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Surface Area of Pyramids

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Presentation on theme: "Surface Area of Pyramids"— Presentation transcript:

1 Surface Area of Pyramids

2 ADDITION TO DIAGRAM – NEW VOCAB
The slant height of a regular pyramid is the distance from the vertex to the midpoint of an edge of the base.

3 The lateral faces of a regular pyramid can be arranged to cover half of a rectangle with a height equal to the slant height of the pyramid. The width of the rectangle is equal to the base perimeter of the pyramid.

4 Add the surface area formula to the notes for pyramids.

5 Example 1 Find the surface area of the regular pyramid. n represents the number of sides of the base, and s represents the length of one side of the base, and l is the slant height. n = 3, s = 14, l = 14 Split the base in half and use triangles to find the height of the base (7√3) SA = .5*14*42+.5*14*7√3 = square units

6 Example 2 Find the surface area of the regular pyramid. n represents the number of sides of the base, and s represents the length of one side of the base, and l is the slant height. n = 6, s = 5.2, l = 13 Use triangles to find the apothem of the base (2.6√3) SA = .5*13* *2.6√3*31.2 = square units

7 Example 3 Find the surface area of the regular pyramid. n represents the number of sides of the base, and s represents the length of one side of the base, and l is the slant height. n = 4, s = 12, l = 13 SA = .5*13*48+12*12 =456 square units

8 Example 4 Work backwards to solve for the missing information.
In a rectangular pyramid, one side of the base is 30 in. The slant height of the pyramid is 29 in, and the SA = 4180 square inches. What is the length of the other side of the rectangular base? 4180=.5*29(30*2+2w)+30w 4180=14.5(60+2w)+30w 4180=870+29w+30w 3310=59w 56.1 in =w

9 Example 5 Work backwards to solve for the missing information.
In a triangular pyramid, the base area is 50 square mm. The slant height of the pyramid is 40 mm, and the SA = 250 square mm. What is the perimeter of the triangular base? 250=.5*40*p+50 200=20p 10 mm =p

10 Example 6 Work backwards to solve for the missing information.
In a square pyramid, the slant height is 5 cm, and the SA = 96 square cm. What is the length of one side of the base? 96=.5*5*4s+s^2 96=10s+s^2 0 = s^2+10s-96 Factor to solve for s. 0 = (s+16)(s-6) This gives s = -16 or 6. Since length can’t be negative, s = 6 cm.


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