1/2h(b1+b2) • (b1+b2) = 5 + 7 = 12 • We don’t know the height • We do have a 30/60/90 right triangle. • We know the short side is 2 (7-5=2) • So the long side (which is also the height) is 2√3 • Therefore the area is ½(2 √3)(5+7) = 12 √3 5 m 2√3 600 7 m 2

  23. Area of a Rhombus or a Kite • The area of a Rhombus or a Kite is ½ x D1 x D2 (diagonal 1 and diagonal 2) • Lets say D1 = 6 and D2= 6 • The area would be ½ x 6 x 6, or 18 • Imagine 1 triangle • The area of this triangle is ½ x D1/2 x D2/2 • This would be ½ x 6/2 x 6/2 or ½ x 3 x 3 = 4.5 • Since there are 4 triangles, we would multiply 4 x 4.5 = 18 • The area of a kite can be proved similarly D1 D2

  24. Example • Find the area of this kite • A = ½ x D1 x D2 • A = ½ x (3+3) x (5+2) • A = ½ x 6 x 7 • A = ½ x 42 • A = 21 2 3 3 5

  25. Example 15 12 • Find the area of this Rhombus • A = ½ x D1 x D2 • We know that D1 = 12 x 2 = 24 • We need D2 • We can either solve using the Pythagorean theorem, or we can recognize the Pythagorean Triple (3,4,5) • So the third side of the triangle is 9 (3 x 3, 4 x 3, and 5 x 3) • Therefore D2 = 9 x 2 = 18 • A = ½ x 24 x 18 • A = 216 9

  26. Assignment • Page 619 8-16 • Page 626 11-25 • Worksheet 7-1 • Worksheet 7-4

  27. Unit 7 Quiz 2 • What is the area equation for a triangle? • What is the area equation for a parallelogram? • What is the area equation for a rectangle? • What is the area equation for a square? • What is the area equation for a Trapezoid? • What is the area equation for a Kite? • What is the area equation for a Rhombus? • When C2>A2+B2 the triangle is _______ • When C2=A2+B2 the triangle is _______ • When C2<A2+B2 the triangle is _______

  28. A look at the Circle Diameter Perimeter . Radius Chord Center Point

  29. Circles • A Circle is a set of points in a plane that are the same distance from the center point • A Radius is a segment of a line which has one endpoint on the center point, and the other endpoint on the circle • A Diameter is a segment (or chord) that passes through the center of the circle, and has an endpoint on each side of the circle • A Chord is any segment whose endpoints are on the circle • The Circumference is the distance around the circle. • To calculate circumference, you multiply the Diameter times pi, or 2 Radii times pi

  30. Circumference • Remember, circumference is the distance around the circle • The ratio of the circumference to the diameter of the circle is represented as • C=Dp or C = 2pR (1 diameter = 2 radii) • This is called “Diameter Pi” or “2 Pi R” • p is the symbol for “pi”, which is an often used number  3.14159. Pi goes on forever, but your calculator can give you a good approximation

  31. Find the Circumference • A circle has a diameter of 10 feet. How far around is it to the nearest tenth of a foot? • C = D x Pi • C = 10 x 3.14 • C = 31.4 FT 10 Feet

  32. Check on Learning • Find the circumference for the following circles: • A circle with a diameter of 2 4/5 inches • 8.79 inches • A circle with a radius of 30 mm • 188.495 mm • A circle with a diameter of 200 miles • 628.318 miles • A circle with a radius of 14 feet • 87.964 feet

  33. Area of a Circle • The area of a circle is calculated as: • pR2 • This is called “pi R squared” • Note: We can take the Diameter (D) and divide it in two to get the Radius (R) • We have to do this operation BEFORE we square it • So we could write this: p(D/2)2

  34. More about π • It’s been calculated for thousands of years:

  35. More about π • William Jones, a self-taught English mathematician born in Wales, is the one who selected the Greek letter for the ratio of a circle's circumference to its diameter in 1706. • is an irrational number. That means that it can not written as the ratio of two integer numbers. For example, the ratio 22/7 is a popular one used for but it is only an approximation which equals about 3.142857143...Another more precise ratio is 355/113 which results in 3.14159292...   • Another characteristic of as an irrational number is the fact that it takes an infinite number of digits to give its exact value, i.e. you can never get to the end of it. • Since 4,000 years ago and up until this very day, people have been trying to get more and more accurate values for pi. Presently supercomputers are used to find the value of with as many digits as possible. Pi has been calculated with a precision containing more than one billion digits, i.e., more that 1,000,000,000 digits!

  36. More about π • Egyptologists and followers of mysticism have been fascinated for centuries by the fact that the Great Pyramid at Gaza seems to approximate pi. The vertical height of the pyramid has the same relationship to the perimeter of its base as the radius of a circle as to its circumference. • The first 144 digits of pi add up to 666 (which many scholars say is “the mark of the Beast”). And 144 = (6+6) x (6+6). • If the circumference of the earth were calculated using π rounded to only the ninth decimal place, an error of no more than one quarter of an inch in 25,000 miles would result. • In 1995, Hiroyoki Gotu memorized 42,195 places of pi and is considered the current pi champion. Some scholars speculate that Japanese is better suited than other languages for memorizing sequences of numbers.

