solid geometry
DESCRIPTION
Solid Geometry. Solid Geometry is the geometry of three-dimensional space It is called three-dimensional , or 3D because there are three dimensions: width, depth and height. Three Dimensions. A geometric object with flat faces and straight edges. each face is a polygon. - PowerPoint PPT PresentationTRANSCRIPT
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Solid Geometry
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Three Dimensions
Solid Geometry is the geometry of three-dimensional spaceIt is called three-dimensional, or 3D because there are three dimensions: width, depth and height.
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Polyhedron
A geometric object with flat faces and straight edges.
each face is a polygon.
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Polyhedron
FACE: Polygon shaped sides of a polyhedron
EDGE: Line segment formed by intersection of two faces
VERTEX: Point where three or more edges meet
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Base
The surface that a solid object stands on – or the bottom line of a shape such as
a triangle or rectangle.
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Polyhedron
Just like a 2D polygon a Polyhedron can be regular
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Polyhedron
…or Irregular
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Polyhedron
A Polyhedron can also be semi-Regular
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Polyhedron
Just like a 2D polygon, a polyhedron can be convex
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Polyhedron
…or concave
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Prism
A solid object that has two identical bases and all flat sides.The shape of the bases give the prism it’s name– "triangular
prism“It is a polyhedron.
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Pyramid
A solid object where:
* The base is a polygon (a straight-sided shape)
* The sides are triangles which meet at the top (the apex).
It is a polyhedron.
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Cylinder
A cylinder is a solid object with:
* two identical flat circular (or elliptical) ends
* and one curved side.
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Cone
A solid (3-dimensional) object that has – a circular base– and one vertex
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Sphere
A solid (3-dimensional) object that has one
curved side
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Prisms & Pyramids
Type Examples Properties
Triangular Prism
● 5 faces 2 triangular bases 3 rectangular faces● 9 edges● 6 vertices
Rectangular Prism
6 faces 2 rectangular bases 4 rectangular faces● 12 edges● 8 vertices
Cube
● 6 faces 2 square bases 4 square faces● 12 edges● 8 vertices
Square Pyramid
● 5 faces 1 square base 4 triangular faces● 8 edges● 5 vertices
Triangular Pyramid
● 4 faces 1 triangular base 3 triangular faces● 6 edges● 4 vertices
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Three Dimensional Figures with Curved Surfaces
Type Example Properties
Cylinder● 2 circular bases
● 1 curved surface
Cone● 1 circular base
● 1 curved surface● 1 vertex
Sphere ● 1 curved surface
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VOLUME
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Prism
V=BhB: Area of the baseh: height/length of the prism
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Prism
V=Bh V= (Area of Triangle) * hV= (½bh) * hV= (½*19*24) *47V=(228) * 47V= 10,716 cm3
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Prism
V=Bh V= (Area of Rect.) * hV= (bh) * hV= (2 * 3) * 6V=(6) * 6V= 36 ft3
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Cylinder
V=BhB: Area of the baseh: height/length of the prism
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Cylinder
V=Bh V= (Area of Circle) * hV= (πr2) * hV= (π * 32) *10V=(π * 9) * 10V= (28.26) * 10V= 282.6 cm3
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Cylinder
V=Bh V= (Area of Circle) * hV= (πr2) * hV= (π * 52) *21V=(π * 25) * 21V= (78.5) * 21V= 1,648.5 ft3
