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Section 9.4 Geometry’s Most Elegant Theorem. “Pythagorean Theorem” Pages 384-391 Rebecca Sproul. Pythagorean Theorem.
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Section 9.4Geometry’s Most Elegant Theorem “Pythagorean Theorem” Pages 384-391 Rebecca Sproul
Pythagorean Theorem As the plays of Shakespeare are to literature, as the Mets are to Mr. Pricci, so is the Pythagorean Theorem to geometry. The Pythagorean is the rule for solving right triangles, it’s widely applied because every polygon can be divided into right triangles by diagonals and altitudes, and it enables many ideas (and objects) to fit together very simply. “Indeed it is elegant in concept and very powerful” (Rhoad 384). • Theorem 69- The square of the measure of the hypotenuse of a right triangle is equal to the sum of the squares of the measures of the legs (Pythagorean Theorem) • [ a2 + b2 = c2] • Sample: Use the Pythagorean Theorem 92 +122 = x2 81 + 144 = x2 225 = x2 225 = x +\- 15 (reject -15) x = 15
Another Look at the Theorem The Pythagorean Theorem must work in any 90 degree triangle. This means that if you know two of the sides, you can always find the third one.
Another Theorem • Theorem 70: If the square of the measure of one side of a triangle is equal to the sum of the squares of the measures of the other two sides, then the angle opposite the longer side is a right angle. If a2 + b2 = c2 Then ABC is a right triangle and C is the right angle
Extension of Theorem 70 If in the diagram on the last slide we increased c while keeping a and b the same, c would become larger. Thus, a valuable extension of Theorem 70 can be stated: • a2 + b2 > c2, then the is acute • a2 + b2 = c2, then the is right • a2 + b2 < c2, then the is obtuse
Extension of Theorem 70 • Examples: Given the following sides in a triangle, classify the triangle as acute, right, or obtuse • 9, 12, 15 • 92 + 122 _ 152 • 81 + 144 _ 225 • 225 = 225 • Right • 11, 13, 7 • 72 + 112 _ 132 • 49 + 121 _ 169 • 170 > 169 • Acute 3. 4 3, 6 3, 5 2 (4 3)2 + (5 2)2 _ (6 3)2 48 + 50 _ 108 98 < 108 Obtuse
Example Problem • 1. Triangle ABC is equilateral with a perimeter of 18. Find x. Since ABC is equilateral, divide its perimeter into 3 equal sides Segment BD is the perpendicular bisector of segment AC, so AC can be divided by 2 BDC is a right angle (perpendicular lines form right angles), so we can use the Pythagorean Theorem to find x. x2 + 32 = 62 x2 + 9 = 36 x2 = 27 x = 3 3
Example Problem 2. RSTU is a rhombus with diagonals of 12 and 24. Find the perimeter of RSTU. • Draw both diagonals in the rhombus, and since diagonal bisects each other in a rhombus, cut each diagonal in half. • Diagonals in a rhombus are perpendicular, so the angles formed in the middle of the diagonals are right angles (perpendicular lines form right angles). • Since we have right angles, use the Pythagorean Theorem to solve for x, one side of the rhombus. Once you find x, multiply it by 4 because all 4 sides in a rhombus are =. 62 + 122 = x2 36 + 144 = x2 169 = x2 x = 13 (reject -13) 4x = 52 = perimeter of rhombus RSTU
Example Problem Draw Domo (1)’s path to Domo (2) and label the distance traveled Notice how the distance apart (x) forms a right triangle when the sides are drawn out Label each side of the right triangle by adding the distance traveled of the corresponding sides of Domo (1)’s path 5 + 10 = 15 2 + 6 = 8 Now use the Pythagorean Theorem 3. Domo (1) traveled 5 km north, 2 km east, 10 km north, and 6 km east to find another Domo (2). How far is Domo (1) from where he started? 82 + 152 = x2 64 + 225 = x2 289 = x2 x = 17 (reject -17) = distance apart from Domos
Example Problem • 4. A square has a perimeter of 36. Find the length of its diagonal. Since this is a square, all sides are =, so divide its perimeter (36) by how many sides it has (4) to get a single side of the square (9). Squares contain right angles, so we can use the right triangle formed by the diagonal drawn in the square by doing the Pythagorean Theorem 92 + 92 = x2 81 + 81 = x2 162 = x2 x = 9 2 (reject the negative) = the diagonal
Example Problem Draw new altitudes in the trapezoid to form a right triangle. The altitudes will have the same length (x) Each altitude forms a right angle Since the top of the trapezoid is parallel to the bottom, subtract the top from the bottom (27 – 17) to find the difference (10) Then divide the difference by 2 to get each of the small legs of each right triangle (5) Now use Pythagorean Theorem • 5. Find x in the trapezoid x2 + 52 = 122 x2 + 25 = 144 x2 = 119 x = 119 (reject the negative) = the altitude
Example Problem Since the ladder against the wall forms a right triangle, the Pythagorean Theorem can be used 6. How far up a wall will an 11m ladder reach, if the foot of the ladder must be 4m from the base of the wall? 42 + x2 = 112 16 + x2 = 121 x2 = 105 x = 105
Sample Problem 1. Find x 46 Click for answer
Sample Problem 2. What type of triangle is demonstrated by the 3 sides in each set: obtuse, acute, or right? 16, 24, 32 17, 9, 15 3, 5, 4 14, 50, 48 obtuse, acute, acute, right Click for answer
Sample Problem 3.Domo walks 15 km north, 4 km west, 3 km north, 9 km west, and 2 km north and got very lost. How far is domo from where is first started? 5 17 Click for answer
Sample Problem 4. The lengths of the diagonals of a rhombus are in the ratio of 2:1. If the perimeter of the rhombus is 60, find the sum of the lengths of the diagonals. 18 5 Click for answer
Sample Problem 5. Find the length of the altitude of an equilateral triangle with a side of 8. 4 3 Click for answer
Sample Problem 100 Click for answer
Sample Problem 7. Find the altitude (length of a segment perpendicular to both bases) of the isosceles trapezoid shown. 8 Click for answer
Sample Problem 8. Domo’s pet bird pecked at a 13-m wooden pole until is cracked and the upper part fell, with the top hitting the ground 7 m from the foot of the pole. Since the upper part had not completely broken off, Domo’s bird pecked away where the pole had cracked. How far was Domo’s bird above the ground? 2 15 Click for answer
Works Cited Page Willis, Bill. "The Pythagorean Theorem." 1999 28 May 2008 <http://www.worsleyschool.net/science/files/pythagoras/pythagoreanth eorem2.html>. Rhoad, Richard, and Milauskas, and Whipple. Geometry for Enjoyment and Challenge. New ed. Boston: McDougal Littell, 1991.