PRISMS. PARTS of a PRISM BASE FACE HEIGHT BASE FACE HEIGHT.

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Presentation transcript:

PRISMS

PARTS of a PRISM BASE FACE HEIGHT BASE FACE HEIGHT

TOTAL SURFACE AREA The sum of the areas of each face. T.A. = ph + 2B p = perimeter of the base h = height of the prism

VOLUME of a PRISM V = Bh B = area of the Base h = height of the prism

A right triangular prism is shown. Find the total surface area since the volume = 315. T.A. = ph + 2B V= Bh = 315 V= (½·10.5·4)h = 315 V= 21h = 315 h = 15 p = h = 15 B =½·10.5·4 = 21 T.A. = 24(15) + 2(21) T.A. = 402

Parts of a Pyramid

SLANT HEIGHT The height of the isosceles triangular lateral face

Examples of Pyramids SLANT HEIGHT

TOTAL SURFACE AREA T.A. = ½pl + B p= perimeter of the base l = slant height B = area of the base

VOLUME of a Pyramid V = ⅓Bh B = Area of the base h = height of the pyramid

Find the total surface area and volume of the pyramid. 25 m 24 m 20 m T.A. = ½pl + B p = 14(4) = 56 p = 14(4) = 56 l = 24 l = 24 B = 14 (14) = 196 B = 14 (14) = 196 T.A. = ½(56)(24) = 868 V = ⅓ Bh B = 14(14) = 196 h = 20 h = 20 V = ⅓ Bh ⅓ (196)(20)

Cylinders Base is always a circle

Parts of a CYLINDER

TOTAL SURFACE AREA T.A. = 2 пrh + 2B r = radius of the base h = height of the cylinder B= area of the base

VOLUME of a Cylinder V = пr 2 h r = radius of the base h = height of the cylinder

Find the total surface area and volume. 7 in 24 in T.A. = 2пrh + 2B h = 24 r = 7 B = πr 2 = π (7) 2 B = πr 2 = π (7) 2 = 49 π T.A. = 2 π(7)(24) + T.A. = 2 π(7)(24) + 2(49 π) T.A. = 336 π + T.A. = 336 π + 98 π = 434π V = π r 2 h V = π (7) 2 (24) V = 1176 π

CONES

TOTAL SURFACE AREA T.A. = пrl + B r = radius of the base l = slant height of the cone B= area of the base

VOLUME of a Cone V = ⅓пr 2 h r = radius of the base h = height of the cone

Find the total surface area and volume. 16 meters 17 meters T.A. = π rl + B T.A. = π (8)(17) + 64 π B = 8 2 π = 64π T.A. = 136 π + 64 π T.A. = 200 π V = ⅓ π r 2 h V = ⅓ π( 8 2 )15 V = 320 π

Spheres

TOTAL AREA T.A. = 4 пr 2 r = radius of the sphere

VOLUME of a Sphere V = 4/3 пr 3 r = radius of the sphere

A basketball has a diameter of about 9 inches. What is the total area and volume of the basketball? T.A. = 4 π r 2 T.A. = 4 π (9/2) 2 T.A. = 81 π V = 4/3 π r 3 V = 4/3 π (9/2) 3 V = π