Cones, Pyramids and Spheres Volume & Surface Area Volume & Surface Area Cones, Pyramids and Spheres 1

Identical isosceles triangles Pyramids A Pyramid is a three dimensional figure with a regular polygon as its base and lateral faces are identical isosceles triangles meeting at a point. Identical isosceles triangles base = quadrilateral base = heptagon base = pentagon

Pyramids

Surface Area of Pyramids Find the surface area of the pyramid. height h = 8 m side s = 6 m apothem a = 4 m Surface Area = area of base + (#of sides of base) (area of one lateral face) What shape is the base? Area of a pentagon h = ½ Pa (P = perimeter) = ½ (5)(6)(4) = 60 m2 l a s

Surface Area of Pyramids What shape are the lateral sides? Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m Area of a triangle = ½ base (height) = ½ (6)(8.9) = 26.7 m2 Attention! the height of the triangle is the slant height ”l ” h l l 2 = h2 + a2 = 82 + 42 = 80 m2 l = 8.9 m a s

Surface Area of Pyramids Surface Area of the Pyramid = 60 m2 + 5(26.7) m2 = 60 m2 + 133.5 m2 = 193.5 m2 Find the surface area of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m h l a s

Pyramids Creamed Coconut 1/3 of the calories 7

= + + Volume of Pyramids Volume of a Pyramid: V = (1/3) Area of the base x height V = (1/3) Ah Volume of a Pyramid = 1/3 x Volume of a Prism = + +

Exercise #2 Find the volume of the pyramid. height h = 8 m apothem a = 4 m side s = 6 m Volume = 1/3 (area of base) (height) = 1/3 ( 60m2)(8m) = 160 m3 h Area of base = ½ Pa a = ½ (5)(6)(4) = 60 m2 s

Pyramids Volume = 1/3 x base area x height Find the volume of this pyramid 10

Cones Pringles A third of the calories A fact: If Pringles came in a cone, which was the same height and diameter as the tall tube, it would contain one third of the calories!!! Why?? Pringles A third of the calories 11

Cones

The Cone + + = Volume of a Cone = A Cone is a three dimensional solid with a circular base and a curved surface that gradually narrows to a vertex. + + = Volume of a Cone =

Exercise #1 = (1/3)(3.14)(1)2(2) = 3.14(1)2(2) = 2.09 m3 = 6.28 m3 Find the volume of a cylinder with a radius r=1 m and height h=2 m. Find the volume of a cone with a radius r=1 m and height h=1 m Volume of a Cylinder = base x height = pr2h = 3.14(1)2(2) = 6.28 m3 Volume of a Cone = (1/3) pr2h = (1/3)(3.14)(1)2(2) = 2.09 m3

Find the area of a cone with a radius r=3 m and height h=4 m. Surface Area of a Cone Find the area of a cone with a radius r=3 m and height h=4 m. r = the radius h = the height l = the slant height Use the Pythagorean Theorem to find l l 2 = r2 + h2 l 2= (3)2 + (4)2 l 2= 25 l = 5 Surface Area of a Cone = pr2 + prl = 3.14(3)2 + 3.14(3)(5) = 75.36 m2

Cones Slant height (l) Volume = Example: find the volume of this cone 16

A sphere is the locus of points in space that are a fixed distance from a given point called the center of a sphere. A radius of a sphere connects the center of the sphere to any point on the sphere. A hemisphere is half of a sphere. A great circle divides a sphere into two hemispheres

Spheres

Example 1A: Finding Volumes of Spheres Find the volume of the sphere. Give your answer in terms of . Volume of a sphere. __ = 2304 in3 Simplify.

Example 1B: Finding Volumes of Spheres Find the diameter of a sphere with volume 36,000 cm3.

Example 1C: Finding Volumes of Spheres Find the volume of the hemisphere.

Example 3A: Finding Surface Area of Spheres Find the surface area of a sphere with diameter 76 cm. Give your answers in terms of . SA = 4r2 Surface area of a sphere SA = 4(38)2 SA = 5776 cm2 Simplify.

Example SA = 4r2 SA = 4(1)2 SA = 4 m2 V = 4 (1)3 3 V = 4 m3 3 Find the surface area and volume of the sphere. SA = 4r2 SA = 4(1)2 SA = 4 m2 V = 4 (1)3 3 V = 4 m3 3

Example Find the surface area and volume of the hemisphere. SA = 2r2 + r2 = 2(11)2 +(11)2 = 363 in2