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Pyramid
V=1/3BhB: Area of the baseh: height/length of the prism
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Pyramid
V=1/3Bh V= 1/3(Area of Tri.) * hV= 1/3 * (½bh) * hV= 1/3 * (½* 6 * 4) *5V= 1/3 * (12) * 5V= 1/3 * 60V= 20 cm3
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Pyramid
V=1/3Bh V= 1/3(Area of Sq.) * hV= 1/3 * (b * h) * hV= 1/3 * (5 * 5) *10V= 1/3 * (25) * 10V= 1/3 * 250V= 250/3 units3
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Cone
V=1/3BhB: Area of the baseh: height/length of the prism
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Cone
V=1/3Bh V= 1/3(Area of Circle)* hV= 1/3(πr2) * hV= 1/3(π * 1.52) *5V=1/3(π * 2.25) * 5V= 1/3(7.065) * 5V= 11.78 in3
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Cone
V=1/3Bh V= 1/3(Area of Circle)* hV= 1/3(πr2) * hV= 1/3(π * 82) * 22V= 1/3(π * 64) * 22V= 1/3(200.96) * 22V= 1,473.71 cm3
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Sphere
V=4/3 πr3
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Sphere
V=4/3 πr3 V= 4/3 πr3
V= 4/3 * π * 143
V= 4/3 * π * 2744V= 4/3 * 8,616.16V= 11,488.21 cm3
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Sphere
V=4/3 πr3 V= 4/3 πr3
V= 4/3 * π * 33
V= 4/3 * π * 27V= 4/3 * 84.78V= 113.04 cm3
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SURFACE AREA
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Surface Area
The sum of the area of the bases and lateral surfaces
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Right Prism
SA=2B+PhB: Area of the baseP: Perimeter of a baseh: height/length of the prism
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Right Prism
SA=2B+Ph SA=2(Area of Rect.)+PhSA= 2(bh) + PhSA=2(3*2)+ (3+2+3+2)5SA=2(6) + (10)5SA= 12 + 50SA= 62 cm2
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Right Prism
SA=2B+Ph SA=2(Area of Tri.)+PhSA= 2(1/2bh) + PhSA=2(1/2*3*8) +
(3+8+√73)7SA=2(12) + (11+√73)5SA= 24 + 55 + 5√73SA= 79 + 5√73 m2
SA = 121.72 m2
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Right Cylinder
SA=2B+ChB: Area of the baseC: Circumference of a baseh: height/length of the cylinder
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Right Cylinder
SA=2B+Ch SA=2(Area of Circ.)+ChSA= 2(πr2) + (2πr)hSA=2(π*152)+(2*π*15)48SA= 2(225π) + (30π)48SA= 1413 + 4521.6SA= 5934.6 cm2
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Right Cylinder
SA=2B+Ch SA=2(Area of Circ.)+ChSA= 2(πr2) + (2πr)hSA=2(π*32)+(2*π*3)*9SA= 2(9π) + (6π)*9SA= 56.52 + 169.56SA= 226.08 cm2
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Right Pyramid
SA=B + ½PsB: Area of the baseP: Perimeter of a bases: slant height of the lateral side
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Right Pyramid (& Cone)
What is “slant height”?
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Right Pyramid
SA=B + ½Ps SA=(Area of Rect.)+½PsSA= (bh) + ½PsSA=(12*12) + ½(12+12+12+12)*√136SA= (144) + ½ (48)*√136SA= 144 + 24√136SA= 423.89 units2
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Right Pyramid
SA=B + ½Ps SA=(Area of Rect.)+½PsSA= (bh) + ½PsSA=(16*10) + ½(16+10+16+10)* 17SA= (160) + ½ (52)*17SA= 160 + 442SA= 602 in210 in
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Right Cone
SA=B + ½CsB: Area of the baseP: Circumference of a bases: slant height of the lateral side
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Right Cone
SA=B + ½Cs SA=(Area of Circ.)+½CsSA= (πr2) + ½(2πr)sSA= (πr2) + (πr)sSA= (π62) + (π6)10SA= 36π + 60πSA= 96πSA= 301.59 in2
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Right Cone
SA=B + ½Cs SA=(Area of Circ.)+½CsSA= (πr2) + ½(2πr)sSA= (πr2) + (πr)sSA= (π0.62) + (π0.6)2SA= .36π + 1.2πSA= 1.56πSA= 4.9 ft2
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Sphere
SA=4πr2
r: radius
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Sphere
SA=4πr2 SA= 4πr2
SA= 4*π*42
SA= 4*π*16SA= 64πSA= 200.96 units2
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Sphere
SA=4πr2 SA= 4πr2
SA= 4*π*322
SA= 4*π*1024SA= 4096πSA= 12,861.44 units2