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Chapter 1: Real Numbers
 
Euclid’s division lemma:
For any given positive integers a and b, there exists unique integers q and r such that
a = bq + r where 0 ≤ r < b
• If b divides a, then r = 0
Example: For a = 15, b = 3, it can be observed that
15 = 3 × 5 + 0
Here, q = 5 and r = 0
• If b does not divide a, then 0 < r < b
Example: For a = 20, b = 6, it can be observed that 20 = 6 × 3 + 2
Here, q = 6, r = 2, 0 < 2 < 6
Results on the basis of Euclid’s division lemma:
• Every positive even integer is of the form 2q, while every positive odd integer is of the form 2q + 1, where q is some integer.
• Any positive integer is of the form 3q, 3q + 1 or 3q + 2, where q is an integer.
Euclid’s division algorithm is a series of well-defined steps based on “Euclid’s division lemma”, to give a procedure for calculating problems.
Finding HCF of two positive integers a and b (a > b) by using Euclid’s division algorithm:
Step 1: Applying Euclid’s division lemma to a and b to find whole numbers q and r, such that
a = bq + r, 0 ≤ r < b
Step 2: If r = 0, then HCF (a, b) = b
If r 0, then again apply division lemma to b and r
Step 3: Continue the same procedure till the remainder is 0. The divisor at this stage will be the HCF of a and b.
Note: HCF (a, b) = HCF (b, r)
Fundamental theorem of arithmetic
Every composite number can be uniquely expressed (factorised) as a product of primes apart from the order in which the prime factors occur.
Example: 1260 can be uniquely factorised as
  2  1260    2   630     3   315     3   105     5    35           7

1260 = 2 × 2 × 3 × 3 × 5 × 7
Result: For any positive integer a, b, HCF (a, b) × LCM (a, b) = a × b
Application of Fundamental theorem of arithmetic
 
Example: Check whether 15n is divisible by 10 or not for any natural number n. Justify your answer.
 
Solution:
A number is divisible by 10 if it is divisible by both 2 and 5.
15n = (3 × 5)n
3 and 5 are the only primes that occur in the factorisation of 15n
By uniqueness of fundamental theorem of Arithmetic, there is no other prime except 3 and 5 in the factorisation of 15n.
2 does not occur in the factorisation of 15n.
Hence, 15n is not divisible by 10.
Rational numbers
Every rational number can be expressed in the form pq, where p, q are integers q 0
Example, 1=11, 0=0q where q 0 is any integer
1.2=1210=65
Irrational numbers
Every irrational number cannot be expressed in the form pq, where p, q are integers and q 0
Example: 3, 11, 12, etc.
Theorem: If a prime number p divides a2, then p divides a, where a is a positive integer.
Some results related to the decimal expansion of rational numbers
• Let x be a rational number whose decimal expansion terminates.
Then ‘x’ can be written in the form x=pq, where p and q are co-primes and the prime factorisation of q is of the form 2n 5n, where n, m are non-negative integers.
• Let x=pq be a rational number such that q = 2n 5m, where n, m are non-negative integers. Then, the decimal expansion of x terminates.
• If x=pq is a rational number such that the prime factorisation of q is not of the form 2n 5m, where n, m are non negative integers, then x has a decimal expansion which is non-terminating repeating.
Chapter 2: Polynomials
 
Polynomial
A polynomial in variable ‘x’ is of the form, p open parentheses x close parentheses equal a subscript n x to the power of n plus a subscript n minus 1 end subscript space x to the power of n minus 1 end exponent plus......... plus a subscript 1 x plus a subscript 0,px=anxn+an-1 xn-1+.........+a1x+a0 where an, an-1, ... a1, a0 are constants and ‘n’ is a positive integer.
For example, px=4x3+3x-1, gt=7t3+t2-8
 Degree of a polynomial
The highest power of x in a polynomial p(x) is called the degree of the polynomial p(x).
A polynomial of degree 1 is called a linear polynomial.
Example: fy=2y-7
A polynomial of degree 2 is called a quadratic polynomial.
Example: gt=2t2-3t+8
A polynomial of degree 3 is called a cubic polynomial.
Example: fu=u3-5u2+10
A polynomial of degree 0 is called a constant polynomial.
Example: ht=-83
The constant polynomial, fx=0, is called zero polynomial.
Degree of zero polynomial is not defined.
Zeroes of a polynomial
A real number ‘k’ is a zero of a polynomial p(x), if p(k) = 0. In this case, ‘k’ is also called the root of the equation, p(x) = 0.
Note: A polynomial of degree n can have at most n zeroes.
 
