Math Formule

8/8/2019 Math Formule

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Area of Triangle

= base x height = a b c = r S = S ( s - a) ( s b) ( s c) = a b sin c4 R

S = a + b + c2

Cos C = a + b - c2 ab

Sin a = Sin b = Sin c = 2 RA B C

Equilateral Triangle

Height = 3/2 ( side)

Area = 3/4 ( side )

R = 2/3 Height

r = 1/3 Height

Isosceles TriangleArea = b/4 ( 4 a - b)

Sum of external angles in any polygon is 360 always A

AO = 2OD 1 F E

Centriod divided the median in the ratio of 2 : 1 B

O is the Centriod

A


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AB + AC = 2(AD+BD)Where AD is the median to BC

B D C

If F represents faces, E represents number ofedges,V represent number ofvertices then

E + 2 = F + V

Sum of internal angles in any polygon = (n 2 ) 180

Rhombus

A D Area of the rhombus is given b = x D1 x D2

Where D1 and D2 are diagonals of therhombus

B E C Area of the rhombus = BC x AE

Where BC= base and AE = height.

Properties of rhombus

1). All sides of the rhombus are equal.

2). Diagonals bisects each otherperpendicular.3). Lengths ofdiagonals are not equal in rhombus.

4). Rhombus is not a symmetrical figure.

5). All four sides are parallel

Trapezium


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A D

Area of the Trapezium

= X ( AD + BC ) X AE

B E C

Properties of Trapezium

1). A Quadrilateral having only two side parallel then it is called trapezium.

Hexagon

F E

A D

B

CArea of the Regular Hexagon is = 3 3 (side)

2

Simple Interest

Simple Interest = P x N x R


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100

P = PrincipalN = Number of YearsR = Rate of Interest

Compound Interest = P ( 1 + R / 100 ) -- PP = Principaln = Number of YearsR = Rate of Interest

Interest Compounded Half yearly

P ( 1 + (R/2) / 100 ) - PRate of Interest for Installment

Interest = 24 x I x 100N ( F + L)

I = Installment ChargeN = Number of Installments

F = Principal left after first InstallmentL = Principal left after last Installment

Co-Ordinate Geometry

A ( X, Y) and B ( X2, Y2 ) are 2 points


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Distance between AB = (Y2 Y ) + (X2 - X )

Slope of Line AB, m = (Y2 Y)(X2 - X)

Equation of the straight line passing through A ( X, Y) and B ( X2, Y2 ) isgiven by

(Y Y ) = m (X - X )where

m = Y2 Y

X2 - XIn Equation of the straight line y = m x + c

. m is the slope of the line. c is the Y intercept .

a is x intercept.b is y intercept

If two lines are parallel then the slopes are equal.M1 = M 2

Where M1 and M 2 are slopes of the lines.

If two lines are perpendicular then the product of the slopes is -1.

M 1 x M 2 = -1.

Where M1 and M 2 are slopes of the lines

Equations of straight line parallel to a x + b y + c 1 = 0 is ab y + c 2 = 0

Equations of straight line perpendicular to a x + b y + c 1 = 0

b x - a y + c 2 = 0


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Perpendicular distancebetween two parallel linesa x + b y + c 1 = 0 is a x + b y + c 2 = 0 is

= c 1 -

a +

LCM of ( a, b ) x HCF of ( a, b) = ( a x b)

Sum offirst N natural numbers = n ( n + 1 )2

Sum ofSquares ofFirst N numbers = n (2n+1)(n+1)6

Sum ofcubes ofFirst N numbers = [ n (n+1)]4

Circles


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Length ofArc ABC = 2r &360

Where is the angle made at theter.

Area of the Sector AOC = r &360

Where is the angle made at the center.

a + c = b + d = 180

Area of the Cyclic Quadrilateral = ( s - a) ( s b) ( s c) ( s d)

Where a, b, c and d are the lengths of the Cyclic Quadrilateral

Length of Direct common tangent


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Length ofDirect common tangent = d - (r1 r2)

Where r1 and r2is radius of the circles.

. d is the distancebetween the centers.

Length of Transverse common tangent

Length ofTransverse common tangent = d - (r1 r2)

Where r1 and r2is radius of the circles. . d is the distancebetween the centers.

(AB) = DB x CB

Length of common chord


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Where o andp are center of the circles..r1 and r2 are radius of the circles with centero and p respectively.

Area of the triangle opa is given by

=> S ( s - a) ( s b) ( s c) = x h x dIn the above equation we know S, a, b, c, and d , so we can find out h.

Length of common chord is given by 2h

S.No Name of the solid Volume Total surfacearea

Lateralsurface area

1 Cube (Side) 6 (side) 4 (side) 2 Cuboids l x b x h 2(lb +bh+lh) 2(bh+ lh)3 Cylinder r h 2 r h + 2 r 2 r h4 Cone 1/3 r h rl + r rl5 Sphere 4/3 r 4 rLength of body diagonal or longest diagonal of a cuboid = ( l + h + b )

Length of body diagonal or longest diagonal of a cube = 3 (side)

Arithmetic Progression.

.a , a +d, a+2d, a+3d, a+4d,..

a is the first term of the series.


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. d is the common difference.

Nth term of the series is given by

tn = a + ( n-1) d.

Sum of N terms is given bySn = n/2 [ 2a + ( n 1) d]

Sn = n/2 [ L + D ]L is the first termD is the last term

Geometric Progression

. a, ar, ar, ar, .

Sum of N terms is given bySn = a (r n - 1)

( r 1)

Sum ofinfinite terms is given byS = a

( 1 r)

Where r is less than 1

Sum of infinite terms is given byS = a

( r - 1)

Where r is greater than 1

Speed and Distance

If a person is traveling from A to B with a km/ hr, and in return fromB to A with b km/hr his average speed is given by

= 2 x a x b km/hr


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( a + b )

If the distance traveled is same with two different speeds then average speedis given by = 2 x a x b km/ hr ( a + b )If the time taken is same with different speeds then average speed is given

by = (a + b) km/ hr.a km/ hr and b km/hr are different speeds.2

Trains

If speed of the train is A km/hr, length of the train is B km.Then time taken to cross a pole in hrs =

B kmA km/ hr.

If speed of the train is A km/hr, length of the train is B km.Then time taken to cross a platform of length C km in hrs =

(B +C) kmA km/ hr.

Two trains A and B are traveling with a km/hr and b km/hr respectively.

Lengths of the train A and B are X km and Y km respectively.Time taken to cross the slower train by faster train = (X +Y) km(If they are traveling in opposite direction) (a+b) km/hr.

Time taken to cross the slower train by faster train(If they are traveling in same direction,Starting point of the faster train is at the end of the slower train)

= (X +Y) kma - b km/hr.

Time taken to cross the slower train by faster train


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(If they are traveling in same direction,Starting point of the faster train and slower train are on the same line)

= (Length of the faster train) kma - b km/hr.

A person is roving the boat at speed ofa km/hr. length of the river he has tocross the river is x km. Water flows in the river at b km/hr.

Then time taken to cross the river= x km(With the current) (a + b) km /hr

Then time taken to cross the river = x km(Against the current) (a - b) km /hr

Conversions

1 mile = 1609 meters

1 mile = 5280 feet.

1mile = 8 furlong1 furlong = 220 feet

1 km/hr = (5 / 18 ) m/s

Math Formule

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