polynomials

polynomials in one variable and its types

Introduction:-

All the expression which consist of variable , coefficient,and constant are called polynomial .

In polynomials the expression like 2x+1 is called  a term, each term of polynomial is called a

coefficient, In this therm x is a variable.

In fact, 2,-5,3 etc. are examples of polynomials. The constant polynomials 0 is called the zero polynomials. 

Degree of a polynomial is the highest power of the polynomial.

There are three types of polynomials on the bases of term:-

i) Monomial:-

 It has only one term.

ii) Binomial:-

 It has two terms.

iii) Trinomial:-

 It has three terms.

There are three types of polynomials on the bases of degree:-

i) linear polynomial :- 

It consist of polynomial with degree 1.

ii) Quadratic polynomial:-

It has polynomial with highest degree 2.

iii) cubic polynomials:-

It consist of polynomial with highest degree 3.

 ~ zeroes  or roots of polynomials refers to there solution.

- A zero of need not be 0 only.

- 0 may be a zero of polynomial.

- Every linear polynomial have only one zero.

- A polynomial can have more then one zero.

~ zeroes of polynomials can be find using factor theorem, reminder theorem, splitting the midterm method, completing the square method etc.

Circle

how find circle area circumference diameter and radius step by step 

Introduction:-

A circle is a collection of points placed at fixed distance from a fixed point  on a plane.

The fixed point is called center and the fixed distance is called radius.

 Radius of the circle:-

The line segment joining the center and any point on the circle is called circle.

 

The three divisions of circle:-

A circle divides the plane on which it is made into three parts.

They are :-

i) interior of the circle

ii) circle itself

iii) exterior of the circle

Chord of the circle:-

The line segment joining the two points distinct point on circle is called chord.

Diameter of the circle:-

The chord which passes through the center is called the diameter of the circle.

Arc of the circle:-

A piece of circle between two points is called an arc .The longer piece is called major arc and the smaller one is called minor arc.

Circumference:-

The complete length of the circle is called circumference.

Segment of the circle:- 

The region between a chord and either of the arc is called a segment of the circle.

There are two types of the segment major segment and minor segment.

Sector of the circle :-

The region between an arc and the two radii , joining the center to the end point of the point of the circle is called a sector.same as segment there are major and minor sector.

Formulas related to circle :-

diameter of circle = 2r

circumference = 2πr

area of circle = πr^2

r = radius of circle

π = consent value 3.14 OR 22/7

Pythagoras theorem

 Pythagoras theorem proof with right angle triangle 

Introduction :-

Pythagoras theorem can be defined as the square of the hypotenuse of a right angle triangle is equal to the square of the base and perpendicular.

i.e. (H)^2 = (B)^2 + (P)^2

It was introduced by Greek philosopher Pythagoras. 

 

Prove of Pythagoras theorem :-

Given :-

triangle ABC right angled at D= 90 degree

to prove :-

(AB) ^2 =  (AC)^2+(CB)^2

construction :-

draw a perpendicular on base AB

Proof :-

triangle ACB ~  triangle ABC  (If a perpendicular is drawn from the vertex of the right triangle to the hypotenuse 

then triangle on both sides of the perpendicular are similar to the whole triangle and to each other.)

AD/AC = AC/AB (sides are proportional as the triangles are similar).

so,   AD.AB = AC^2 .....(1)

Similarly,

 triangle CDB ~ triangle ABC

so,

DC/CB = CB/AB

now,  DC.AB= CB^2......(2)

adding (1) and (2),

AD.AB+ DC.AB = AC^2 +CB^2

AB(AD+DC) = AC^2 +CB^2

AB.AB = AC^2 + CB^2

AB^2 = AC^2 + CB^2

Hence proved !!

 

Question on Pythagoras;-

(1) A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground . Find  the length of the ladder.

 sol)  Let AB be the ladder and CA be the wall with window at A.

also,  BC = 2.5 and CA= 6 m

 from Pythagoras theorem,we have ;

        AB^2 = BC^2 + CA^2

                   = 2.5^2 + 6^2

                   = 42.25

so,         AB  = 6.5

thus, length of the ladder is 6.5 m.

triangles

Area of triangles its properties & formulae

Introduction:- 

A triangle is polygon  with three edges  three vertices and three angle. It is a plan 2D figure .

