1. Measuring Segments

2. 1) Complete: _________ points determine a line. _________ points determent a plane. 2) Give three names for a line that contains points A and B 3) Find the perimeter and area of a rectangle that has a width of 6.2 meters and a length of 9.5 meters. 4) A rectangle has a perimeter of 56 inches. Find the maximum area of the rectangle. State the length and width of the rectangle. 5) Use the formula A = ½ Pa to find A if P = 72 and a = 374.4

3. 1) Complete: _________ points determine a line. _________ points determent a plane. 2 points determine a line. 3 points determine a plane. 2) Give three names for a line that contains points A and B Line AB, Line BA, AB, and BA 3) Find the perimeter and area of a rectangle that has a width of 6.2 meters and a length of 9.5 meters. Perimeter = 2l + 2w P = 2(9.5) + 2(6.2) P = 19 + 12.4 P = 31.4 m Area = lw A = 9.5 * 6.2 A = 58.9 m^2

4. 4) A rectangle has a perimeter of 56 inches. Find the maximum area of the rectangle. State the length and width of the rectangle. 56/4 = 14 14 * 14 = 196 in^2 So the maximum area is 196 in^2 and the length and width are 14 in by 14 in. 5) Use the formula A = ½ Pa to find A if P = 72 and a = 374.4 A = ½Pa A = ½(72)(374.4) A = ½ (26956.8) A = 13478.4

5. Measure- The length of line AB, written AB, is the distance between A and B. (Always positive) Congruent- The same size or measure. Postulate- A statement that is assumed to be true. Ruler Postulate: The points on any line can be paired with real numbers so that, given any two points P and Q on the line, P corresponds to zero, and Q corresponds to a positive number. x 5-4-2024-53-3-51 PQ

6. Segment Addition Postulate: If Q is between P and R, then PQ + QR = PR. If PQ + QR = PR, then Q is between P and R. Pythagorean Theorem: In a right triangle, the sum of the squares of the measures of the legs equals the square of the measure of the hypotenuse. a^2 + b^2 = c^2 Distance Formula: The distance d between any two points with coordinates (x 1, y 1 ) and (x 2, y 2 ) is given by the formula d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) QPR a b c

7. Example 1: Find RS, ST, and RT on the number line shown below. 0 RST 108642-2-8-6-4-10 Example 2: Find PQ, QW, and PW on the number line. WQP RS R = -8, S = -2 -8 - -2 = -6 So RS = 6 ST S = -2, T = 6 -2 - 6 = -8 So ST = 8 RT Segment Addition Postulate RS = 6, ST = 8 RS + ST = RT 6 + 8 = 14 So RT = 14 QW Q = 0, W = -5 0 - -5 = 5 So QW = 5 PQ P = 10, Q = 0 10 - 0= 10 So PQ = 10 PW Segment Addition Postulate PQ = 10, QW = 5 PQ + QW = PW 10 + 5 = 15 So PW = 15

8. Example 3: Find the measure of MN if N is between M and P, MN = 3x + 2, NP = 2x -1, and MP = 46. NMP Use the segment addition Postulate. MN + NP = MP 3x + 2 + 2x – 1 = 46 5x + 1 = 46 5x = 45 x = 9 Now plug 9 in for x in the equation for MN MN = 3x + 2 MN = 3(9) + 2 MN = 27 + 2 MN = 29

9. Example 4: If W is between R and S, RS = 7n + 8, RW = 4n – 3, and WS = 6n + 2, find the value of n and WS. WRS Use the segment addition Postulate. RW + WS = RS 4n – 3 + 6n + 2 = 7n + 8 10n -1 = 7n + 8 10n = 7n + 9 3n = 9 n = 3 Now plug 3 in for n in the equation for WS WS = 6n + 2 WS = 6(3) + 2 WS = 18 + 2 WS = 20

10. Example 5: Find the distance between points H(2, 3) and K(-3, -1) by using the Pythagorean theorem. y x K H a b The Pythagorean Theorem a^2 + b^2 = c^2 a = 4, b = 5 a^2 + b^2 = c^2 4^2 + 5^2 = c^2 16 + 25 = c^2 41 = c^2 √41 = c So the distance between the two points is √41 or about 6.403. c

11. Example 6: Find the distance between points R(4, -3) and T(-2, 5) by using the Pythagorean theorem. y x R T a b The Pythagorean Theorem a^2 + b^2 = c^2 a = 8, b = 6 a^2 + b^2 = c^2 8^2 + 6^2 = c^2 64 + 36 = c^2 100 = c^2 10= c So the distance between the two points is 10. c

12. Example 7: Use the distance formula to find JK for J(9, -5) and K(-6, 12). Distance Formula d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) Pick one point to be x 1 and y 1 and the other point will be x 2 and y 2. Let point J be x 1 and y 1 and point K be x 2 and y 2. d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((-6 – 9) 2 + (12 – -5) 2 ) d=√((-6 – 9) 2 + (12 + 5) 2 ) d=√((-15) 2 + (17) 2 ) d= √((225) + (289)) d= √(514) So the distance between the two points is √(514) or about 22.6.

13. Example 8: Use the distance formula to find JK for J(-3, -5) and K(4, -6). Distance Formula d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) Pick one point to be x 1 and y 1 and the other point will be x 2 and y 2. Let point J be x 1 and y 1 and point K be x 2 and y 2. d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((4 – -3) 2 + (-6 – -5) 2 ) d=√((4 + 3) 2 + (-6 + 5) 2 ) d=√((7) 2 + (-1) 2 ) d= √((49) + (1)) d= √(50) So the distance between the two points is √(50) or about 7.07.