• opposite sides are equal

• each angle is of 900

• Diagonals are equal &bisect each other

SQUARE

• All sides are equal

• each angle is of 900

• Diagonals are equal & bisect each other at 900

CICRLE

RequirementEdge/side

Cube

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

Side 2

Bottom

Back

Top

Side 1Front

Side 2

Bottom

Back

Top

Side 1Front

Length (L)Breadth (B)

Height (H)

Cuboid

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

Lateral surface area=2(L+B)H unit2

Volume=area of base×height =L×B×H unit3

Diagonal of cuboid=√L2+B2+H2

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

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

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

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)

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

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

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

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

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

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

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.

SphereRequirementRadius(r)

Height = diameter

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

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

Hemisphere

RequirementRadius/ diameter

Height=radius

Frustum of Cone

VolumeV = π/12 h (D2 + D d + d2)   (9a)

m = ( ( (D - d) / 2 )2 + h2)1/2    (9c)

Cone shape

Cylindrical shape

Bottle

VolumeV = π/4 h (D2 - d2)   (5)

Hollow Cylinder

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.

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

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

FRUSTUM OF PYRAMID

VOLUMEV = h/3 ( A1 + A2 + (A1 A2)1/2)   (7)

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)

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)

V = 2/3 π r2 h    (13a)A0 = π/2 r (4 h +

s)   (13b)

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)

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)

V = π2/4 D d2    (16a)

A0 = π2 D d   (16b)

V = 2/3 r2 h   (18a)

Am = 2 r h (18b)

A0 = Am + π/2 r2 + π/2 r

(r2 + h2)1/2 (18c)

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)

V ≈ π/12 h (2 D2 + d2)   (19a)

VolumeV = A1 h  (3a)whereA1 = side area (m2, ft2)