Example: 2 and –3 are the zeroes of the quadratic polynomial, x2+x-6.
 22+2-6 =0, -32+-3-6 =0
Geometrical meaning of zeroes of a polynomial
The zero of a polynomial, y=px, (if it exists) is the x-coordinate of the point where the graph of y = p(x) intersects the x-axis.
For example:

In the above graph, the graph intersects the x-axis at exactly two points.
The number of zeroes of the corresponding polynomial is 2.
Relationship between zeroes and coefficients of a polynomial
Linear Polynomial
The zero of the linear polynomial, ax + b, is -ba=-Constant termCoefficient of x
Quadratic polynomial
If a and β are the zeroes of the quadratic polynomial, px=ax2+bx+c, , then x-α, x-β are the factors of p(x).

px=ax2+bx+c=k x2-α + βx+ α β, where k ≠ 0 is constant
Sum of zeroes=α+β=-ba=-Coefficient of xCoefficient of x2

Product of zeroes=αβ=ca=Constant termCoefficient of x2

Cubic polynomial
If α,β, γ, are the zeroes of the cubic polynomial, fx=ax3+bx2+cx+d, then x-α, x-β, x-γ are the factors of f (x).
fx=ax3+bx2+cx+d=kx3-α+β+γx2+αβ+βγ+γαx-αβγ, where k is a non-zero constant
α+β+γ=-ba=-Coefficient of x2Coefficient of x3αβ+βγ+γα=ca=Coefficient of xCoefficient of x3αβγ=-da=-Constant termCoefficient of x3

Division algorithm for polynomials
It states that for any two polynomials, p(x) and g(x) 0, there exists polynomials, q(x) and r(x), such that
px=gxqx+rx, where r(x) = 0 or degree r(x) < degree g(x)
Here, p (x) is called dividend, g (x) is called divisor, q (x) is called quotient, and r (x) is called remainder.
Example: Divide x4-x3+3x2-x+3 by x2-x+1 and verify the division algorithm.
Solution:
It is given that,
Dividend =x4-x3+3x2-x+3, Divisor = x2-x+1
 
Divisor × Quotient + Remainder
=x2-x+1x2+2+x+1=x4+2x2-x3-2x+x2+2+x+1=x4+x3+3x2-x+3=Dividend
Division algorithm is verified.Chapter 3: Pair of Linear Equations in Two Variables
 
Linear equation in two variables
Linear equation in two variables x and y is of the form ax + by + c = 0, where a, b, and c are real numbers, such that both a and b are not zero.
Example: 6x + 3y = 9
• A linear equation in two variables has infinitely many solutions.
• Equations of x-axis and y-axis are respectively y = 0 and x = 0.
• The graph of x = a is a straight line parallel to the y-axis, and is at a distance of ‘a’ units from the y-axis.
• The graph of y = a is a straight line parallel to the x-axis, and is at a distance of ‘a’ units from the x-axis.
• Every point on the graph of a linear equation in two variables is a solution of the linear equation and vice versa.
Example: Consider the linear equation 6x + y = 12  … (1)
(1, 6) is a solution of (1) [LHS = 6 × 1 + 6 = 6 + 6 = 12 = RHS]
But (2, 3) is not a solution of (1) since LHS = 6 × 2 + 3 = 12 + 3 = 15 ≠ RHS
Point (1, 6) lies on the line representing the equation (1), whereas point (2, 3) does not lie on the line.
Pair of linear equation in two variables
• Two linear equations in the same two variables are called a pair of linear equations in two variables.
• The general form of a pair of linear equations is a1x + b1y + c1 = 0, a2x +b2y + c2 = 0, where a1, a2, b1, b2, c1, c2 are real numbers such that a12+b120, a22+b220
Graphical representation of linear equations:
Example: Represent the following system of linear equations graphically.
x + y + 2 = 0, 2x – 3y + 9 = 0
 
Solution:
The given equations are
x + y + 2 = 0       (1)
2x – 3y + 9 = 0   (2)
Table for the equations x + y + 2 = 0
x   0 –2
y –2   0
 
Table for the equation 2x – 3y + 9 = 0
x 0 –4.5
y 3  0
 
By plotting and joining the points (0, –2) and (–2, 0), the line representing equation (1) is obtained.
By plotting and joining the points (0, 3) and (–4.5, 0), the line representing equation (2) is obtained.
 
