1. 3.3 How Can I Use a Square? Pg. 11 Squares and Square Roots

2. 3.3 – How Can I Use a Square?___________ Squares and Square Roots You now have the tools to find the area of many complex shapes and are also able to use transformations to create new shapes. In this lesson, you will combine these skills to explore the area of a square and to develop a method for finding the length of the longest side of a right triangle.

3. 3.13 –SIDE OF A SQUARE What do you notice abut the square at right? a. Eunice does not know how to solve for x. Explain to her how to find the missing dimension. = 10

4. b. What if the area of the shape above is instead 66ft 2 ? What would x be in that case?

5. Between 5 and 6

6. Between 4 and 5

7. Find what integers the square root is between. Between 11 and 12

8. Find what integers the square root is between. Between 10 and 11

9. d. You can simplify square roots by using a factor tree. Pairs come out of the square root and "loners" are stuck inside. If possible, simplify the square roots.

10. 3.14 –LENGTH OF THE LONG SIDE OF A TRIANGLE While Alexandria was doodling on graph paper, she made the design at right. She started with the shaded right triangle. She then rotated it 90  clockwise and translated the result so that the right angle of the image was at B. She continued this pattern until she completed the square.

12. a. Draw Alexandria's design below. Be sure it is to scale. It is already started for you. What is the shape of quadrilateral ABCD? How can you tell?

13. 3 73 7 7 3

14. b. What is the shape of the inner quadrilateral? How do you know? square

15. c. What is the area of the inner quadrilateral? Start by finding the area of the large square and the areas of the four equal triangles. Show all work that leads to your conclusion.

16. 3 73 7 7 3 10.5 100 – 4(10.5) 100 – 42 58 10

17. 3 73 7 7 3 10.5 100 – 4(10.5) 100 – 42 58 d. What's the length of the longest side of the shaded triangle?

18. 3.15 – DONNA'S DILEMMA Donna needs help! She needs to find PQ (the length of in ∆PQR shown.

19. c. Find the perimeter of her triangle.

20. 2 5 2 2 2 5 5 5 5 5 5 5

21. 2 5 2 2 2 5 5 5 5 5 5 5 49–20 =29 7 7 Perimeter =

22. 3.16 – ANOTHER WAY Robert complained that while the method from problem 2.50 works, it seems like too much work! He decides to use the area of the triangles, the inner square, and the large outer square to look for a short cut.

23. a. Find the area of the large square around the entire shape. (a+b)(a+b) a 2 + 2ab + b 2

24. b. Find the area of the inner square. Then find the TOTAL area of the four triangles. inner = c 2 4(½bh) = 2ab

25. c. Since the large square has the same area as the 4 triangles and inner square, set them equal to each other and reduce. a 2 + 2ab + b 2 = 2ab + c 2 a 2 + b 2 = c 2

26. 3.17 – A SHORTER WAY Build a square off of each side of the following triangles (the first one is done for you.) Then find the area of each square. What is the relationship between the areas of the squares?

27. 9 16 25 9 + 16 = 25

28. 144 169 25 + 144 =169

29. 64 225 289 64 + 225 =289

30. http://www.cpm.org/flash/technology/pyt hagoreanv1.2.swf

31. 3.18 – PYTHAGOREAN THEOREM The relationship you found in problem 2.65 (between the square of the lengths of the legs and the square of the length of the hypotenuse in a right triangle) is known as the Pythagorean Theorem (pronounced Peh-Tha-Gore-Ian). This relationship is a powerful tool because once you know the lengths of any two sides of a right triangle, you can find the length of the third side.

32. For each triangle below, find the value of the variable. Write answers in simplified square root form.

33. 49 121 y2y2 y 2 + 49 = 121 y 2 = 72

34. 25 16 x2x2 25 + 16 = x 2 31 = x 2

35. x2x2 64 196 x 2 + 64 = 196 x 2 = 132

36. 3.19 – APPLY THE PYTHAGOREAN THEOREM Use the Pythagorean theorem to find the following. a. Examine the rectangle shown below. Find its perimeter and area. Show all work.

37. x 2 + 36 = 225 x 2 = 189

38. b. Graph with A(2, 6) and Then draw a slope triangle. Use the slope triangle to find the length of A B 7 3 7 2 + 3 2 = x 2 49 + 9 = x 2 58 = x 2