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Area of Parallelograms

Area of Parallelograms. Bell Ringer-Monday. 9-3. Area of Parallelograms. Course 2. The area of a figure is the number of unit squares needed to cover the figure. Area is measured in square units. AREA OF A RECTANGLE.

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Area of Parallelograms

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  1. Area of Parallelograms

  2. Bell Ringer-Monday

  3. 9-3 Area of Parallelograms Course 2 The area of a figure is the number of unit squares needed to cover the figure. Area is measured in square units. AREA OF A RECTANGLE The area A of a rectangle is the product of its length l and its width w. w A = lw l

  4. 9-3 Area of Parallelograms Course 2 Additional Example 1: Finding the Area of a Rectangle Find the area of the rectangle. 4.5 in. 7.4 in. A = lw Use the formula. Substitute for l and w. A = 7.4 · 4.5 Multiply. A = 33.3 The area of the rectangle is 33.3 in2.

  5. 9-3 Area of Parallelograms Course 2 Check It Out: Example 1 Find the area of the rectangle. in. A = lw Use the formula. Substitute for l and w. Multiply.

  6. 9-3 Area of Parallelograms Course 2 Additional Example 2: Finding Length or Width of a Rectangle The area of a playing field is 1,470 ft2 and the length is 504 in. What is the width of the field? Use the formula for the area of a rectangle. A = lw 1,470 = 42 · w Substitute 1,470 for A and 42 for l. 1,470 42 42 42 = · w Divide both sides by 42 to isolate w. 35 = w The width of the playing field is 35 ft.

  7. 9-3 Area of Parallelograms Course 2 Check It Out: Example 2 The area of a rectangular parking lot is 3,570 ft2 and the width is 840 in. What is the length of the parking lot? Use the formula for the area of a rectangle. A = lw 3,570 = l · 70 Substitute 3,570 for A and 70 for w. 3,570 70 70 70 = · w Divide both sides by 70 to isolate w. 51 = l The length of the parking lot is 51 ft.

  8. 9-3 Area of Parallelograms Course 2 Definition:

  9. 9-3 Area of Parallelograms AREA OF A PARALLELOGRAM h b Course 2 The area A of a parallelogram is the product of its base b and its height h. A = bh

  10. 9-3 Area of Parallelograms Course 2 Additional Example 3: Finding the Area of a Parallelogram Find the area of the parallelogram. A = bh A = 16 · 8 8 m A = 128 16 m The area of the parallelogram is 128 m2.

  11. 9-3 Area of Parallelograms Course 2 Check It Out: Example 3 Find the area of the parallelogram. A = bh A = 12 · 6 6 cm A = 72 12 cm The area of the parallelogram is 72 cm2.

  12. 9-3 Area of Parallelograms Course 2 Additional Example 4: Measurement Application A carpenter is covering a 150 ft2 floor with square tiles that are each 2 ft in length. What is the least number of tiles the carpenter will need to cover the floor? First find the area of each tile. A = lw Use the formula for the area of a square. A = 2 · 2 Substitute 2 for l and 2 for w. A = 4 Multiply. The area of each square tile is 4 ft2.

  13. 9-3 Area of Parallelograms Course 2 Additional Example 4 Continued To find the number of tiles needed, divide the area of the floor by the area of one tile. 150 ft2 4 ft2 = 37.5 Since covering the floor requires more than 37 tiles, the carpenter would need at least 38 tiles.

  14. 9-3 Area of parallelograms Course 2 Insert Lesson Title Here Check It Out: Example 4 Amanda is covering a 200 in2 mosaic frame with square tiles that are each 1.5 inches in length. What is the least number of tiles Amanda will need to cover the frame? First find the area of each tile. A = lw Use the formula for the area of a square. A = 1.5 · 1.5 Substitute 1.5 for l and 1.5 for w. A = 2.25 Multiply. The area of each square tile is 2.25 in2.

  15. 9-3 Area of Parallelograms Course 2 Insert Lesson Title Here Check It Out: Example 4 Continued To find the number of tiles needed, divide the area of the frame by the area of one tile. 200 ft2 2.25 ft2 ≈ 88.9 Since covering the frame requires more than 88 tiles, Amanda would need at least 89 tiles.

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