 

 
 System of simultaneous linear equations
Consistent system
A system of simultaneous linear equations is said to be consistent if it has at least one solution.
Inconsistent system
A system of simultaneous linear equations is said to be inconsistent if it has no solution.
A pair of linear equations in two variables can be solved by
1) Graphical method or
2) Algebraic method
Nature of solution of simultaneous linear equations
Based on graph:
Case (i): The lines intersect at a point.
The point of intersection is the unique solution of the two equations.
In this case, the pair of equations is consistent.
Case (ii): The lines coincide
The pair of equations has infinitely many solutions – each point on the line is a solution. In this case, the pair of equations is dependent (which is consistent).
Case (iii): The lines are parallel.
The pair of equations has no solution. In this case, the pair of equations is inconsistent.
Based on the coefficients:
Let a1x + b1y + c1 = 0, a2x +b2y + c2 = 0 be a system of linear equations.
Case (i) a1a2b1b2
In this case, the given system is consistent.
This implies that the system has a unique solution.
Case (ii) a1a2=b1b2c1c2
In this case, the given system is inconsistent.
This implies that the system has no solution.
Case (iii) a1a2=b1b2=c1c2
In this case, the given system is dependent and consistent.
This implies that the system has infinitely many solutions.
Algebraic method for solving simultaneous linear equations
Simultaneous linear equations can be solved algebraically by the following methods.
Substitution method
Example: Solve the following system of equations by substitution method.
 x – 4y + 7 = 0
3x + 2y = 0
Solution:
The given equations are
x – 4y + 7 = 0  … (1)
3x + 2y = 0      … (2)
From equation (2), 3x = – 2y
x=-23y
Put x=-23y in equation (1)
-23y-4y+7=0-2y-12y3=-7-14y=-21y=-21-14=32 x=-2332=-1
Therefore, the required solution is -1,32.
Elimination method
Solve the following pair of linear equations by elimination method.
7x – 2y = 10
5x + 3y = 6
Solution:
7x – 2y = 10  … (1)
5x + 3y = 6    … (2)
Multiplying equation (1) by 5 and equation (2) by 7, we get
35x – 10y = 50  … (3)
35x + 21y = 42 … (4)
Subtracting equation (4) from (3), we get
-31y=8y=-831
Now, using equation (1):
7x = 10 + 2y
x=1710+2×-8  31=4231
Required solution is 4231,-831
Cross-multiplication method
The solution of the system of linear equations a1x + b1y + c1 = 0, a2x +b2y + c2 = 0 can be determined by the following diagram.
 
 
That is,
xb1c2-b2c1=yc1a2-c2a1=1a1b2-a2b1x=b1c2-b2c1a1b2-a2b1, y =c1a2-c2a1a1b2-a2b1         a1b2-a2b10

Certain pairs of equations that are not linear can be reduced to linear form by suitable substitutions.
 
Example: Consider the following system of equations.
2x-2-1y-1=15x-2-6y-1=20
Let x – 2 = u, y – 1 = v. Then, the given system of equations reduces to
2uv = 1
5u – 6v = 20
Chapter 6: Triangles
 
Congruent and similar figures
• Two geometric figures having the same shape and size are said to be congruent figures.
• Two geometric figures having the same shape, but not necessarily the same size, are called similar figures.
• All congruent figures are similar. However, the converse is not true.
• Two polygons with the same number of sides are similar, if
a) their corresponding angles are equal
b) their corresponding sides are in the same ratio (or proportion)
Similarity of triangles
Two triangles are similar, if
• their corresponding angles are equal
• their corresponding sides are in the same ratio (or proportion)
Basic proportionality theorem
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Corollary: If D and E are points on the sides, AB and AC, respectively of DABC such that DEBC, then
 a.ABAD=ACAEb.ABDB=ACEC
 Converse of basic proportionality theorem
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Criteria for similarity of triangles
AAA similarity criterion
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence, the two triangles are similar.
AA similarity criterion
If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.
SSS similarity criterion
If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence, the two triangles are similar.
SAS similarity criterion
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Areas of similar triangles
• The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
• If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
Pythagoras theorem
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Converse of Pythagoras theorem
In a triangle, if the square of one side is equal to the sum of the squares of other two sides, then the angle opposite to the first side is a right angle.
Chapter 8: Introduction to Trigonometry
 
