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5-Minute Check on Lesson 11-3

Transparency 11-4. 5-Minute Check on Lesson 11-3. Find the area of each regular polygon. Round to the nearest tenth if necessary. A hexagon with side length 8 cm. A square with an apothem length of 14 in. A triangle with a side length of 18.6 m.

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5-Minute Check on Lesson 11-3

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  1. Transparency 11-4 5-Minute Check on Lesson 11-3 • Find the area of each regular polygon. Round to the nearest tenth if necessary. • A hexagon with side length 8 cm. • A square with an apothem length of 14 in. • A triangle with a side length of 18.6 m. • Find the area of each shaded region. Assume all polygons are regular. Round to the nearest tenth if necessary. • 4. 5. • 6. Find the area of a circle with a diameter of 8 inches. A = 166.3 cm² A = 784 in² A = 149.8 m² 9 A = 51.4 units² 5 A = 92.5 units² Standardized Test Practice: 4π 8π 16π 64π A B C C D Click the mouse button or press the Space Bar to display the answers.

  2. Lesson 11-4 Areas of Irregular Figures

  3. Objectives • Find areas of irregular figures • Find areas of irregular figures on the coordinate plane

  4. Area of Irregular Shapes Irregular Shapes Area:Sum of Separate Parts h r x Example Area:A = ½ circle + triangle + square A = ½ * πr2 + ½ x * h + x * x

  5. Irregular Shapes Example 1 12 12 6 8 Area of shape = Area of semi-circle + Area of square + Area of triangle semi-circle area = ½ πr² = ½ π6² = 18π square area = s² = 12² = 144 triangle area = ½ bh = ½ 12(8) = 48 Area of shape = 18π + 144 + 48 = 192 + 18π = 248.55

  6. Irregular Shapes Example 2 120 yards 12 8 30 yds Area of shape = 2 Areas of semi-circle + Area of rectangle 2*semi-circle area = 2(½ πr²) = π30² = 900π rectangle area = l·w = 120·60 = 7200 Area of shape = 900π + 7200 = 10,027.43 square yards

  7. Example 4-1a Find the area of the figure in square feet. Round to the nearest tenth if necessary. The figure can be separated into a rectangle with dimensions 16 feet by 32 feet, a triangle with a base of 32 feet and a height of 15 feet, and two semicircles with radii of 8 feet.

  8. Example 4-1a Area formulas Substitution Simplify. Use a calculator. Answer: The area of the irregular figure is 953.1 square feet to the nearest tenth.

  9. Example 4-1b Find the area of the figure in square feet. Round to the nearest tenth if necessary. Answer:

  10. rose garden Example 4-2a A rectangular rose garden is centered in a border of lawn. Find the area of the lawn around the garden in square feet. The length of the entire lawn is 25 + 100 + 25 or 150 feet. The width of the entire lawn is 25 + 20 + 25 or 70 feet. The length of the rose garden is 100 feet and the width is 20 feet.

  11. Example 4-2a area of irregular figure = area of entire lawn – area of rose garden Area formulas Substitution Simplify. Simplify. Answer: The area of the lawn around the garden is 8500 sq feet.

  12. Example 4-2b INTERIOR DESIGNCara wants to wallpaper one wall of her family room. She has a fireplace in the center of the wall. Find the area of the wall around the fireplace. Answer:

  13. Example 4-3a Find the area of polygon MNPQR. First, separate the figure into regions. Draw an auxiliary line perpendicular to QR from M (we will call this point S) and an auxiliary line from N to the x-axis (we will call this point K). This divides the figure into triangle MRS, triangle NKM, trapezoid POKN and trapezoid PQSO.

  14. Example 4-3a Now, find the area of each of the figures. Find the difference between x-coordinates to find the lengths of the bases of the triangles and the lengths of the bases of the trapezoids. Find the difference between y-coordinates to find the heights of the triangles and trapezoids.

  15.  Area formulas Substitution Simplify. Example 4-3a Answer: The area of polygon MNPQR is 44.5 sq units.

  16. Example 4-3b Find the area of polygon ABCDE. Answer:

  17. Summary & Homework • Summary: • The area of an irregular figure is the sum of the areas of its nonoverlapping parts • Homework: • pg 619-621; 3, 8-13

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