1 / 8

Objective: 1) To find perimeters of rectangles and squares, and circumferences of circles.

Ch 1.9 Standard 8.0 : Students know, derive, and solve problems involving perimeter, circumference, and area. Standard 10.0 : Students compute areas of polygons, including rectangles. Objective: 1) To find perimeters of rectangles and squares, and circumferences of circles.

rene
Download Presentation

Objective: 1) To find perimeters of rectangles and squares, and circumferences of circles.

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ch 1.9Standard 8.0: Students know, derive, and solve problems involving perimeter, circumference, and area.Standard 10.0:Students compute areas of polygons, including rectangles. Objective: 1) To find perimeters of rectangles and squares, and circumferences of circles. 2) To find areas of rectangles, squares, and circles.

  2. Perimeter The perimeterP of a plane figure is the sum of the side lengths of the figure.

  3. Example (Perimeter) 1) Find the perimeter of the rectangle. = 2(6) + 2(4) = 12 + 8 = 20 in. 2) Find the perimeter of a square with s = 3.5 in. P = 4s = 4(3.5) = 14 in.

  4. Circumference The circumferenceC of a circle is the distance around the circle which can be calculated using the formula: C = πd or C = 2πr

  5. Example (Circumference) Find the circumference of a circle with radius 14m. = 2π(14) = 28π  88.0 m

  6. Area The areaA of a plane figure is the sum of the areas of its non-overlapping parts.

  7. Example (Area) 1) Find the area of the rectangle. = (6 in)(4 in) = 24 in2 2) Find the area of a square with s = 3.5 in. A = s2 A = (3.5)2 A = 12.25 in2

  8. Area of a Circle The areaA of a circle is the sum of the areas of its non-overlapping parts which can be calculated using the formula: A = πr2 Example: Find the area of a circle with radius 8 cm. A = πr2 = π(8)2 = 64π  201.1 cm2

More Related