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12-5 Surface Area of Pyramids

12-5 Surface Area of Pyramids. Objectives. Find lateral areas of regular pyramids Find surface areas of regular pyramids. Characteristics. All faces except the base intersect at one point called the vertex The base is always a polygon

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12-5 Surface Area of Pyramids

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  1. 12-5 Surface Area of Pyramids

  2. Objectives • Find lateral areas of regular pyramids • Find surface areas of regular pyramids

  3. Characteristics • All faces except the base intersect at one point called the vertex • The base is always a polygon • The faces that intersect at the vertex are called lateral faces and form triangles • The edges of the lateral faces that have the vertex as an endpoint are all lateral edges • The altitude is the segment from the vertex perpendicular to the base

  4. Parts of a Pyramid Vertex Lateral Edge Lateral Face Base Altitude

  5. FYI Lateral area of Regular pyramids can be found by adding the area of all its congruent triangular faces

  6. Formula for Lateral Area of a Pyramid If a regular pyramid has a lateral area of L square units, a slant height of l units, and its base has a perimeter of P units then L=1/2(P)(l )

  7. A regular octagonal pyramid has a side of 25 kilometers and a slant height of 75 kilometers. Find the lateral area of this figure. Example #1 75 Km 25 Km Top-Down view

  8. The Work L=1/2(P)(l) P=200 Slant height (l)=75 L=1/2(200)(75) 75 Km L=1/2(15000) 25 Km Lateral Area=7500 km² Top-Down view

  9. Example #1 Explained Find the slant height Side of base = 25 so the perimeter (P) is 200 km Slant height = 75 km Formula for Lateral Area is L=1/2 (P)(l) P=200 and l=75 L=1/2(200)(75) L=1/2(15000) L=7500km²

  10. Formula For Surface Area of a Regular Pyramid If a regular pyramid has a surface area of T square units, a slant height of l units and its base has a perimeter of P units and area of B square units, then T=1/2(P)(l) +B

  11. The slant height of the pyramid is the hypotenuse of a right triangle with legs that are the altitude from the vertex and a segment with a length that is one half of the side measure of the base. Because of this you can use the Pythagorean Theorem to find a missing side. Slant Height (the hypotenuse) Altitude (a side of the triangle) Segment (like a radius of a circle because it is half the length of the side of a square-also a side of the triangle)

  12. A square pyramid has an altitude of 72 fathoms and a length of one side of the base being 54 fathoms. Before finding the surface area find the slant height. Example #2 l 72 Fathoms 54 Fathoms

  13. L=1/2(P)(l)+B The Work 27² + 72² = c² 729 + 5184 = c² Perimeter=54x4 or 216 fa √5913 = √c² Area=54x54 or 2916 fa² c≈76.89 so l=76.89 fa T=1/2(216)(76.89)+2916 l T=1/2(16608.24)+2916 72 Fathoms T=8304.12+2916 T=11220.2 Fathoms² 54 Fathoms

  14. Example #2 Explained First Find the Slant Height The segment from the center of the pyramid to the side is like a radius so it = 27 fm Use Pythagorean theorem 27² + 72² = c² C=√3913 or 76.9 fathoms which is the slant height (l) Second Find the Surface Area Find the Area and perimeter of base Area (B)=2916 Perimeter (P)= 216 Use the Formula T=1/2(P)(l)+B T=1/2(216)(76.9)+2916 T=1/2(16610.4)+2916 T=8305.2+2916 T=11221.2 fathoms²

  15. example #3-Surface Area of a Pentagonal Pyramid 13 ft 20 ft

  16. 1st-find the missing segment2nd-find ½ the length of one side of the base using Trigonometry3rd-find the area and perimeter of the base4th-use this to find the surface area a²+13²=20² A=15.2 ft 13 ft Tan(36)=x/15.2 X=11 ft 20 ft Perimeter=5(2·11) Area=1/2(22)(15.2)·5 Length of segment from Pyramid (A) Area=836 P=110 ft T=1/2(P)(l)+B T=1/2(110)(20)+836 36º (Half of the central angle) T(Surface Area)=1936 ft² 11 ft

  17. Example #3 Explained First Use Pythagorean Theorem to find the missing side Pythagorean Theorem = 13² + b² = 20² 169 + b² = 400 b²=231 b=15.2 ft Now Find the Length of the sides of the base using Trig The central angle of a pentagon is 72º so half of that is 36º The side of the triangle is the missing segment you found earlier (b) Using this information you can set up a trig equation Tan 36º =x/15.2 15.2 (tan 36º)=x X= 11 ft Next find the perimeter and area of the base P=5(11·2) P=110 ft A=1/2(15.2)(22)·5 A=836 Finally find the surface area The Formula for S.A. is T=1/2Pl+B T=1/2(110)(20)+836 Surface Area (T)=1936 15.2 ft 36º (Half of the central angle) x

  18. Assignment Pg 663-667 7-39 evens 42-45 Pg. 663 7-15, 18-23 Omit 20

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