Geometry Unit 7 (Textbook Chapter 9) Solving a right triangle: Find all missing sides and all missing angles

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1 Geometry Unit 7 (Textbook Chapter 9) Name Objective 1: Right Triangles and Pythagorean Theorem In many geometry problems, it is necessary to find a missing side or a missing angle of a right triangle. Finding all the missing sides and angles is called solving a triangle. Solving a right triangle: Find all missing sides and all missing angles 1) Find all missing angles: Use Triangle Sum Theorem: in a right triangle, all angles add to 180. Use any angle information given to find angles (remember that you already know one angle is 90 2) Find all missing sides: If two sides are given: Use Pythagorean theorem a 2 + b 2 = c 2 to find the missing side. 9.1 The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs equals the square of the hypotenuse. Or in other words: If a triangle is a right triangle, then the sides form the relationship a 2 + b 2 = c 2. c is the length of the side opposite the right angle, called the hypotenuse. 9.2 Converse of Pythagorean Theorem The Pythagorean theorem can be used to determine whether the sides given for a triangle are the sides of a right triangle. If the Pythagorean relationship holds, the triangle is right The Pythagorean theorem will work only for right triangles: Acute Triangles: a 2 + b 2 > c 2 The side that would have been the hypotenuse is extra short Right Triangles: a2 + b 2 = c 2 Obtuse Triangles: a 2 + b 2 < c 2 The side that would have been the hypotenuse is extra long Determine if the following triangles are acute, right, or obtuse: Example 1: Example 2: Page 1

2 Section 9.1 Practice Problems Give all answers in exact simplified form and rounded to the nearest tenth of a centimeter. 3) 4) 5) 6) Is a triangle with sides 5.0, 1.4, and 4.8 a right triangle? 7) Show your work to justify your answer. Page 2

3 Assignment #1 Find the missing side of each triangle below. Show your work and solve properly! Round your answers to the nearest tenth if necessary. 1) 2) Find the missing side of each triangle. Leave your answers in simplest radical form. Sides a and b are legs, and c is the hypotenuse. 3) 4) a = 11 m, c = 15 m Determine if each triangle is a right triangle. Show your work. 5) 6) State if the three sides form a right triangle. Show your work. 7) 10 cm, 49.5 cm, 50.5 cm 8) 9 in, 12 in, 15 in Page 3

4 State if each triangle form an acute, obtuse, or right triangle. 9) 10) 11) Additional Book Problems 9.2 p #9-17 odds Page 4

5 9.3 Special right triangles There are 2 special right triangles that are found in everyday shapes and many geometry problems. It is helpful to know the relationships of the sides of these special right triangles so they can be easily solved. Special Triangle #1: (isosceles right triangle) The ratio of side- side- hypotenuse of an isosceles right triangle is always 1:1: 2 Use the Pythagorean theorem and triangle sum theorem to fill in the table below. Special Triangle #2: The ratio of side- side- hypotenuse of a right triangle is always 1: 3 :2 Note: The hypotenuse of a right triangle is twice as long as the side opposite the 30 angle. See the equilateral triangle at the right. Use the Pythagorean theorem and the hypotenuse- short leg relationship in the table below to show the given ratio of 1: 3 :2 of side- side- hypotenuse is true. To summarize: Page 5

6 To summarize: To solve for any side of a special right triangle: Get x alone on the side of the triangle you are trying to find. Use the new ratio to find the missing side. 9.3 Practice Problems: Give your answers in exact form (as a simplified root) unless otherwise indicated. All measurements are in centimeters. All measurements are in centimeters. In Exercises 1 3, find the unknown lengths. 1. a 2. a, b 3. a, b a b 30 a b 6 a Find the area of rectangle 5. Find the perimeter and 6. AC, AB, ABCD. area of KLMN. and area ABC. A B K 45 N 7 M L A 45 C B D C 7) Find the area of an isosceles trapezoid if the bases have lengths 12 cm and 18 cm and the base angles have measure 60 degrees. Page 6

7 Assignment #3 9.3 p #1-11 all, 16, 17 Page 7

8 9.4 Story Problems Use what you have learned to answer the questions in section 9.4 of your textbook. Assignment #4 9.4 p #1, 4-7 all, 13, 14 Page 8

9 9.5 Distance Formula and Circle Equations ( ) 2 + ( y 2 y 1 ) 2 Distance Formula between 2 points on a graph: d = x 2 x 1 On a grid, you can find the length of each segment by using the segment as the hypotenuse of a right triangle. Create a right triangle that has the segment shown as its hypotenuse Count the squares on the horizontal legs Count the squares on the vertical legs Use the Pythagorean Theorem to find the length of the hypotenuse. a) d= b) d= c) d= d) d= If there is no grid (or something is impossible to graph) you can: Subtract the x- values of the points to get the horizontal leg distance ( x 2 x 1 ) Subtract the y- values of the points to get the vertical leg distance. ( y 2 y 1 ) Let d=distance of segment (hypotenuse) Finding the Equation of a circle on a graph r 2 = ( x a) 2 + ( y b) 2 with radius r and center (a,b) Circle: The set of all points on a plane that are a fixed distance from a center. Let us put that center at (a,b). So the circle is all the points (x,y) that are "r" away from the center (a,b). Now we can work out exactly where all those points are! We make a right-angled triangle (as shown), and then use Pythagorean theorem a 2 + b 2 = c 2 (x-a) 2 + (y-b) 2 = r 2 And that is the "Standard Form" for the equation of a circle! (You can see all the important information at a glance: the center (a,b) and the radius r Page 9

10 9.5 Practice Problems In Exercises 1 3, find the distance between each pair of points. 1. ( 5, 5), (1, 3) 2. ( 11, 5), (5, 7) 3. (8, 2), ( 7, 6) In Exercises 4 and 5, use the distance formula and the slope of segments to identify the type of quadrilateral. Explain your reasoning. 4. A( 2, 1), B(3, 2), C(8, 1), D(3, 4) 5. T( 3, 3), U(4, 4), V(0, 6), W( 5, 1) 6) Find the equation of the circle with a) center ( 1,5) and radius 2. b) center (2,3) and radius 10. 7) Find the center and radius of the circle whose equation is a) ( x 3) 2 + ( y 2) 2 = 81 b) ( x 3) 2 + ( y + 2) 2 = 36 c) x 2 + ( y + 2) 2 = 40 8). P is the center of the circle. What s wrong with this picture? y A(4, 6) P(10, 1) B(5, 5) C(16, 3) Page 10

11 Assignment #5 9.5 p #1-11 odds Page 11

12 9.6 Circles and the Pythagorean Theorem Remember Chapter 6? The tangent to a circle is perpendicular to the radius drawn to the point of tangency Angles inscribed in semi- circles are right angles. Use what we have learned so far to answer the following practice problems In Exercises 1 and 2, find the area of the shaded region in each figure. All measurements are in centimeters. Write your answers in terms of and rounded to the nearest 0.1 cm AO 5. AC Tangent PT, QM 12, m P 30 C T A O B P S M Q 3. AP 63 cm. Radius of circle O 37 cm. How far is A from the circumference of the circle? O P A 4. Two perpendicular chords with lengths 12.2 cm and 8.8 cm have a common endpoint. What is the area of the circle? 5. ABCD is inscribed in a circle. AC is a diameter. If AB 9.6 cm, BC 5.7 cm, and CD 3.1 cm, find AD. Page 12

13 Assignment #6 9.6 p #1-5 all Page 13

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