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Chapter 3 Lesson 7 Using Formulas pgs. 131-136

In this lesson, students will learn to solve problems using mathematical formulas, focusing on the perimeters and areas of rectangles. Key concepts include understanding the distance formula (d = rt), calculating the perimeter (P = 2(l + w)), and finding the area (A = lw) of rectangles. Through practical examples, students will apply these formulas to determine unknown dimensions and verify their calculations. This lesson empowers learners to confidently tackle geometry problems and enhances their algebraic skills.

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Chapter 3 Lesson 7 Using Formulas pgs. 131-136

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  1. Chapter 3 Lesson 7Using Formulaspgs. 131-136 What you will learn: Solve problems by using formulas Solve problems involving the perimeters & areas of rectangles

  2. Vocabulary: • Formula (131): an equation that shows a Relationship among certain quantities. A formula Usually contains two or more variables. • Perimeter (132): the distance around a geometric figure • Area (132): the space inside a geometric figure expressed as square units

  3. Key Concept (131) Distance Formula Words: Distance equals the rate multiplied by the time Variables: Let d= distance, r = rate, & t = time Equation: d = rt What is the rate in miles per hour of a dolphin that travels 120 miles in 4 hours? d = rt Make it look pretty! 4r = 120 120 = r4 4r = 120 4 r = 30 The dolphin travels at a rate of 30 mph

  4. Key Concept (132): Perimeter of a Rectangle Words: The perimeter of a rectangle is twice the sum of the length and width Symbols: P = l + l + w + w Use the way you or P = 2l + 2w prefer! or P = 2 (l + w) l Model: w

  5. Key Concept (132): Area of a Rectangle Words: The area of a rectangle is the product of the length and width Symbols: A =lw l Remember area is Model: w expressed in square units

  6. Apply what you know! Find the perimeter of a rectangle. Check: 24 = 2(8+4) 24 = 2(12) 24 = 24  P = 2(l + w) P = 2(8+4) P = 2(12) P = 24 ft. 8 ft. 4 ft The perimeter of this rectangle is 28 meters. Its width is 8 meters. Find the length. 28 = 2(6+8) 28 = 2(14) 28 = 28  P = 2(l+w) 28 = 2(l+8) 28 = 2l + 16 28 - 16 = 2l +16 - 16 12 = 2l 2 2 6 = l 8m

  7. Apply more of what you know! Find the area of the rectangle: Check: 120= 12(10) 120 = 120  A = lw A = 12(10) A = 120 in2 12 in. 10 in. The area of a rectangle is 323 square yards. Its length is 17 yards. Find the width. Check: A = lw 323 = 17w 323 = 17w w = 19 yds 17 17 A = lw 323 = 17(19) 323 = 323 

  8. Find the perimeter and area of each rectangle: 6 mi A= lw A = 6(4) A = 24 mi2 P = 2(l+w) P = 2(6 + 4) P = 2(10) P = 20 mi Check: 20= 2(6+4) 20= 2(10) 20 = 20  4 mi 18 in. P = 2(l+w) P = 2(18 + 50) P = 2(68) P = 136 in. A= lw A = 18(50) A = 900 in2 Check: 136 = 2(18+50) 136 = 2(68) 136 = 136  50 in.

  9. Find the missing dimension: Check: 24=2(7+5) 24=2(12) 24=24  P = 2(l+w) 24 = 2(7+w) 24 = 14 + 2w 24-14 = 14-14+2w 10 = 2w 2 2 w= 5m 7m w P = 24m w A = 176 yd2 Check: 176=16(11) 176=176  A = lw 176 = 16w 176 = 16w 16 16 w = 11yd 16yd

  10. Homework Practice Page is by the door! Quiz Monday!! Test Wednesday!!

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