1. Bell Ringer 1-31-11 An architect designed a rectangular room with an area of 925 square feet. 1. What equation can be used to find the width of the room? 2. What is the width of the room if the length is 37 feet? 3. What is the perimeter of the room? 37 ft 925 ft 2

2. An architect designed a rectangular room with an area of 925 square feet. 1.What equation can be used to find the width of the room? 37 ft 925 ft 2

3. An architect designed a rectangular room with an area of 925 square feet. 2. What is the width of the room if the length is 37 feet? 37 ft 925 ft 2

4. An architect designed a rectangular room with an area of 925 square feet. 3. What is the perimeter of the room? 37 ft 925 ft 2

5. Quiz 1-31-11 Rectangle STUV is similar to rectangle LMNO. If the area of rectangle STUV is 72 square units, what is the area of rectangle LMNO? T S U V 12 M L N O 6

6. What is the area of LMNO? I first had to find the width of rectangle STUV. I divided the area by the length to get a width of 6 units. Since the figures are similar, I set up a proportion to find the missing dimension of rectangle LMNO. I multiplied by a scale factor of.5 to get the missing dimension. I multiplied the length and width to get the area. If the area of rectangle STUV is 72 square units, what is the area of rectangle LMNO? The area of LMNO is 18 square units. Area STUV=72 units 2 Length STUV = 12 units Length LMNO = 6 units 2 Similar Rectangles Reduction A LMNO=(3)(6) A LMNO=18 units 2

7. Rectangle STUV is similar to rectangle LMNO. If the area of rectangle STUV is 72 square units, what is the area of rectangle LMNO? T S U V 12 M L N O 6 6 3 A=(3)(6) A=18 units 2

8. Speed Test 1. Get out a dry erase marker. 2. You have 1 minute to complete as many problems as you can. 3.We will grade in 1 minute. 4. Graph your results. Keep the graph in your notebook. 5. We will do this every day.

9. Problem of the Week & Word Problem #4-3 1. You have 5 minutes to work on the problem of the week and word problem. 2. The problem of the week must follow the Read, Think, Solve, Justify format. 3. When you are finished, turn them in. 4. They are due Friday.

10. Reach for the Stars Monday The height of the flap of an envelope is 8 inches. The width of the envelope is 12 inches. Calculate the area of the envelope. 8 in 12 in 9 in A. 204 in 2 B. 156 in 2 C. 172 in 2 D. 140 in 2

11. Glencoe pg. 374: 6-15 6.V = 30 in 3 rectangular prism; rectangle 7.V = 216 mm 3 cube or rectangular prism; square 8.V = 525 yd 3 triangular prism; triangle 9.V = 768 m 3 triangular prism; triangle 10.V = 2407.3 cm 3 cylinder; circle 11.V = 55.4 m 3 cylinder; circle 12. V = 408 in 3 rectangular prism; rectangle 13.V = 297.5 ft 3 triangular prism; triangle 14.V = 39,250 m 3 cylinder; circle 15.V = 236.1 cm 3 cylinder; circle 4 pts each 1 pt/label 10 pts: not identifying type of figure and shape of the base

12. M. Up pg. 192: 2,4 pg. 196: 2,4,5 Pg. 192: 2. square pyramid; square A = 39.06 in 2 V = 104 in 3 4. rectangular pyramid; rectangle A = 143 ft 2 V = 477 ft 3 Pg. 196 2. sphere V = 904.3 in 3 4. cone; circle A = 50.24 in 2 V = 150.7 in 3 5. cone; circle A = 4.52 m 2 V = 2.1 m 3

13. Class Work: Volume of Pyramids, Cones, and Spheres You need your notes. Title the notes: Volume of Pyramids, Cones, and Spheres I will check your work at the end of class.

14. Volume The amount of “stuff” a container can hold. Area is measured in Square units = units 2 Volume is measured in Cubic Units = units 3

15. Volume of Rectangular Pyramids Volume = Area of Rectangle (Height) Volume = Area of the Base (Height)

16. Volume of Pyramids 80 = ⅓(48h) h = 5 ft 80 = ⅓(8)(6)h 16 Rectangular Pyramid Base is a rectangle A rectangular pyramid can hold 80 cubic feet. The length of the base is8 ft, and the width of the base is 6 feet. Calculate the height of the pyramid. V = ⅓lwh

17. Volume of Triangular Pyramids Volume = Area of Triangle (Height) Volume = Area of the Base (Height)

18. Volume Identify the type of figure Identify the shape of the base Calculate the area of the base Identify the height Multiply the area of the base by the height Divide by 3 Label (remember to cube 3 ) Triangular Pyramid Triangle B = 10 ft 2 h = 12 ft 40 ft 3 B = Area of the Base Volume = Area of the Base (Height) 10 12

19. Volume of Cones Volume = Area of Circle (Height) Volume = Area of the Base (Height)

20. Volume of Cones V = ⅓ Bh (B is the area of the base of the solid figure) Cone Circle Area: πr 2 V = ⅓πr 2 h V = ⅓(3.14)(3 2 )(7½) V = ⅓(3.14)(9)(7.5) V = ⅓(28.26)(7.5) V = ⅓(210) V = 70 in 3 Lucia made a funnel out of a piece of paper. The radius of the funnel is 3 inches and the height is 7 ½ inches. What is the approximate volume of the funnel to the nearest cubic inch?

21. Volume of Spheres

22. V = 4/3 πr 3 Sphere V = 4/3 πr 3 V = 4/3 (3.14)(5 3 ) V = 4/3 (3.14)(125) V = 4/3 (392.5) V = 524 in 3 The diameter of Sam’s beach ball is 10 inches. What is the approximate volume of the beach ball to the nearest cubic inch?

23. Class Work Glencoe Pg. 373: 1,2,3,9 Pg. 382: 2,3,4,6,11,16 You may use a calculator, but remember to show me all the steps.