Trigonometric ratios


sinθ=Opposite sideHypotenuse=ABACcosθ=Adjacent sideHypotenuse=BCACtanθ=Opposite sideAdjacent side=ABBCcosecθ=1sinθ=HypotenuseOpposite side=ACABsecθ=1cosθ=HypotenuseAdjacent side=ACBCcotθ=1tanθ=Adjacent sideOpposte side=BCABAlso, tanθ=sinθcosθ, cotθ=cosθsinθ 
If one of the trigonometric ratios of an acute angle is known, then the remaining trigonometric ratios of the angle can be calculated.
Trigonometric ratios of some specific angles
 
θ 0 30° 45° 60° 90°
sinθ 0 12 12 32 1
cosθ 1 32 12 12 0
tanθ 0 13 1 3 Not defined
cosecθ Not defined 2 2 23 1
secθ 1 23 2 2 Not defined
cotθ Not defined 3 1 13 0
 
Trigonometric ratios of complementary Angles
sin90°-θ=cosθ        cos90°-θ=sinθtan90°-θ=cotθ        cot90°-θ=tanθcosec90°-θ=secθ        sec90°-θ=cosecθ
Where θ is an acute angle.
Trigonometric identities
• cos 2 A + sin 2A = 1
• 1 + tan 2 A = sec 2 A
• 1 + cot 2 A = cosec 2 A
Chapter 14: Statistics
 
Mean of ungrouped data
If x1, x2 …, xn are observations with respective frequencies f1, f2, … fn for a given data, then the mean x of the data is given by x=f1x1+f2x2+...+fnxnf1+f2+...+fn
x=fixifi, where i varies from 1 to n
Mean of grouped data
Direct method
x=fixifi, where fi is the frequency corresponding to the class mark xi.
Assumed-mean method
x=a+d=a+fidifi, where ‘a’ is the assumed mean, di = xia, and fi is the frequency corresponding to the class mark xi
Step-deviation method
x=a+hu=a+hfiuifi, where ui=xi-ah, fi is the frequency corresponding to the class mark xi , a is the assumed mean and h is the class size.
• The assumed-mean method and the step-deviation method are simplified forms of the direct method.
• The mean obtained by all the three methods is the same.
• Step-deviation method is convenient to apply if all di’s have a common factor.
Mode
Mode of ungrouped data
The mode or modal value of a distribution is the observation for which the frequency is the maximum.
Mode of grouped data
Mode of a grouped data is given by:
Mode=lf1-f02f1-f0-f2×h
Where, l = Lower limit of the modal class
h = Size of the class interval (assuming all class sizes to be equal)
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2= Frequency of the class succeeding the modal class
Median
Median of ungrouped data
If n (number of observations) is an odd number, then median = value of the n+12thobservation
If n (number of observations) is an even number, then median = mean of the values of the n2th and n2+1thobservations
Median of grouped data
Median of a grouped data is given by:
Median=l+n2-cff×h
Where l = Lower limit of median class
n = Number of observations
cf = Cumulative frequency of the class preceding the median class
f = Frequency of the median class
h = Class size (assuming class size to be equal)
Empirical relationship between the three measures of central tendency
3 Median = Mode + 2 Mean
Graphical representation of cumulative frequency distribution
Ogive (of the less- than type)

Draw ogive of the less-than type for the given distribution.
Class interval 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120
Frequency 7 8 6 8 6 5
 
Solution:
The cumulative frequency distribution for the given data can be found as:
Class interval Upper class limit Frequency Cumulative frequency
0 – 20 20 7 7
20 – 40 40 8 15
40 – 60 60 6 21
60 – 80 80 8 29
80 – 100 100 6 35
100 – 120 120 5 40

By taking the horizontal axis as the upper class limit and the vertical axis as the corresponding cumulative frequency, we can plot the cumulative frequency for each upper class limit.
Then, the required ogive (of the less-than type) is obtained as:

 
 
