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Ratio and Proportion 7-1.

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Presentation on theme: "Ratio and Proportion 7-1."— Presentation transcript:

1 Ratio and Proportion 7-1

2 EXAMPLE 1 Simplify ratios Simplify the ratio. b. 5 ft 20 in. 64 m : 6 m a. SOLUTION a. Write 64 m : 6 m as 64 m 6 m . Then divide out the units and simplify. 64 m 6 m = 32 3 = 32 : 3 b. To simplify a ratio with unlike units, multiply by a conversion factor. 5 ft 20 in. = 5 ft 20 in. 12 in. 1 ft = 60 20 = 3 1

3 GUIDED PRACTICE for Example 1 Simplify the ratio. yards to 3 yards SOLUTION Write 24 yards : 3 yards as 24 3 Then divide out the units and simplify. 24 3 = 8 1 = 8 : 1

4 GUIDED PRACTICE for Example 1 Simplify the ratio. cm : 6 m SOLUTION To simplify a ratio with unlike units, multiply by a conversion factors. 150cm 6m = 150 6 1m 100cm = 1 4 = 1 : 4

5 EXAMPLE 4 Solve proportions Solve the proportion. ALGEBRA a. 5 10 x 16
= SOLUTION a. 5 10 x 16 = Write original proportion. = 10 x 5 16 Cross Products Property = 10 x 80 Multiply. = x 8 Divide each side by 10.

6 EXAMPLE 4 Solve proportions b. 1 y + 1 = 2 3y SOLUTION 1 y + 1 2 3y b.
Write original proportion. = 2 (y + 1) 1 3y Cross Products Property = 2y + 2 3y Distributive Property = y 2 Subtract 2y from each side.

7 GUIDED PRACTICE for Example 4 5. 2 x 5 8 = SOLUTION 2 x 5 8 = = 5 x
Write original proportion. = 5 x 2 8 Cross Products Property = 5 x 16 Multiply. = x 16 5 Divide each side by 5 .

8 GUIDED PRACTICE for Example 4 6. 1 x – 3 4 3x = SOLUTION 1 x – 3 4 3x
Write original proportion. 3x 4(x – 3) = Cross Products Property 3x 4x – 12 = Multiply. 3x – 4x – 12 = Subtract 4x from each side. – x = – 12 x = 12

9 GUIDED PRACTICE for Example 4 7. y – 3 7 y 14 = SOLUTION y – 3 7 y 14
Write original proportion. = 14(y – 3) 7 y Cross Products Property 14y – 42 7y = Multiply. Subtract 7y from each side and add 42 to each side. 14y – 7y 42 = y = 6 Subtract , then divide

10 EXAMPLE 2 Use a ratio to find a dimension Painting You are planning to paint a mural on a rectangular wall. You know that the perimeter of the wall is 484 feet and that the ratio of its length to its width is 9 : 2. Find the area of the wall. SOLUTION Write expressions for the length and width. Because the ratio of length to width is 9 : 2, you can represent the length by 9x and the width by 2x. STEP 1

11 Use a ratio to find a dimension
EXAMPLE 2 Use a ratio to find a dimension STEP 2 Solve an equation to find x. = 2l + 2w P Formula for perimeter of rectangle = 2(9x) + 2(2x) 484 Substitute for l, w, and P. = 484 22x Multiply and combine like terms. = 22 x Divide each side by 22. Evaluate the expressions for the length and width. Substitute the value of x into each expression. STEP 3 Length = 9x = 9(22) = 198 Width = 2x = 2(22) = 44 The wall is 198 feet long and 44 feet wide, so its area is 198 ft 44 ft = 8712 ft . 2

12 The angle measures are 30 , 2(30 ) = 60 , and 3(30 ) = 90. ANSWER
EXAMPLE 3 Use extended ratios ALGEBRA The measures of the angles in CDE are in the extended ratio of 1 : 2 : 3. Find the measures of the angles. SOLUTION Begin by sketching the triangle. Then use the extended ratio of 1 : 2 : 3 to label the measures as x° , 2x° , and 3x° . = 180 o x + 2x + 3x Triangle Sum Theorem = 6x 180 Combine like terms. = 30 x Divide each side by 6. The angle measures are 30 , 2(30 ) = 60 , and 3(30 ) = 90. o ANSWER

13 GUIDED PRACTICE for Examples 2 and 3 3. The perimeter of a room is 48 feet and the ratio of its length to its width is 7 : 5. Find the length and width of the room. SOLUTION Write expressions for the length and width. Because the ratio of length is 7 : 5, you can represent the length by 7x and the width by 5x. STEP 1

14 Solve an equation to find x.
GUIDED PRACTICE for Examples 2 and 3 STEP 2 Solve an equation to find x. = 2l + 2w P Formula for perimeter of rectangle = 2(7x) + 2(5x) 48 Substitute for l, w, and P. = 48 24x Multiply and combine like terms. = 2 x Evaluate the expressions for the length and width. Substitute the value of x into each expression. STEP 3 Length = 7x + 7(2) = 14 ft Width = 5x + 5(2) = 10 ft

15 The angle measures are 20 , 3(20 ) = 60 , and 5(20 ) = 100. ANSWER
GUIDED PRACTICE for Examples 2 and 3 4. A triangle’s angle measures are in the extended ratio of 1 : 3 : 5. Find the measures of the angles. x 3x 5x SOLUTION Begin by sketching the triangle. Then use the extended ratio of 1 : 3 : 5 to label the measures as x° , 2x° , and 3x° . = x + 3x + 5x 180 o Triangle Sum Theorem = 9x 180 Combine like terms. = 20 x Divide each side by 9. The angle measures are 20 , 3(20 ) = 60 , and 5(20 ) = 100. o ANSWER


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