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4-6. Estimating Square Roots. Warm Up. Problem of the Day. Lesson Presentation. Course 3. 3. 8+ 144 4. 7 289. Warm Up Find the two square roots of each number. Evaluate each expression. 1. 144 2. 256. ± 12. ± 16. 20. 119. Problem of the Day
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4-6 Estimating Square Roots Warm Up Problem of the Day Lesson Presentation Course 3
3. 8+ 144 4. 7 289 Warm Up Find the two square roots of each number. Evaluate each expression. 1. 144 2. 256 ±12 ±16 20 119
Problem of the Day A pyramid of blocks is built in layers. The bottom layer has 62, or 36, blocks. The next layer has 52 blocks, and so on until the top layer has 1 block. How many blocks are there in all? 91 blocks
4-6 Learn to estimate square roots to a given number of decimal places and solve problems using square roots. Course 3
55 is between 7 and 8 because 55 is between 49 and 64. Additional Example 1A: Estimating Square Roots of Numbers Each square root is between two integers. Name the integers. Explain your answer. Think: What are perfect squares close to 55? 55 49 < 55 72 = 49 64 > 55 82 = 64
– is between –9 and –10 because 90 is between 81 and 100. 90 Additional Example 1B: Estimating Square Roots of Numbers Continued Each square root is between two integers. Name the integers. Explain your answer. Think: What are perfect squares close to 90? – 90 81 < 90 –92 = 81 100 > 90 –102 = 100
80 is between 8 and 9 because 80 is between 64 and 81. Check It Out: Example 1A Each square root is between two integers. Name the integers. Explain your answer. Think: What are perfect squares close to 80? 80 64 < 80 82 = 64 81 > 80 92 = 81
– is between –6 and –7 because 45 is between 36 and 49. 45 Check It Out: Example 1B Each square root is between two integers. Name the integers. Think: What are perfect squares close to 45? – 45 36 < 45 –62 = 36 49 > 45 –72 = 49
Understand the Problem Additional Example 2: Problem Solving Application You want to sew a fringe on a square tablecloth with an area of 500 square inches. Calculate the length of each side of the tablecloth and the length of fringe you will need to the nearest tenth of an inch. First find the length of a side. Then you can use the length of a side to find the perimeter, the length of fringe around the tablecloth.
The length of a side, in feet, is the number that you multiply by itself to get 500. To be accurate, find this number to the nearest tenth. If you do not know a step-by-step method for finding 500, use guess and check. Make a Plan Additional Example 2 Continued
Solve Guess 22.5 Guess 22.2 Guess 22.4 Guess 22.3 22.52 = 506.25 22.22 = 492.84 22.42 = 501.76 22.32 = 497.29 Too high Too low Too high Too low Square root is between 22 and 22.5 Square root is between 22.2 and 22.5 Square root is between 22.2 and 22.4 Square root is between 22.3 and 22.4 3 4 2 1 22.0 22.2 22.4 22.6 Additional Example 2 Continued Because 500 is between 222 and 232, the square root of 500 is between 22 and 23. The square root is between 22.3 and 22.4.
Solve To the nearest tenth, 500 is about 22.4. Additional Example 2 Continued The square root is between 22.3 and 22.4. To round to the nearest tenth, look at the next decimal place. Consider 22.35. Too low 22.352= 499.5225 The square root must be greater than 22.35, so round up. The length of each side of the table is about 22.4 in.
Additional Example 2 Continued Solve The length of a side of the tablecloth is 22.4 inches, to the nearest tenth of an inch. Now estimate the length around the tablecloth. 4 • 22.4 = 89.6 Perimeter = 4 • side You will need about 89.6 inches of fringe.
Look Back Additional Example 2 Continued The length 90 inches divided by 4 is 22.5 inches. A 22.5-inch square has an area of 506.25 square inches, which is close to 500, so the answers are reasonable.
Understand the Problem Check It Out: Example 2 You want to build a fence around a square garden that is 250 square feet. Calculate the length of one side of the garden and the total length of the fence, to the nearest tenth. First find the length of a side. Then you can use the length of a side to find the perimeter, the length of the fence.
The length of a side, in feet, is the number that you multiply by itself to get 250. To be accurate, find this number to the nearest tenth. If you do not know a step-by-step method for finding 250, use guess and check. Make a Plan Check It Out: Example 2 Continued
Solve 15.5 15.7 15.9 16.1 Check It Out: Example 2 Continued Because 250 is between 152 and 162, the square root of 250 is between 15 and 16. 2 4 1 3 The square root is between 15.8 and 15.9.
Solve To the nearest tenth, 250 is about 15.8. Check It Out: Example 2 Continued To round to the nearest tenth, look at the next decimal place. 15.852= 251.2225 Consider 15.85. The square root is lower than 15.85, so round down. The length of each side of the garden is about 15.8 ft.
Check It Out: Example 2 Continued Solve The length of a side of the garden is 15.8 feet, to the nearest tenth of a foot. Now estimate the length around the garden. 4 • 15.8 = 63.2 Perimeter = 4 • side You will need about 63.2 feet of fence.
Look Back Check It Out: Example 2 Continued The length 63.2 feet divided by 4 is 15.8 feet. A 15.8 foot square has an area of 249.64 square feet, which is close to 250, so the answers are reasonable.
Additional Example 3: Using a Calculator to Estimate the Value of a Square Root Use a calculator to find 600. Round to the nearest tenth. Using a calculator, 600 ≈ 24.49489742…. Rounded, 600 is 24.5.
Using a calculator, 800 ≈ 28.2842712…. Check It Out: Example 3 Use a calculator to find 800. Round to the nearest tenth. Rounded, 800 is 28.3.
Lesson Quiz Each square root is between two integers. Name the two integers. 1. 27 2. – 456 5 and 6 –22 and –21 Use a calculator to find each value. Round to the nearest tenth. 9.4 35.0 3. 89 4. 1223 5. A square field has an area of 2000 square feet. To the nearest foot, how much fencing would be needed to enclose the field? 179 ft