  37. Assignment • Page 64 10-13, 23-33 • Worksheet 1-7 • Circles Worksheet • Discovering Pi

  38. Unit 7 Quiz 3 • A kite has a diagonal of 12 and a diagonal of 11. What is it’s area? • A rhombus has a diagonal of 2 and a diagonal of 8. What is it’s area? • A square has a side = 20 inches. What is it’s area? • A rectangle has a base of 16, and a height of 5. What is it’s area? • A parallelogram has a base of 8 and a height of 8. What is it’s area? • A triangle has a base of 18 and a height of 10. What is it’s area? • A triangle has a base of 11 and a height of 8. What is it’s area? • A trapezoid has a top base of 8, a bottom base of 9, and a height of 6. what is it’s area? • A trapezoid has a top base of 4, a bottom base of 20, and a height of 23. What is it’s area? • A parallelogram has a base of 20 and a height of 23. What is it’s area?

  39. Unit 7 Quiz 4 10 Points • How do you calculate pi? • Explain in your own words

  40. Similarity • Polygons are similar if they have the exact same measure of degree for all angles, and that all sides are proportional –they have the same ratio • The similarity ratio is found by writing the length of one side over the length of the same, corresponding side from the second polygon • For example, the similarity ratio here is 2/3 (3/4.5 = 2/3) 3 4.5 2 3 2 3 4 6

  41. Perimeters and Area of Similar Polygons • If you have a similarity ratio between two polygons  for example a/b, then: • The similarity ratio for the perimeter is the same  that is a/b • The similarity ratio for the area is a little different  it is a2/b2 • We can remember this because when we label area it is always “squared” –for example cm2

  42. Example 6 m • These polygons are similar • What is the similarity ratio? • 6/9, which reduces to 2/3 • If the perimeter of the small polygon is 20 m, what is the perimeter of the large polygon? • 2/3 = 20/P, 2P = 60, P = 30 • If the area of the small polygon is 60 m2, what is the area of the large polygon? • 22/32 = 60/A, 4/9 = 60/A, • Cross multiply and divide: 4A = 60x9, • A = 135 9 m

  43. Example • The area of the small pentagon is about 27.5 cm2 • What is the area, A, of the large pentagon? • 4/10 = 2/5 • 22/52 = 27.5/A • 4/25 = 27.5/A • 4A = 25(27.5) • 4A = 687.5 • A = 171.875 cm2 4 cm 10 cm Notice, that you need to reduce the ratio (4/10 = 2/5) before you square the top and bottom, or you will get the wrong answer…

  44. Find Similarity Ratios • The area of two similar triangles are 50 cm2 and 98 cm2 • Find the similarity ratio A/B • A2/B2 = 50/98 now simplify • A2/B2 = 25/49 now square root all • A/B = 5/7 • This is the similarity ratio, and the perimeter ratio

  45. Assignment • Page 638 9-16,19-29 • Worksheet 8-6

  46. Unit 7 Quiz 5 –All answers to the nearest 10th • A circle has a radius of 5. • What is it’s circumference? • What is it’s area? • A circle has a radius of 8. • What is it’s circumference? • What is it’s area? • A circle has a diameter of 12. • What is it’s circumference? • What is it’s area? • A circle has a diameter of 30. • What is it’s circumference? • What is it’s area? • A circle has a circumference of 12π • What is it’s diameter? • What is it’s area?

  47. A Toss of a Coin I inch Radius • You are at a carnival, where they have a coin toss • They have an 8 inch square, with a one inch radius circle on the square • If you toss a quarter and the entire quarter is on the circle, you win a prize • What is the probability of winning? 8 inches

  48. The Solution • A quarter is about 1 inch in diameter • Therefore, to be completely on the 2 inch circle, the quarter has to be at least ½ in from the edge of the circle • This means the desired area is a circle 1 inch in diameter (2 inch total, minus ½ inch on each side, = 1 inch) • The total area is the 8 inch square • To calculate Probability, we calculate the ratio of the desired outcome divided by the total outcome • Therefore: π(.50)2/82 = .012 or 1.2% • Not very good ODDS!! 8 inches

  49. Probability • The probability of an event occurring is the ratio of the number of favorable outcomes to the number of possible outcomes • A geometric probability model is one in which we use points to represent outcomes. (We will also use area). You find probabilities by comparing measurements of sets of points. For example: if points of segments (like a number line) represent possible outcomes, then  • P(event) = length of favorable segment length of entire segment

  50. Example • Suppose a fly lands on a 12 inch ruler’s edge. What is the probability that the fly lands on a point between 3 and 7? P(landing between 3 and 7) = Length of favorable segment = 4 or 1 Length of total segment 12 3 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 4 12

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