 

There are four most important types of triangles :-

1) Equilateral triangle :-

These triangles have all three sides equal in length . It also have all three angle equal .

2) Scalene  triangle :-

This type of triangles  don't have  any equal sides.

3) Isosceles triangle :-

This type of triangle have two equal sides .

 

4)Right angle triangle :-

this type of triangles have one angle as 90 degree .

 

 Areas Related To Triangle :- (Formulae)

Area of triangle is defined as total space that is enclosed by any particular triangle.

To find the area of triangle there are some formulae :-

1) area of triangle = 1/2 *b*h

2) area of triangle = under root s (s-a) (s-b) (s-c)

3) area of equilateral triangle = under root  3/4 *(a^2)

Properties Of Triangle :-

1) The sum of  all angles of a triangle is 180 degree.

2) An exterior angle of a triangle  is equal  to the sum of its two interior opposite angle .

3) The sum of the length of any two sides of triangle is always greater than the third side.

4)  Angle opposite to longer side is larger.

5)  Side equal to greater angle is longer.

6) The exterior angle of a triangle always add up to 360 degree.

Maths

 Arithmetic progressions

INTRODUCTION:-

An arithmetic progression is a sequence in which terms increase or decrease regularly by the same constant. This constant is called the common difference of the retrogression (series).

In other words an arithmetic progression is the list of number in which the first term is given  and each

term is obtained by adding a fixed number d to the preceding term.

This fixed number d is called the common difference. This can be positive, negative, or zero.

EXAMPLE:-

1) 1,2.3,4,5......20

here d =1 and a (first term)=1

2) -1.0,-1.5, -2.0,-2.5

here d= -0.5 and a=-1.0

GENERAL FORM :-

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

It represents an arithmetic progression where a is the first term and d is common difference . This is called the general form of an AP.

FORMULAE :-

1) d = a(2)-a(1)

2) a(2) = a + (2-1)d  , a(3) = a + (3-1)d

3) a(n) = a +( n- 1)d

To find  the sum:-

1) s= n/2[2a + (n-1)d]

2) s= n/2[a +l ]

QUESTIONS:-

Q1) write first four terms of the AP , when the first term a and the common difference d are given as 

follows ?

1) a = 10 , d = 10

2) a = 4 ,  d = -3

Q2) which term of the AP : 3,8,13,18,...., is 78 ?

  

Q3)  200 logs are stacked in the following manner ; 20 logs in the bottom row , 19 in the next row 18 

in the row next to it and so on . In how many rows are the 200 logs can be arranged and how many logs are in the  top row? 

SOLUTION:-

sol1) we know that if the first term is a and the common difference is d. then a.a+d.a+2d .a+3d....

represent an AP for different values of a and d.

(1) putting a = 10 and d = 10 in a,a+d , a+2d , a+3d.....we get the required AP as 10,10+10,10+2*10...

i.e.,10.20.30,40....

(2)putting a = 4 and a = -3 in a,a+d,a+2d, a+3d.....we get the required AP as  -2,-2+0,-2+2*0, -2+3*0

i.e. -2,-2.-2.-2.....

sol2) we have: a = 3, d = 5

let 78 be the nth term of the given AP . Then,

a(n) = 78

therefore  3 + (n-1)5 = 78

                 5(n-1)= 75

                  n= 15+1

                  n= 16

Thus, 78 is the 16th  term of the given AP .

sol3) clearly,log stacked in each row from a sequence 20, 19,18,17,.....it is an AP with a= 20 and d = -1

let s(n) = 200. Then,

n/2[2*20+(n-1)(-1)]= 200

n^2-41n+400= 0

(n-16) (n-25) =0

i.e., n= 16 or 25

n = 25 is not valid for this problem 

therefore  no. of logs in the 16th row =a (16)

                                                             = a + 15d =20 + 15 (-1)

                                                             = 20-15=5