Ogive (of the more-than type)
 Example: Draw ogive of the more-than type for the following distribution.
Class interval 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120
Frequency 7 8 6 8 6 5

Solution:
The cumulative frequency for the given data can be found as:
Class interval Lower class limit Frequency Cumulative frequency
0 – 20 0 7 40
20 – 40 20 8 33
40 – 60 40 6 25
60 – 80 60 8 19
80 – 100 80 6 11
100 – 120 100 5 5
 
By taking the horizontal axis as the lower class limit and the vertical axis as the corresponding cumulative frequency, we can plot the cumulative frequency for each lower class limit.
Then, the required ogive (of the more-than type) is obtained as:


Note:
The x-coordinate of the point of intersection of the “more-than ogive” and “less-than ogive” of a given grouped data gives its median.
 

  Chapter 4: Quadratic Equations
 
General form of quadratic equations
The general form of quadratic equation in the variable ‘x’ is ax2+bx+c=0, where a, b, c are real numbers and a 0.
For example, 3x2+6x+2=0, x2-2=0
Roots of quadratic equations
A real number ‘k’ is said to be the root of the quadratic equation, ax2+bx+c=0, if ak2+bk+c=0
For example, 3 and –10 are the roots of the quadratic equation,x2+7x-30=0, because
32+7×3-30=9+21-30=30-30=0 = R.H.S.-102+7×-10-30=100-70-30=0 = R.H.S.
Note: x = α (α may or may not be real) is a solution of the quadratic equation, ax2+bx+c=0, if it satisfies the quadratic equation.
Solution of quadratic equation by factorization method
If we can factorise ax2+bx+c=0, where a 0, into a product of two linear factors, then the roots of this quadratic equation can be calculated by equating each factor to zero.
Example: Find the roots of the equation, 2x2-73x+15=0,, by factorisation.
Solution:
2x2-73x+15=02x2-23x-53x+15=02xx-3-53x-3=0x-32x-53=0x-3=0 or 2x-53=0x=3 or x=532
Therefore, 3 and 532are the roots of the given quadratic equation.
Solution of quadratic equation by completing the square
A quadratic equation can also be solved by the method of completing the square.
Example: Find the roots of the quadratic equation, 5x2+7x-6=0, by the method of completing the square.
Solution:
5x2+7x-6=05x2+75x-65=0x2+2·x·710+7102-7102-65=0x+7102-49100-65=0x+7102=169100x+710=±169100=±1310x+710=1310or x+710=-13   10x=1310-710=35 or x=-13   10-710=-2
Therefore, –2 and 35 are the roots of the given quadratic equation.
Quadratic formula
The roots of the quadratic equation, ax2+bx+c=0,, are given by, -b±b2-4ac2a, where b2-4ac0
Nature of roots of quadratic equations
For the quadratic equation, ax2+bx+c=0, where a ≠ 0, the discriminant ‘D’ is defined as
D=b2-4ac
The quadratic equation, ax2+bx+c=0,, where a ≠ 0, has
(i) two distinct real roots, if D=b2-4ac>0
(ii) two equal real roots, if D=b2-4ac=0
(iii) no real roots, if D=b2-4ac<0
Chapter 5: Arithmetic Progressions
 
Arithmetic progression (AP)
• An arithmetic progression is a list of numbers in which the difference between any two consecutive terms is equal.
• In an AP, each term, except the first term, is obtained by adding a fixed number called common difference to the preceding term.
• The common difference of an AP can be positive, negative or zero.
Example:
1 is an AP whose first term and common difference are 3 and 3 respectively.
General form of AP
• The general form of an AP can be written as a, a + d, a + 2d, a + 3d …, where a is the first term and d is the common difference.
• A given list of numbers i.e., a1, a2, a3 … forms an AP if ak+1ak is the same for all values of k.
nth term of an AP
The nth term (an) of an AP with first term a and common difference d is given by
an = a + (n – 1) d
Here, an is called the general term of the AP.
Sum of first n terms of an AP
The sum of the first n terms of an AP is given by
S=n22a+n-1d, where a is the first term and d is the common difference.
If there are only n terms in an AP, then S=n2a+d, where d = an is the last term.
Chapter 7: Coordinate Geometry
 
