Volume of Pyramids and Cones

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12.5 Volume of Pyramids and Cones

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Presentation transcript:

Volume of Pyramids and Cones 10.3 There is a simple relationship between the volumes of prisms and pyramids with congruent bases and the same height, and between cylinders and cones with congruent bases and the same height. Volume of Pyramid. Volume of Cone. JRLeon Geometry Chapter 10.2 HGSH

Volume of Pyramids and Cones 10.3 Find the volume of a regular hexagonal pyramid with a height of 8 cm. Each side of its base is 6 cm. Find the area of the base. 3 6 3 3 𝑩= 𝟏 𝟐 𝒃𝒉 (𝒏) 𝑩= 𝟏 𝟐 (𝟔)(𝟑 𝟑 )(𝟔) 𝑩= 𝟏 𝟐 (𝟑𝟔)(𝟑 𝟑 ) 𝑩= 𝟓𝟒 𝟑 Find the volume. V = 𝟏 𝟑 BH V = 𝟏 𝟑 (𝟓𝟒 𝟑 )(8) V = 𝟏𝟒𝟒 𝟑 𝒄𝒎𝟑 JRLeon Geometry Chapter 10.2 HGSH

Volume of Pyramids and Cones 10.4 JRLeon Geometry Chapter 10.2 HGSH

Volume of Pyramids and Cones 10.4 JRLeon Geometry Chapter 10.2 HGSH

Displacement and Density 10.5 What happens if you step into a bathtub that is filled to the brim? If you add a scoop of ice cream to a glass filled with root beer? In each case, you’ll have a mess! The volume of the liquid that overflows in each case equals the volume of the solid below the liquid level. This volume is called an object’s displacement. EXAMPLE: An important property of a material is its density. Density is the mass of matter in a given volume. You can find the mass of an object by weighing it. You calculate density by dividing the mass by the volume: JRLeon Geometry Chapter 10.5 HGSH

Displacement and Density 10.5 EXAMPLE: First, find the volume of displaced water. Then divide the mass by the volume to get the density of the metal. JRLeon Geometry Chapter 10.5 HGSH

Volume of Pyramids and Cones 10.4 10.3 pg. 540: 2-10 even 10.4 pg. 548: 1-3 10.5 pg. 552 : 1-8 JRLeon Geometry Chapter 10.2 HGSH