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Section 12-2 Pyramids Section 12-3 Cones

Section 12-2 Pyramids Section 12-3 Cones. Regular Pyramid –. 1) Base is a regular polygon. Vertex. 2) Faces are congruent isosceles triangles. 3) Altitude meets the base at its center. Slant Height -. Not the same as lateral edge. Altitude.

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Section 12-2 Pyramids Section 12-3 Cones

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  1. Section 12-2 Pyramids Section 12-3 Cones

  2. Regular Pyramid – 1) Base is a regular polygon Vertex 2) Faces are congruent isosceles triangles 3) Altitude meets the base at its center Slant Height - Not the same as lateral edge Altitude

  3. Lateral Area – area of one face multiplied by number of sides Lateral Area can also be found by taking half the perimeter of the base times the slant height. V = Bh L.A. = ½pl T.A. = L.A. + B

  4. Find the L.A., T.A. and Volume. 10 8in. 10 8in. 6in. = (12·12)(8) V = Bh T.A. = L.A. + B 6in. 6 12 T.A. = 240 + (12·12) L.A. = ½pl T.A. = 384 in² L.A. = ½(12·4)(10) L.A. = 240 in² V = 384 in³

  5. Find the L.A.,T.A. and Volume. 5 4 5in. 3 V = Bh = (6·6)(4) T.A. = L.A. + B T.A. = 60 + (6·6) 6in. T.A. = 96 in² L.A. = ½pl L.A. = ½(6·4)(5) V = 48 in³ L.A. = 60 in²

  6. Cone – Just like a pyramid, but with a circular base L.A. = ½·2πr·l Slant Height T.A. = πrl + πr² Altitude V = πr²h

  7. Find the L.A., T.A. and Volume. 13m 13m 5 5 12m 24m T.A. = πrl + πr² L.A. = ½·2πr·l T.A. = π(12·13) + π12² L.A. = π(12)·13 T.A. = 156π + 144π L.A. = 156π m² T.A. = 300π m²

  8. 13m 5 24m V = πr²h V = π(12)²(5) V = 240π m³

  9. Find the L.A., T.A. and Volume. 3²+ 6²= x² x²= 45 T.A. = πrl + πr² T.A. = π(3· ) + π3² 6 ft T.A. = 9π + 9π ft² 3ft L.A. = ½·2πr·l V = πr²h V = π(3)²(6) L.A. = π(3)· V = 18π ft³ L.A. = 9π ft²

  10. A solid metal cylinder with radius 6cm and height 18cm is melted down and recast as a solid cone with radius 9cm. Find the height of the cone. V = πr²h V = π(6)²(18) 648π = π(9)²h 648π = 27πh V = 648π cm³ V = πr²h h = 24cm

  11. A soup can has a height of 5 inches and a diameter of 6 inches. Determine the area of the label for the can. L.A. = 2πrh L.A. = 30π in² L.A. = 2π(3)·5 If 1 oz = 11.5 cubic inches, about how many ounces are in the soup can? V = πr²h V = π(3)²·5 12.3 oz. 141.3 in³ V = 45π =

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