Axes and coordinates
• The distance of a point from the y-axis is called its x-coordinate or abscissa, and the distance of the point from the x-axis is called its y-coordinate or ordinate.
• If the abscissa of a point is x and the ordinate is y, then (x, y) are called the coordinates of the point.
• The coordinates of a point on the x-axis are of the form (x, 0) and the coordinates of the point on the y-axis are of the form (0, y).
• The coordinates of the origin are (0, 0).
• The coordinates of a point are of the form (+, +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant, where + denotes a positive real number and – denotes a negative real number.
Distance formula
The distance between the points P(x1, y1) and Q (x2, y2) is given by PQ=x2-x12+y2-y12.
Cor: The distance of a point (x, y) from the origin O (0, 0) is given by OP=x2+y2.
Section formula
2
 
The co-ordinates of the point P (x, y), which divides the line segment joining the points A (x1, y1) and B (x2, y2) internally in the ratio m:n, are given by:
Px, y=mx2+nx1m+n, my2+ny1m+n
Cor: The mid-point of the line segment joining the points A (x1, y1) and B (x2, y2) is x1+x22,y1+y22 [Note: Here, m = n = 1]
Area of a triangle
The area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by the numerical value of the expression 12x1y2-y3+x2y3-y1+x3y1-y2
Chapter 9: Some Applications of Trigonometry

Line of sight
It is the line drawn from the eye of an observer to a point on the object viewed by the observer.
Angle of elevation
1
Let P be the position of the eye of the observer. Let Q be the object above the horizontal line PR.
Angle of elevation of the object Q with respect to the observer P is the angle made by the line of sight PQ with the horizontal line PR. That is, QPR is the angle of elevation.
Angle of depression
 
2
Let P be the position of the eye of the observer. Let Q be the object below the horizontal line PX.
Angle of depression of the object Q with respect to the observer P is the angle made by the line of sight PQ with the horizontal line PX. That is, XPQ is the angle of depression. It can be seen that
PQR = XPQ    [Alternate interior angles]
The height or length of an object or the distance between two distant objects can be calculated by using trigonometric ratios.
 
Example: Two wells are located on the opposite sides of a 18 m tall building. As observed from the top of the building, the angles of depression of the two wells are 30° and 45°. Find the distance between the wells. Use 3=1.732
Solution: The given situation can be represented as
4
Here, PQ is the building. A and B are the positions of the two wells such that:
XPB = 30°, X¢PA =45°
Now, PAQ = X¢PA = 45°
PBQ = XPB = 30°
In PAQ, we have
PQAQ= tan 45°18AQ=1AQ=18 m
In PBQ, we have
PQQB= tan 30°18BQ=13QB=183  AB=AQ+QB=18+183m                         =181+3m                         =181+1.732m                         =18 ×2.732 m                         =49.176 m
Chapter 10: Circles
 
Secant
A line that intersects a circle in two points is called a secant of the circle.
Here, PQ is a secant of the circle with centre O.
1
 
Tangent
A tangent to a circle is a line that intersects the circle at exactly one point. The point is called the point of contact of the tangent.
2
Here, l is tangent to the circle with centre O and point P is the point of contact of the tangent l.
• Only one tangent can be drawn at a point on the circle.
• The tangent to a circle is a particular case of the secant, when the two end points of its corresponding chord coincide.
The tangent at any point on a circle is perpendicular to the radius through the point of contact.
• No tangent can be drawn to a circle passing through a point lying inside the circle.
• One and only one tangent can be drawn to a circle passing through a point lying on the circle.
• Exactly two tangents can be drawn to a circle through a point lying outside the circle.
Length of the tangent
The length of the segment of the tangent from an external point P to the point of contact with the circle is called the length of the tangent from the point P to the circle.
The lengths of tangents drawn from an external point to a circle are equal.
Chapter 11: Constructions
 
Division of a line segment in a given ratio
Example: Draw PQ = 9 cm and divide it in the ratio 2:5.
Steps of construction:
(1) Draw PQ = 9 cm
(2) Draw a ray PX, making an acute angle with PQ.
(3) Mark 7 (= 2 + 5) points A1, A2, A3 … A7 along PX such that
PA1 = A1A2 = A2 A3 = A3 A4 = A4 A5 = A5 A6 = A6 A7
(4) Join QA7
(5) Through the point A2, draw a line parallel to QA7 by making an angle equal to PQA7 at A2, intersecting PQ at point R.
PR:RQ = 2:5
1
 
