Transcript Slide 1
8. Bottom left corner at (0,0), rest of coordinates at (2, 0), (0, 2) and (2, 2) 9. Coordinates at (0,0), (0, 1), (5, 0) or (0,0), (1, 0), (0, 5) 10. Using Midpt Formula: E (0, 5) and F (6, 5) Distance Formula BC = 6 = EF 11. (0, 0), (0, 2m), (2m, 2m), (2m, 0) 12. (0, 0), (0, x), (3x, x), (3x, 0) 17. P = 2s + 2t units A = st square units 18. (n, n) 19. (p, 0) 24. Distance Formula: KL = 2; MP = 2; LM = 3; PK = 3 25. You’re assuming figure has rt angle 26. a. BD = 38 in. CE = 24 in. b. B (24, 0) C (24, 28) 38. 112 39. 22 D(24, 38) E(0, 38) KM = KM (Reflex) SSS Warm Up 1. Find each angle measure. 60°; 60°; 60° True or False. If false explain. 2. Every equilateral triangle is isosceles. True 3. Every isosceles triangle is equilateral. False; an isosceles triangle can have only two congruent sides. Recall that an isosceles triangle has at least two congruent sides. The congruent sides are called the legs. The vertex angle is the angle formed by the legs. The side opposite the vertex angle is called the base, and the base angles are the two angles that have the base as a side. 3 is the vertex angle. 1 and 2 are the base angles. Reading Math The Isosceles Triangle Theorem is sometimes stated as “Base angles of an isosceles triangle are congruent.” Example 1: The length of YX is 20 feet. Explain why the length of YZ is the same. The mYZX = 180 – 140, so mYZX = 40°. Since YZX X, ∆XYZ is isosceles by the Converse of the Isosceles Triangle Theorem. Thus YZ = YX = 20 ft. Example 2A: Finding the Measure of an Angle Find mF. mF = mD = x° Isosc. ∆ Thm. mF + mD + mA = 180 ∆ Sum Thm. Substitute the x + x + 22 = 180 given values. Simplify and subtract 2x = 158 22 from both sides. x = 79 Divide both sides by 2. Thus mF = 79° Example 2B: Finding the Measure of an Angle Find mG. mJ = mG Isosc. ∆ Thm. (x + 44) = 3x 44 = 2x Substitute the given values. Simplify x from both sides. Divide both sides by 2. Thus mG = 22° + 44° = 66°. x = 22 The following corollary and its converse show the connection between equilateral triangles and equiangular triangles. Example 3A: Using Properties of Equilateral Triangles Find the value of x. ∆LKM is equilateral. Equilateral ∆ equiangular ∆ (2x + 32) = 60 2x = 28 x = 14 The measure of each of an equiangular ∆ is 60°. Subtract 32 both sides. Divide both sides by 2. Example 3B: Using Properties of Equilateral Triangles Find the value of y. ∆NPO is equiangular. Equiangular ∆ equilateral ∆ 5y – 6 = 4y + 12 y = 18 Definition of equilateral ∆. Subtract 4y and add 6 to both sides. Remember! A coordinate proof may be easier if you place one side of the triangle along the x-axis and locate a vertex at the origin or on the y-axis. Example 4: Using Coordinate Proof Prove that the segment joining the midpoints of two sides of an isosceles triangle is half the base. Given: In isosceles ∆ABC, X is the mdpt. of AB, and Y is the mdpt. of AC. 1 Prove: XY = 2 AC. Example 4 Continued Proof: Draw a diagram and place the coordinates as shown. By the Midpoint Formula, the coordinates of X are (a, b), and Y are (3a, b). By the Distance Formula, XY = √4a2 = 2a, and AC = 4a. 1 Therefore XY = AC. 2 12. Angle add. mATB = 40°. mBAT = 40° (Alt Int s) ATB BAT (def of ) Since ΔABT is isos by converse of isos Δ thm, BT = BA = 2.4 mi 13. 69° 33. 20 14. 33° 34. 3 15. 130° or 172° 40. Two sides of a Δ are iff the s 16. 31 ° 17. 92 18. 48 19. 26 20. 20 28. m1= 58°; m2 = 64°; m3 = 122° 29. m1= 127°; m2 = 26.5°; m3 = 53° 30. opp. those sides are