Chapter12_Solid Geometry II

8/9/2019 Chapter12_Solid Geometry II

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CHAPTER 12

Math

Form 2

12.1 Geometric Properties

12.2 Net Of Geometric Solids

12.3 Surface Area

Prepared By:

Norisah btMustaffa


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12.1 Geometric Properties12.1 Geometric Properties

Geometric solids are objects that

have three dimensions.

Example:

cylinder

Cube

Prisms cuboid

cones

sphere


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12.1a Prisms12.1a Prisms

Base

Base

Vertex

Lateral edge

Prisms is a solid that has two parallel faces calledbases.

The bases are congruent polygons whereas otherfaces are parallelograms.

Prisms do not have a fixed number ofedges, vertices

and faces.


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A prisms is named by the shape of itsbases.

Hexagonal

prisms

Triangular prism


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Cubes and cuboids are consideredprismsbecause they have all the properties

of aprism.

Two parallel bases arecongruent

Other faces are parallelograms


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12.1b Pyramids12.1b Pyramids

base

Lateral

edge

vertexSquarepyrami

d Triangularpyramid

heigh

t

A pyramid is a solid that has a base in the shape of a polygon.

The other faces are triangles that have a common vertexcalled apex.

A pyramid named by the shape of its base.

The perpendicular distance from the common vertex to thebase is theheight of the pyramid.

Pyramid do not have a fixed number of vertices, edges, and


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12.1c Cylinders12.1c Cylinders

base

base

A cylinder is a solid that hastwo parallel bases which arecongruent circles.

A cylinder has three faces, twoflat faces and one curved face,two curved edges and novertices.

For a right cylinder, theperpendicular distance betweenthe two faces is the height of

the cylinder.

height


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12.1d Cones12.1d Cones

base

vertex

A cone is a solid thathas circular base and a

curved face which meetsat a common vertex

called apex.The perpendiculardistance from the vertexto the base is theheight of the cone.

A cone has two faces,one edge and one

vertex.

height


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12.1e Spheres12.1e Spheres

A sphere is a solid where all points on itssurfaceare equidistant from a central point.

The distance from the central point to anypointon the surface is call the radius of the

sphere.

centre

radius


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12.2 Nets of Geometric Solids12.2 Nets of Geometric Solids

A net is a flat diagramobtainedwhen a geometric solid isopened

up and unfolded.

A net is also known aslayout.

The following diagrams showsthe

nets of some geometric

solids.

12.2a Drawing Nets for Geometric Solids12.2a Drawing Nets for Geometric Solids


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12.2a Nets of Geometric Solids12.2a Nets of Geometric Solids

Number ofshape:

2 triangles3 rectangles

Triangular prism

Net


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Number ofshape:

6 squares

CubeNet


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Number of shape:6 rectangles or 2squares and 4rectangles.

Cuboid Net


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Pyramid

Number of shape:

1 square, 4triangles

Net


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Cylinder

Number of shape:

1 rectangle, 2circles

Net


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Cone

Number of shape:

1 circle, 1 sector of

Net

or


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12.2b Stating the Types of Solids Given Their Nets12.2b Stating the Types of Solids Given Their Nets

We can determine the type

ofsolid represented by a netbycomparing the shapes in

the

net with the shapes inknownsolids.

We can also imaginefolding upthe net to form a hollow

solid.

If you are still not be ableto

pyramid

From the netshown below, state

the name of thesolid.

Solution:


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12.2c Constructing Models of Solids Given Their Nets12.2c Constructing Models of Solids Given Their Nets

To construct a

solidmodel from anet,

follow the stepsbelow:

1) Cut out thenet

along its outer

sides.

2) Fold its alongthe

lines


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12.3a Finding the Surface Area of Prisms,12.3a Finding the Surface Area of Prisms,

Pyramids,Pyramids,Cylinders and ConesCylinders and ConesThe surface area of a solid is the total area ofall

the outer surfaces ofthe solid.

