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Review of Formulas. TAKS WARM UP. 1. Find the circumference and area: A. B. 3.3 cm. 4.5. 8 in. Find the volume of a box whose dimensions are 15 feet wide, 14feet long, and 44 feet high. How man square feet are in a parallelogram whose base
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TAKS WARM UP • 1. Find the circumference and area: • A. B. 3.3 cm 4.5 8 in • Find the volume of a box whose dimensions are 15 feet wide, • 14feet long, and 44 feet high. • How man square feet are in a parallelogram whose base • measures 8 meters and has a height of 6.4 meters? • 4. Paul is 6 ft tall and casts a 9 ft shadow. At the same time • of day a telephone pole casts a 54 ft shadow. Write a • proportion to find the height of the telephone pole.
Review perimeters, circumference, and area of rectangles, parallelograms, triangles, and circles.
Perimeter of a rectangle. The perimeter of a rectangle is found by? 36” 16”
Perimeter of a rectangle. The perimeter of a rectangle is found by? Adding the lengths of the four sides, or add the lengths of two adjacent sides and multiply by 2. In symbols this is P = 2 x (l + w), or P = 2l + 2w 36” 16”
Area of a rectangle. The area of a rectangle is found by? 36” 16”
Area of a rectangle. The area of a rectangle is found by? Multiplying the length by the width. In symbols this is A = l x w 36” 16”
Perimeter of a parallelogram The perimeter of a parallelogram is found by? 30” 10”
Perimeter of a parallelogram The perimeter of a parallelogram is found by? Adding the lengths of the four sides, or add the lengths of two adjacent sides and multiply by 2. In symbols this is P = 2 x (l + w), or P = 2l + 2w 30” 10”
Area of a parallelogram The area of a parallelogram is found by? Height = 8” 30” = Base
Area of a parallelogram The area of a parallelogram is found by? Multiply the base times the height. In symbols this is A = b x h. Height = 8” 30” = Base
Area of a parallelogram Multiply the base times the height. In symbols this is A = b x h. This is why it works. Height = 8” 30” = Base
Area of a parallelogram Multiply the base times the height. In symbols this is A = b x h. This is why it works. If we cut the parallelogram at ½ of the base and rearrange the pieces we have a rectangle Part B Part A Height = 8” Part A Part B 30” = Base
Perimeter of a triangle The perimeter of a triangle is found by? b = 30” c = 25” a = 40”
Perimeter of a triangle The perimeter of a triangle is found by? Adding the lengths of the three sides. In symbols this is P = a + b + c b = 30” c = 25” a = 40”
Area of a triangle The area of a triangle is found by? Height = 20” base= 40”
Area of a triangle The area of a triangle is found by? Multiply the base by the height and take half the result. In symbols this is A = ½ x b x h. height = 20” base= 40”
Area of a triangle In symbols this is A = ½ x b x h. This is why it works. If we cut the triangle at ½ the height and rearrange pieces A,B, and C we create a rectangle that is ½ the height of the triangle by the base or length by width. “B” “C” 10” “A” “B” “C” base= 40” “A” base= 40”
Circumference of a circle The circumference of a circle is found by? 10”
Circumference of a circle The circumference of a circle is found by? Multiply the diameter by ∏ (approximately 3.14) In symbols this is C = ∏ x d. 10”
Area of a circle The area of a circle is found by? 10”
Area of a circle The area of a circle is found by? Formula: ∏r2 Multiply 3.14 X (r x r) {Remember PMDAS, make sure you do your operations in the correct order.} 10” 3.14 X (5 X 5) = 3.14 x 25 = 3.14 X 25
Area of a circle The area of a circle is found by? Formula: ∏r2 Multiply 3.14 X (r x r) {Remember PMDAS, make sure you do your operations in the correct order.} 10” 3.14 X (5 X 5) = 3.14 x 25 = 3.14 X 25 1550 +6280
Area of a circle The area of a circle is found by? Formula: ∏r2 Multiply 3.14 X (r x r) {Remember PMDAS, make sure you do your operations in the correct order.} 10” 3.14 X (5 X 5) = 3.14 x 25 = 3.14 X 25 1550 +6280 7830
Area of a circle The area of a circle is found by? Formula: ∏r2 Multiply 3.14 X (r x r) {Remember PMDAS, make sure you do your operations in the correct order.} 10” 3.14 X (5 X 5) = 3.14 x 25 = 3.14 X 25 1550 +6280 78.30
6.8A (6.8) Measurement. The student solves application problems involving estimation and measurement of length, area, time, temperature, volume, weight, and angles. The student is expected to: (A) estimate measurements (including circumference) and evaluate reasonableness of results 27
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