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Course 3 4-6 Estimating Square Roots 4-6 Estimating Square Roots Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson.

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Presentation on theme: "Course 3 4-6 Estimating Square Roots 4-6 Estimating Square Roots Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson."— Presentation transcript:

1 Course 3 4-6 Estimating Square Roots 4-6 Estimating Square Roots Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

2 Course 3 4-6 Estimating Square Roots Warm Up Find the two square roots of each number. Evaluate each expression. 1216 20 119 1. 144 2. 256 3. 8+ 144 4. 7 289

3 Course 3 4-6 Estimating Square Roots Problem of the Day A pyramid of blocks is built in layers. The bottom layer has 6 2, or 36, blocks. The next layer has 5 2 blocks, and so on until the top layer has 1 block. How many blocks are there in all? 91 blocks

4 Course 3 4-6 Estimating Square Roots Learn to estimate square roots to a given number of decimal places and solve problems using square roots. Course 3 4-6

5 Course 3 4-6 Estimating Square Roots Additional Example 1A: Estimating Square Roots of Numbers Each square root is between two integers. Name the integers. Explain your answer. 55 Think: What are perfect squares close to 55? 7 2 = 49 49 < 55 8 2 = 64 64 > 55 55 is between 7 and 8 because 55 is between 49 and 64.

6 Course 3 4-6 Estimating Square Roots 90 – Additional Example 1B: Estimating Square Roots of Numbers Continued Think: What are perfect squares close to 90? –9 2 = 81 81 < 90 –10 2 = 100 100 > 90 Each square root is between two integers. Name the integers. Explain your answer. is between –9 and –10 because 90 is between 81 and 100. 90 –

7 Course 3 4-6 Estimating Square Roots Check It Out: Example 1A Each square root is between two integers. Name the integers. Explain your answer. 80 Think: What are perfect squares close to 80? 8 2 = 64 64 < 80 9 2 = 81 81 > 80 80 is between 8 and 9 because 80 is between 64 and 81.

8 Course 3 4-6 Estimating Square Roots 45 – Each square root is between two integers. Name the integers. Think: What are perfect squares close to 45? –6 2 = 36 36 < 45 –7 2 = 49 49 > 45 Check It Out: Example 1B is between –6 and –7 because 45 is between 36 and 49. 45 –

9 Course 3 4-6 Estimating Square Roots 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. 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. Understand the Problem

10 Course 3 4-6 Estimating Square Roots Additional Example 2 Continued Make a Plan 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.

11 Course 3 4-6 Estimating Square Roots Because 500 is between 22 2 and 23 2, the square root of 500 is between 22 and 23. Solve Square root is between 22 and 22.5 Too high 22.5 2 = 506.25 Guess 22.5 1 Square root is between 22.2 and 22.5 Too low 22.2 2 = 492.84 Guess 22.2 2 Square root is between 22.2 and 22.4 Too high 22.4 2 = 501.76 Guess 22.4 3 Square root is between 22.3 and 22.4 Too low 22.3 2 = 497.29 Guess 22.3 4 22.022.222.422.6 The square root is between 22.3 and 22.4. Additional Example 2 Continued

12 Course 3 4-6 Estimating Square Roots Consider 22.35. 22.35 2 = 499.5225 Too low The square root must be greater than 22.35, so round up. To the nearest tenth, 500 is about 22.4. Solve The length of each side of the table is about 22.4 in. The square root is between 22.3 and 22.4. To round to the nearest tenth, look at the next decimal place. Additional Example 2 Continued

13 Course 3 4-6 Estimating Square Roots 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. Solve 4 22.4 = 89.6 Perimeter = 4 side You will need about 89.6 inches of fringe. Additional Example 2 Continued

14 Course 3 4-6 Estimating Square Roots 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. Look Back Additional Example 2 Continued

15 Course 3 4-6 Estimating Square Roots 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. 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. Understand the Problem

16 Course 3 4-6 Estimating Square Roots 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

17 Course 3 4-6 Estimating Square Roots Because 250 is between 15 2 and 16 2, the square root of 250 is between 15 and 16. Solve 1 2 3 4 15.515.715.916.1 The square root is between 15.8 and 15.9. Guess 15.5 15.5 2 = 240.25 Too Low Square root is between 15.5 and 16 Guess 15.9 15.9 2 = 252.81 Too high Square root is between 15.5 and 15.9 Guess 15.7 15.7 2 = 246.49 Too Low Square root is between 15.7 and 15.9 Guess 15.8 15.8 2 = 249.46 Too Low Square root is between 15.8 and 15.9 Check It Out: Example 2 Continued

18 Course 3 4-6 Estimating Square Roots Consider 15.85. 15.85 2 = 251.2225 The square root is lower than 15.85, so round down. To the nearest tenth, 250 is about 15.8. Solve The length of each side of the garden is about 15.8 ft. To round to the nearest tenth, look at the next decimal place. Check It Out: Example 2 Continued

19 Course 3 4-6 Estimating Square Roots 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. Solve 4 15.8 = 63.2 Perimeter = 4 side You will need about 63.2 feet of fence. Check It Out: Example 2 Continued

20 Course 3 4-6 Estimating Square Roots 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. Look Back Check It Out: Example 2 Continued

21 Course 3 4-6 Estimating Square Roots Additional Example 3: Using a Calculator to Estimate the Value of a Square Root Using a calculator, 600 ≈ 24.49489742 …. Rounded, 600 is 24.5. Use a calculator to find 600. Round to the nearest tenth.

22 Course 3 4-6 Estimating Square Roots Check It Out: Example 3 Rounded, 800 is 28.3. Using a calculator, 800 ≈ 28.2842712 …. Use a calculator to find 800. Round to the nearest tenth.

23 Course 3 4-6 Estimating Square Roots Lesson Quiz Each square root is between two integers. Name the two integers. 1. 27 2. – 456 Use a calculator to find each value. Round to the nearest tenth. 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? 5 and 6–22 and –21 9.435.0 179 ft


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