Construction of a triangle similar to a given triangle as per the given scale factor
Case I: Scale factor less than 1
Example: Draw a ABC with sides BC = 8 cm, AC = 7 cm, and B = 70°. Then, construct a similar triangle whose sides are 35thof the corresponding sides of ABC.
Steps of construction:
(1) Draw BC = 8 cm
(2) At B, draw XBC = 70°
(3) With C as centre and radius 7 cm, draw an arc intersecting BX at A.
(4) Join AB, and ABC is thus obtained.
(5) Draw a ray BY, making an acute angle with BC.
(6) Mark 5 points, B1, B2, B3, B4, B5, along BY such that BB1 = B1B2 = B2B3 = B3B4 = B4B5
(7) Join CB5
(8) Through the point B3, draw a line parallel to B5 C by making an angle equal to BB5C, intersecting BC at C′.
(9) Through the point C′, draw a line parallel to AC, intersecting BA at A′. Thus,
2
∆A′BC′ is the required triangle.
 
Case II: Scale factor greater than 1
Example: Construct an isosceles triangle with base 5 cm and equal sides of 6 cm. Then, construct another triangle whose sides are 43rdof the corresponding sides of the first triangle.
Steps of construction:
(1) Draw BC = 5 cm
(2) With B as the centre and C as the radius 6 cm, draw arcs on the same side of BC, intersecting at A.
(3) Join AB and AC to get the required ABC.
(4) Draw a ray BX, making an acute angle with BC on the side opposite to the vertex A.
(5) Mark 4 points B1, B2, B3, B4, along BX such that BB1 = B1B2 = B2B3 = B3 B4
(6) Join B3 C. Draw a line through B4 parallel to B3 C, making an angle equal to BB3 C, intersecting the extended line segment BC at C′.
(7) Through point C′, draw a line parallel to CA, intersecting extended BA at A′.
3
The resulting ∆A′BC′ is the required triangle.
 
Construction of tangents to a circle
Example: Draw a circle of radius 3 cm. From a point 5 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.
Steps of construction:
(1) First draw a circle with centre O and radius 3 cm. Take a point P such that OP = 5 cm, and then join OP.
(2) Draw a perpendicular bisector of OP. Let M be the mid point of OP.
(3) With M as the centre and OM as the radius, draw a circle. Let it intersect the previously drawn circle at A and B.
(4) Joint PA and PB. Therefore, PA and PB are the required tangents. It can be observed that PA = PB = 4 cm.
4
Chapter 12: Areas Related to Circles
 
Area of a circle = πr2
Circumference of a circle = 2πr; where r is the radius of a circle.
Area of sector of a circle
• Area of the sector of angle θ=θ360°×πr2, where r is the radius of the circle
3
 
• Area of a quadrant of a circle with radius r=πr24For quadrant θ=90°.

 
Length of arc of a circle
Length of the arc of a sector of angle θ=θ360°×2πr, where r is the radius of the circle
5 
Area of segment of a circle
Area of segment APB = Area of sector OAPB – Area of ∆OAB
=θ360×πr2-area of ΔOAB
6Chapter 13: Surface Areas and Volumes
 
Cuboid
• Surface area = 2 (lb + bh + hl)
• Volume = l × b × h, where l, b, h are respectively length, breadth and height of the cuboid
1
 
Cube
• Surface area = 6a2
• Volume = a3, where a is the edge of the cube
2
 
Cylinder
• Curved surface area (CSA) = 2πrh
• Total surface area (TSA) =2πr2+2πrh=2πrr+h
• Volume = πr2h, where r is the radius and h is the height of the cylinder
3
 
Cone
• Curved surface area (CSA) = πrl
• Total surface area (TSA) =πr2+πrl=πr r+l
• Volume=13πr2h, where r is the radius and h is the height of the cone
4
 
Sphere
• Surface area = 4πr2
• Volume=43πr3, where r is the radius of the sphere
5
 
Hemisphere
• Curved surface area (CSA) = 2πr2
• Total surface area (TSA) = 3πr2
• Volume =23πr3, where r is the radius of the hemisphere
6
Note: Volume of the combination of solids is the sum of the volumes of the individual solids
Conversion of a solid from one shape into another
When a solid is converted into another solid of a different shape, the volume of the solid does not change.
 
