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CCGPS Unit 5 Overview Area and Volume MCC6.G.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. Examples: • Area of Right Triangles • Area of Triangles • Area of Squares • Area of Kites • Area of Parallelogram • Area of Trapezoid • Real-World Problems with Area Vocabulary Words • Right Triangle • Square • Kite • Parallelogram • Trapezoid • Area A L T I T U D E Base A triangle that has ◦ exactly one 90 angle 4 ft Substitute the values into the equation. • b = 3ft • h = 4ft We are solving for A. 3 ft A = ½ (4 ft)×(3 ft) A = ½ (12 ft2) A = 6 ft2 8.4 ft We are solving for A. A = ½ (8.4 ft)×(6.2 ft) A = ½ (52.08 ft2) 6.2 ft A = 26.04 ft2 A L T I T U D E A polygon having three sides. Base We are solving for A. A = ½ (6 ft)×(4 ft) 4 ft A = ½ (24 ft2) 6 ft A = 12 ft2 9 ft We are solving for A. A = ½(6 ft)×(9 ft) + ½(6 ft)×(9 ft) 6 ft A = ½(54 ft2) + ½(54 ft2) A = 27 ft2 + 27 ft2 9 ft A = 54 ft2 9 ft We are solving for A. 6 ft A = (6 ft.)×(9 ft.) A = 54 ft2 4 ft. 4 ft. A Rhombus is a four-sided Polygon where all sides have equal length (It looks like a someone sat on a square) 4 ft We are solving for A. 3 ft 4 ft A = (4 ft)×(3 ft) A = 12 ft2 15 ft We are solving for A. A = (15 ft)×(10 ft) 10 ft A = 150 ft2 We are solving for A. A = (10 ft)×(10 ft) 10 ft. 10 ft. A = 100 ft2 12 in. Suppose you were asked to find the area of this kite. Using what we already know about triangles, how can we find the area of the kite? We are solving for A. 36 in. A = ½(24in×12 in) + ½(24 in×36 in) A = ½(288 in2) + ½(864 in2) A = 144 in2 + 432 in2 A = 576 in2 24 in. How can we find the area of this trapezoid? 12 in. We are solving for A. 24 in. 12 in. 36 in. 12 in. 12 in. A= ½(12 in× 24 in) ½(12 in× 24 in) + (12 in× 24 in) A= ½(288 in2) ½(288 in2) + (288 in2) A = 144 in2 + 144 in2 + 288 in2 A = 576 in2 Area = ½ Base x Height A = ½ (9 feet) x (25 feet) A = ½ (225 ft2) A = 112.5 ft2 Mr. and Mrs. Brady purchased this home in Marietta, Georgia. Mrs. Brady is tired of looking at the brown dirt patch in the median of the road in front of their home. She has asked her husband to plant something in that area which will give it a more attractive curb appeal. Mr. Brady knows that the soil will need to be treated with fertilizer in order for it to be able to grow beautiful flowers. Mr. Brady needs to find the area of the ground to be treated. He has provided the length of two sides that make a right angle for you to use to determine the area. One side is twenty five feet. The other side is 9 feet. What is the Area? Area = Base x Height A = (2.5 meters) x (50 meters) A = (125 meters2) If one lane is 125 m2. and we have eight lanes than we need to multiple the area of one lane by the total number of lanes 8 125 m2 x 8 = 1000 m2 The Beijing National Aquatics Center has request your help. They require a cover for the Olympic swimming pool that is housed within this beautiful building. The pool has eight lanes that are each 2.5 meters wide. The length of each lane is 50 meters long. How long will the cover need to be if you wish to cover only the surface of the pool? MCC6.G.2 Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate fraction edge lengths, and show that the volume is the same as it would be found by multiplying the edge lengths of the prism. Apply the formulas V = lwh and V = Bh to find the volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems. Examples: • Use unit cubes to pack a right rectangular prism with fractional edge lengths • Find the volume of right rectangular prisms using the volume formulas • Real-world problems Vocabulary Words • Prism • Right Rectangular Prism • Edge • Base of a Prism • Volume • Unit Cubes Find the volume of a rectangular prism with fractional edge lengths You cannot use a whole unit block as a visual because when you try the 2 units block will extend past the boundary of the original shape. To fill a fractional side we will need a fractional unit cube. 2 units 1.5 units 1 unit Now we can insert the ½ unit blocks onto the bottom row of our figure . 6 x ½ = 3 whole unit blocks on bottom row 6 X ½ = 3 whole unit blocks on the top row 3 + 3 = 6 total whole unit blocks 1 unit 1 unit1/2 in. MCC6.G.4 Represent threedimensional figures using nets made up of rectangles and triangles using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems. Examples: • Nets of triangular prisms • Nets of rectangular prisms • Nets of square pyramids • Nets of rectangular pyramids • Surface area of each of the prisms and pyramids • Real-world problems Vocabulary Words • Nets • Triangular Prisms • Pyramid • Surface Area 6 ft. By looking at a net of this square pyramid determine the surface area. Surface are of the square in the middle: 2 SA (5ftft.)×(5 ft.) A ==25 5 ft. Surface are of one 2 of the triangle segments: SA = 25 ft Surface are of one of the triangle segments: A = 15 ft2 SA = ½ (5 ft.)×(6 ft.) Total Surface Area: 1 square & 4 Triangles 2 2 A SA==25 15ftft2+ 4 (15 ft ) A = 85 ft2 3 ft. C A 1 ft. B C A 3 ft. B 5 ft. By looking at a net of this rectangular prism determine the surface area. Similar rectangles are By looking at a net of this labeled. rectangular prism determine the Surface area of all six rectangles. surface area. Similar rectangles are Rectangle labeled. A - SA = 3 ft2 2. Rectangle A SA = + 3 ft Surface area of the rectangle C. B. A. Rectangle B - SA = + 15 ft2 SA = (5 (3 ft.)×(1 ft.)×(5 ft.)= + 15 ft2 Rectangle B - SA Rectangle C - SA = + 5 ft2 2 2 SA = 15 5 ftft 3 Rectangle C - SA = + 5 ft2 Total = SA = 46 ft2