1 / 27

2.2 Surface Area

2.2 Surface Area. Focus on …. • solving problems involving the surface area of three dimensional objects • finding an unknown dimension of a three dimensional object given its surface area. Calculate Surface Areas. Example 1 Calculate the Surface Area of a Right Cone.

ricardojohnson
Download Presentation

2.2 Surface Area

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. 2.2 Surface Area

  2. Focus on … • solving problems involving the surface area of three dimensional objects • finding an unknown dimension of a three dimensional object given its surface area

  3. Calculate Surface Areas

  4. Example 1 Calculate the Surface Area of a Right Cone • A right cone has a circular base with diameter 27 cm and slant height 40 cm. • Calculate the surface area of the cone, to the nearest tenth of a square centimetre.

  5. Solution • The surface area of a right cone is the sum of the area of the single circular base, B, and the lateral area. • SA = πr2 + πrs • SA = π(13.5)2 + π(13.5)(40) • SA = 2269.015… • The surface area of the cone is approximately 2269.0 cm2.

  6. Your Turn • Sketch a right cone with diameter 16 cm and slant height 12 cm. • What is its surface area?

  7. SA = πr2 + πrs • SA = 3.14(8)2 + 3.14(8)(12) • SA = 200.96 + 301.44 • SA = 502.4 cm2

  8. Example 2 Calculate the Surface Area of a Right Pyramid • A right rectangular pyramid has a rectangular base measuring 16 cm by 20 cm. The slant height of the triangular face with the shorter base is 18 cm, while the slant height of the triangular face with the longer base is 17 cm. What is the surface area of the pyramid?

  9. Break it into pieces… • You have 1 rectangle, and 2 triangles of one size, and 2 triangles of the other size. • Area of rectangle: A = lw A = (16)(20) Area of rectangle = 320 cm2

  10. Triangles: • Triangle 1&2 • Area = ½(base)(height) • Area = ½ (20)(17) • Area = 170 cm2 x 2 (2 triangles) = 340 cm2

  11. Triangles 3&4 • Area = ½(base)(height) • Area = ½(16)(18) • Area = 144 (x2 triangles) • Area = 288 cm2

  12. Now, add it all together… • Rectangle = 320 • Triangles = 340 + 288 • Total = 948 cm2

  13. Your Turn • Sketch a right rectangular pyramid with a square base measuring 10 cm on each side. The slant height of each face is 8.5 cm. • What is the surface area of the pyramid?

  14. Example 3 Calculate the Surface Area of a Sphere • A spherical satellite is wrapped with polyester film to protect it during transportation. How much film is required to cover the Echo Satellite that has a circumference of 95.8 m? Express your answer to the nearest tenth of a square metre.

  15. Solution • The formula for the surface area of a sphere is SA = 4πr2, where r is the radius. • Calculate the radius • C = 2πr • 95.8 = 2πr • r = 95.8/2π • r = 15.247…

  16. Substitute the radius into the formula for the surface area. • SA = 4πr2 • SA = 4π(15.247…)2 • SA = 2921.3168 • Approximately 2921.3 m2of film is required to cover the satellite.

  17. Example 4 Determine a Dimension When the Surface Area is Known • The surface area of an official 5-pin bowling ball varies from approximately 459.96 cm2 to 506.71 cm2. • What is the variation in the diameter of the bowling ball?

  18. Solution

  19. Solution

  20. Solution

  21. Solution

  22. Solution

  23. Reread the question • Find the diameter • Find the difference of diameters • Include correct units.

  24. Your Turn • To the nearest millimetre, calculate the radius of a sphere with a surface area of 1 m2.

  25. Assignment • P.74 #1-17 odd

More Related