surface area and volume
DESCRIPTION
maths powerpoint on surface area and volumeTRANSCRIPT
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• opposite sides are equal
• each angle is of 900
• Diagonals are equal &bisect each other
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SQUARE
• All sides are equal
• each angle is of 900
• Diagonals are equal & bisect each other at 900
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CICRLE
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RequirementEdge/side
Cube
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Cube
• Volume = Base area x height
= L x L x L
= L3
L
L
L
• Total surface area = 2LxL + 2LxL + 2LxL
= 6L2
•Diagonal of cube=√3×L unit
•Lateral surface area=perimeter of base ×height =4×L3 unit3
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Side 2
Bottom
Back
Top
Side 1Front
Side 2
Bottom
Back
Top
Side 1Front
Length (L)Breadth (B)
Height (H)
Cuboid
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H
B
LH
B
H
L
L
L
H
B
L
H
B
H
L
L
Total surface Area = L x H + B x H + L x H + B x H + L x B + L x B
= 2 LxB + 2BxH + 2LxH
= 2 ( LB + BH + LH )
Total surface Area
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Lateral surface area=2(L+B)H unit2
Volume=area of base×height =L×B×H unit3
Diagonal of cuboid=√L2+B2+H2
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Example:
7 cm
4 cm
8 cm
Here:
l=8cm
B=4cm
H=7cm
SA=2(lb+bh+lh)=2(8X4+4x7+8x7)
=2(64+56+112)
=232cm²
V = lwh
V = 8(4)(7)
V = 224 cm³
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Curved surface area of cylinder
Curved surface area of an object is the area of outer covering of it. If a rectangular paper is folded ,the length becomes the circumference and the breadth becomes the height. So csa of a cylinder is 2πr x height.
Curved surface area of cylinder=2πrh
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Circumference of circle = 2 π r
Area covered by cylinder = Surface area of of cylinder = (2 π r) x( h)
r h
Outer Curved Surface area of cylinder
It is the area covered by the outer surface of a cylinder.
Formation of Cylinder by bangles
Circumference of circle = 2 π r
r
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Total Surface area of a solid cylinder
=(2 π r) x( h) + 2 π r2
Curved surface
Area of curved surface +area of two circular surfaces=
circular surfaces
= 2 π r( h+ r)
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2πr
h
r
h
Surface area of cylinder = Area of rectangle= 2 πrh
Other method of Finding Surface area of cylinder with the help of paper
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Here:
r=3.1
H=12
SA = 2πrh + 2πr²
SA = 2π(3.1)(12) + 2π(3.1)²
SA = 2π (37.2) + 2π(9.61)
SA = π(74.4) + π(19.2)
SA = 233.7 + 60.4
SA = 294.1 in²
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Right Circular Cone
RequirementRadius(r)Height(h)Slant height(l)At least 2 must be given
(l)2= (h)2 + (r)2
l =√ h2 + r2
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l2πr
l
Area of a circle having sector (circumference) 2π l = π l 2
Area of circle having circumference 1 = π l 2/ 2 π l
So area of sector having sector 2 π r = (π l 2/ 2 π l )x 2 π r = π rl
Surface area of cone
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3( V ) = π r2h
r
h h
r
Here the vertical height and radius of cylinder & cone are same.
3( volume of cone) = volume of cylinder
V = 1/3 π r2h
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if both cylinder and cone have same height and radius then volume of a cylinder is three times the volume of a cone ,
Volume = 3V Volume =V
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Mr. Mohan has only a little jar of juice he wants to distribute it to his three friends. This time he choose the cone shaped glass so that quantity of juice seem to appreciable.
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SphereRequirementRadius(r)
Height = diameter
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V1
r
V=1/3 πr2h
If h = r thenV=1/3 πr3
rr
If we make a cone having radius and height equal to the radius of sphere. Then a water filled cone can fill the sphere in 4 times.
V1 = 4V = 4(1/3 πr3)
= 4/3 πr3
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4( 1/3πr2h ) = 4( 1/3πr3 ) = V
h=r r
Volume of a Sphere
Click to See the experiment
Here the vertical height and radius of cone are same as radius of sphere.
4( volume of cone) = volume of Sphere
V = 4/3 π r3
r
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Hemisphere
RequirementRadius/ diameter
Height=radius
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Frustum of Cone
VolumeV = π/12 h (D2 + D d + d2) (9a)
m = ( ( (D - d) / 2 )2 + h2)1/2 (9c)
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Cone shape
Cylindrical shape
Bottle
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VolumeV = π/4 h (D2 - d2) (5)
Hollow Cylinder
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TRIANGULAR PRISM
To find the surface area of a triangular prism you need to be able to imagine that you can take the prism apart like so:Notice there are TWO congruent triangles and THREE rectangles. The rectangles may or may not all be the same.
Find each area, then add.
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Example:
8mm
9mm
6 mm 6mm
Find the AREA of each SURFACE
1. Top or bottom triangle:
A = ½ bh
A = ½ (6)(6)
A = 18
2. The two dark sides are the same.
A = lw
A = 6(9)
A = 54 ADD THEM ALL UP!
18 + 18 + 54 + 54 + 72
SA = 216 mm²
3. The back rectangle is different
A = lw
A = 8(9)
A = 72
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PYRAMID
Volume
V = 1/3 h A1 (6)
whereA1 = area of base (m2, ft2)
h = perpendicular height of pyramid (m, ft)
SurfaceA = ∑ sum of areas of triangles forming sides
+ Ab (6b)
wherethe surface areas of the triangular faces will have different formulas for different shaped bases
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FRUSTUM OF PYRAMID
VOLUMEV = h/3 ( A1 + A2 + (A1 A2)1/2) (7)
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Zone of SPHERE
V = π/6 h (3a2 + 3b2 + h) (11a)
Am = 2 π r h (11b)
A0 = π (2 r h + a2 + b2) (11c)
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V = π/6 h (3/4 s2 + h2) = π h2 (r - h/3) (12a)
Am = 2 π r h = π/4 (s2 + 4 h2) (12b)
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V = 2/3 π r2 h (13a)A0 = π/2 r (4 h +
s) (13b)
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V = π/6 h3 (14a)
A0 = 4 π ((R + r)3 (R -
r))1/2 = 2 π h (R + r) (14b)h = 2 (R2 - r2)1/2 (14c)
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V = 2/3 π R2 h (15a)
A0 = 2 π R (h + (R2 -
h2/4)1/2) (15b)h = 2 (R2 - r2)1/2 (15c)
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V = π2/4 D d2 (16a)
A0 = π2 D d (16b)
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V = 2/3 r2 h (18a)
Am = 2 r h (18b)
A0 = Am + π/2 r2 + π/2 r
(r2 + h2)1/2 (18c)
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V = π/4 d2 h (17a)Am = π d h (17b)
A0 = π r (h1 + h2 + r
+ (r2 + (h1 - h2)2/4)1/2) (17c)
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V ≈ π/12 h (2 D2 + d2) (19a)
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VolumeV = A1 h (3a)whereA1 = side area (m2, ft2)
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