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Warm-Up

Warm-Up. 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.

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Warm-Up

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  1. Warm-Up • 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

  2. 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

  3. 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

  4. Chapter 1Section 4 Measuring Segments

  5. x -5 -5 -4 -3 -2 -1 0 1 2 3 4 5 Vocabulary 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. P Q

  6. Vocabulary Cont. 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 (x1, y1) and (x2, y2) is given by the formula d=√((x2 – x1)2 + (y2 – y1)2) P Q R c a b

  7. Example 1: Find RS, ST, and RT on the number line shown below. R W S Q T P -10 -8 -6 -4 -2 0 2 4 6 8 10 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 Example 2: Find PQ, QW, and PW on the number line. PQ P = 10, Q = 0 10 - 0= 10 So PQ = 10 QW Q = 0, W = -5 0 - -5 = 5 So QW = 5 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. M N P 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. R W S 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. y x Example 5: Find the distance between points H(2, 3) and K(-3, -1) by using the Pythagorean theorem. 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. H c a K b

  11. y x Example 6: Find the distance between points R(4, -3) and T(-2, 5) by using the Pythagorean theorem. T 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 a R b

  12. Example 7: Use the distance formula to find JK for J(9, -5) and K(-6, 12). Distance Formula d= Pick one point to be x1 and y1 and the other point will be x2 and y2. Let point J be x1 and y1 and point K be x2 and y2. d=√((x2 – x1)2 + (y2 – y1)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=√((x2 – x1)2 + (y2 – y1)2) Pick one point to be x1 and y1 and the other point will be x2 and y2. Let point J be x1 and y1 and point K be x2 and y2. d=√((x2 – x1)2 + (y2 – y1)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.

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