1 / 11

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.

Download Presentation

Section 12-2 Pyramids Section 12-3 Cones

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. 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 =

More Related