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Chapter 1

Chapter 1 . Square Roots and Surface Area. 1.1 – square roots of perfect squares 1.2 – square roots of non-perfect squares. Chapter 1. review. The Pythagorean Theorem:. Area = . L 2. Remember, the opposite of squaring is take the square root. Volume = . L 3. handout.

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Chapter 1

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  1. Chapter 1 Square Roots and Surface Area

  2. 1.1 – square roots of perfect squares1.2 – square roots of non-perfect squares Chapter 1

  3. review The Pythagorean Theorem: Area = L2 Remember, the opposite of squaring is take the square root. Volume = L3

  4. handout Complete the questions in the handout to the best of your ability, because this will be a summative assessment.

  5. Perfect squares Fractions can be perfect squares too, if they have a square root that is also a fraction (or a terminating or repeating decimal). • L = 9/10 • A = (9/10)2 • A = 92/102 • A = 81/100 units2 1 unit 9/10 units • What happens when you square a fraction? • What do you think happens when you take the square root of a fraction?

  6. example What is the area? The area is .

  7. example What’s the side length?

  8. Example - Decimals You can also use decimals. A fraction in simplest form is a perfect square if it can be written as a product of two equal fractions. If a decimal can be written as a fraction that is a perfect square, then the decimal is also a perfect square.

  9. Perfect squares An easy way to tell if a fraction is a perfect square is by putting it in lowest terms and then checking if the numerator and denominator are both perfect squares. Example: Since 4 and 9 are perfect squares, we know that 8/18 is a perfect square. Example: It’s already in lowest terms, and 2 isn’t a perfect square, so 2/9 isn’t a perfect square either. When dealing with a decimal, you should turn it into a fraction and check to see if it is a perfect square using this method.

  10. Approximating square roots Khan Academy on Approximating Square Roots • For fractions and decimals, the techniques are mostly the same: • If you have a fraction, you should try to find a fraction close to it that is a perfect square, and use it to estimate. • Ex. • Benchmarking is when you find an upper limit and lower limit with square roots you know. • It may be easiest to turn it into a decimal, especially if the denominator is 10 or 100. • 0.30 is between 0.25 and 0.36, so it’s square root is between 0.5 and 0.6, approximately 0.56 • Sometimes you may need to use a combination of these tricks.

  11. P. 11-13 # 6, 7, 8, 11, 15, 17. P. 18-20 # 10, 11, 15. Independent Practice

  12. 1.3 – Surface area of objects made from right rectangular prisms Chapter 1

  13. Right rectangular prisms • Assume each side length is 1 unit, and the surface area of each face is 1 unit2. • What is the surface area of 1 cube? • Put two cubes together to make a “train.” • What’s the area of this train? • Make a table! • What pattern do you see? Would this go on forever?

  14. Challenge Use five cubes to build the object with the lowest surface area that you can. • What’s the highest surface are you can find? • What kind of object will always have the highest surface area?

  15. Unit cubes What are some of the different ways that we can calculate the surface area when we’re working with these unit cubes? • Count the squares that we can see on each side/view. • Count the sides of all the squares, and subtract the ones that are touching one another. These shapes that are made up of a combination of other shapes are called composite objects. Composite Object Creator How would it change the surface area if I told you that the side length of each cube was 2 units, instead of just 1 unit?

  16. example What is the surface area of the whole building? We can either add up everything that we see, or we can add everything, and just subtract the overlap. In these cases, it may be easiest just to subtract the areas where the shapes overlap. Overlap: 20 x 10 = 200 x 2 = 400 Subtract: 6600 – 400 = 6200 Prism 1: 20 x 30 = 600 x 2 = 1200 (sidewalls) 20 x 60 = 1200 x 2 = 2400 (front/back walls) 60 x 30 = 1800 (roof) Prism 2: 10 x 20 = 200 x 4 = 800 (walls) 20 x 20 = 400 (roof) Prism 1 + Prism 2 = 1200 + 2400 + 1800 + 800 + 400 = 6600 m2 Where is the overlap? The surface area is 6200 m2.

  17. Example • A contractor quotes to paint the exterior of the building at a rate of $2.50/m2. These parts of the building are not to be painted: • the 2 roofs • the office door with area of 2 m2 • 3 loadingdoors, eachmeasuring 10 m by 15 m • 4 windows, each with an area of 1 m2. • How much would it cost to paint the building. • We should try to figure out the area of what we aren’t painting. • The two roofs: 400 +1800 = 2200 • The office door = 2 • The 4 windows = 4 • 3 loading doors: 10 x 15 x 3 = 450 • 2200 + 2 + 4 + 450 = 2656 m2 • 6200 – 2656 = 3544 m2 Each m2 costs $2.50, so:  3544 x 2.50 = $8860.00 The building will cost $8860.00 to paint.

  18. Pg. 30-32, # 6, 8, 10, 11, 13, 15, 17, Independent Practice

  19. 1.4 – Surface areas of other composite objects Chapter 1

  20. Triangular prisms and Cylinders S1 S2 S3 H 5 bh + (S1 + S2 + S3)H 3 2πr2 + 2πrh # of Faces = SA = # of Faces = SA =

  21. example Two round cakes have diameters of 14 cm and 26 cm, and are 5 cm tall. They are arranged as show. The cakes are covered in frosting. What is the area of frosting? • Top cake: • π(7)2 + 2π(7)(5) = 49π + 70π =373.85 cm2 • Bottom cake: • π(13)2 + 2π(13)(5) = 530.93 + 408.41 • = 939.34 cm2 • The overlap is the area between: • π(7)2 = 153.94 • 939.34 + 373.85 – 153.94 = 1158.91 Will the bottom of the cake be frosted? The area of the frosting will be 1159 cm2.

  22. Example • How many faces are there to paint? • 1 triangle (the other is against the house) • 3 rectangles (on the triangular prism) • 2 columns • 4 rectangles on the base • We need the height of the triangle, so we need to use Pythagorean: • c2 = a2 + b2 • 22 = 12 + h2 • h2 = 4 – 1 = 3 • h = 1.732 2.0 m The roof, columns, and base of this porch are to be painted. The radius of each column is 20 cm. What is the area to be painted, to the nearest square metre? 1.0 m Triangle:  (1/2)bh = (1/2)(2)(1.732) = 1.732 m2

  23. Example (continued) • How many faces are there to paint? • 1 triangle (the other is against the house) • 3 rectangles (on the triangular prism) • 2 columns • 4 rectangles on the base Triangle = 1.732 cm2 Rectangles = 3 x (2.0 x 2.2) = 13.2 cm2 Columns = 2 x 2π(0.20)(2.5) = 6.283 cm2 Base = (2 x 2.2) + 2 x (2.2 x 0.15) + (2.0 x 0.15) = 5.36 cm2 There are four circles not being painted on the base/roof:  π(.20)2 x 4 = 0.504 cm2  1.732 + 13.2 + 6.283 + 5.36 – 0.504 = 26.071 cm2 The roof, columns, and base of this porch are to be painted. The radius of each column is 20 cm. What is the area to be painted, to the nearest square metre? The area to be painted is 26 cm2.

  24. Pg. 40-43, # 4, 6, 8, 9, 10, 12, 13, 15, 16 Independent Practice

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