The surface area of a solid can be found fromits net.

To find the surface area of a solid, follow the

steps given.

Determine the shapes of allsurfaces of

solid.Find the number of each shape.

Calculate the area of each shape.

Add up all areas. The sum is the

surfacearea of the solid.


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The following table shows theways of finding the surface area

of some solids.

Solid Net Surface Area

1. CubeSurface area

= 6 x area ofa

square= 6 x h2

= 6h2

h

h


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2. Cuboid

Area of rectangle A =hl

Area of rectangle B=

bl

Surface area= 2 x area of A+

2 x area of B +

2 x area of C

= 2(hlxblx

bh)

hh

h

b

h

A

B

B

A

C C

l

l

h

b


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3. Prism

Area of triangle =

bh

Area of rectangle= bl

Surface area= 2 x area triangle+

3 x area of

rectangle

= 2 x bh+ 3 xbl

=bh + 3bl

h

b

b

b

l

l

bh


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4. Pyramid

Area of rectangle =

abArea of triangle P= as

Area of triangle Q=

Surface area

= 6 x area ofrectangle

+

2 x area ofrectangle P+

2 x area of

triangle Q

= ab + 2 x as + 2 x bh

=ab + as +

a

h

b

s

s a

b

h

Q

Q

PQ


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5. Cylinder

l =circumference ofthe

base circle= 2 r

Area of rectangle P=

2 rh

Surface area

= area ofrectangle +

2 x area ofcircle= 2 rh + 2 r2

r

h

hl

r


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6. Cone

Area of sector = rsArea of circle = r2

Surface area

= area of sector

+ area ofcircle= rs + r2

s

r r

s

2 3b i di h S f f S h


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12.3b Finding the Surface Area of Sphere12.3b Finding the Surface Area of Sphere

The surface area of asphere can be calculated byusing the following formula.

r

Surface area of a sphere =4 r2

r is theradius of the

sphere.

i di h i i f lid12 3 Fi di h Di i f S lid


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12.3c Finding other Dimensions of a Solid12.3c Finding other Dimensions of a Solid

Given the Surface Area and Other RelevantGiven the Surface Area and Other Relevant

InformationInformationOther dimensions such as length of sides, height,

radius or diameter of solids can be found if we knowits surface area and other relevant information.rExample: Find the length of each side of acube if its surface area is 150cm2.

Let p be the length of eachside of the cube.The area of a surface is =p2 Each cube has 6surfaces.

So, 6p2 = 150

p2 = 150 6

= 25

p = 25 = 5cm

The length ofeach side is 5

cm.

2 3d S l i bl l i S f12 3d S l i P bl I l i S f


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12.3d Solving Problems Involving Surface12.3d Solving Problems Involving Surface

AreaArea

7cm

6cm

6cm

Find the surface area of the shapeas shown in the diagram. (Take =22)

7

Solution:The diagram consists of 2solids, a cone and a cylinder.

Surface area of the cone:= rs= 22 x 3 x 7

7= 66 cm2

Surface area of the cylinder:= 2 rh= 2 x 22 x 3 x 6

7= 113.14 cm2

Thebaseofconeisnotincluded

Theareaoftwocirclesarenotinclud

ed

Total surfacearea= 66 + 113.14

= 179.14 cm2.

12 3d S l i P bl I l i S f12 3d S l i P bl I l i S f


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12.3d Solving Problems Involving Surface12.3d Solving Problems Involving Surface

AreaAreaIf the surface area of the cuboid shown in thediagram is 288 cm2, find the value of y?

ycm

9cm

6cm

A cuboid has 6 surfaces.2(y x 9) + 2(y x 6) + 2(6 x 9) = 288cm2

2(9y + 6y + 54) = 288

15y + 54 = 144

15y = 144 - 54

y = 90 15

Solution:

The value

of y is 6cm.


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Chapter12_Solid Geometry II

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