Frustum of a cone
• Volume of the frustum of a cone=13πr12+r22+r1 r2h
• CSA of the frustum of a cone=πr1 r2l, where l=h2+r1-r22
• TSA of the frustum of a cone=πr1+ r2l+πr12+πr22, where l=h2+r1-r22
Chapter 15:  Probability
 
The theoretical probability (also called classical probability) of an event E, denoted as P(E) is given by
PE=Number of outcomes favourable to ENumber of all possible outcomes of the experiment
 
Elementary events:
• An event having only one outcome of the experiment is known as elementary event.
• The sum of the probabilities of all the elementary events of an experiment is 1.
 
Example: A dice is thrown once. What is the probability of getting 1 on the dice?
Solution:
When a dice is thrown once, the possible outcomes are 1, 2, 3, 4, 5, 6.
Let A be the event of getting 1 on the dice.
 PA=Number of outcomes favourable to ANumber of all possible outcomes=16
Complementary events
For an event E of an experiment, the event E represents ‘not E’, which is called the complement of the event E. We say, E and E are complementary events.
PE+PE=1PE=1-PE
The probability of an impossible event of an experiment is 0.
The probability of a sure (or certain) event of an experiment is 1.
∴ 0 ≤ P (E) ≤ 1
Chapter 4: Determinants
 
Determinant of a square matrix A is denoted by Aor det (A).
 
Determinant of a matrix A=a1×1 is A=a=a
 
Determinant of a matrix A=a11a12a21a22 is given by,
1
 
Expansion of a determinant of a matrix
Determinant of a matrix A=a1a2a3b1b2b3c1c2c3 is given by (expanding along R1):
A=a1a2a3b1b2b3c1c2c3=-11+1a1b2b3c2c3+-11+2a2b1b3c1c3+-11+3a3b1b2c1c2                          =a1b2b3c2c3-a2b1b3c1c3+a3b1b2c1c2
Similarly, we can find the determinant of A by expanding along any other row or along any column.
 
If A = kB, where A and B are square matrices of order n, then A=knB, where n = 1, 2, 3
 
Properties of determinants
• If the rows and the columns of a square matrix are interchanged, then the value of the determinant remains unchanged.
This property is same as saying, if A is a square matrix, then A=A'
• If we interchange any two rows (or columns), then sign of determinant changes.
• If any two rows or any two columns of a determinant are identical or proportional, then the value of the determinant is zero.
• If each element of a row or a column of is muldeterminant tiplied by a constant a, then its determinant value gets multiplied by a.
• If A=aij3×3, then kA=k3A
• If elements of a row or a column in a determinant can be expressed as sum of two (or more) elements, then the given determinant can be expressed as sum of two (or more) determinants.
Example:
a1+d1b1c1a2+d2b2c2a3+d3b3c3=a1b1c1a2b2c2a3b3c3+d1b1c1d2b2c2d3b3c3
• If the equi-multiples of corresponding elements of another row or column are added to each element of any row or column of a determinant, then the value of the determinant remains unchanged.
a1+λa3b1+λb3c1+λc3a2           b2           c2       a3           b3            c3       =a1b1c1a2b2c2a3b3c3
 
Area of a triangle
Area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by,
Δ=12x1     y1    1x2       y2       1x3    y3   1
Since area is always positive, we take the absolute value of the above determinant.
 
Minors and cofactors
• Minor of an element aij (denoted by Mij) of the determinant of matrix A is the determinant obtained by deleting its ith row and jth column.
Example:
If A=a11a12a13a21a22a23a31a32a33, then M23=a11a12a31a32
• Co-factor of an element aij is defined by Aij=-1i+j Mij, where Mij is the minor of aij.
Example:
If A=a11a12a13a21a22a23a31a32a33, then A23=-12+3 M23=-a11a12a31a32
• If element of a row (or column) are multiplied with co-factors of any other row (or column), then their sum is zero.
For example: If A is square matrix of order 3, then
a21 A11+a22 A12+a23 A23= 0
• If elements of one row or column of a determinant is multiplied with its corresponding co-factors, then their sum is equal to the value of determinant.
Example:
If A is a square matrix of order 3, then a31 A31+a32 A32+a33 A33=A
 
Adjoint and inverse of a matrix
• If A = a11a12a13a21a22a23a31a32a33, then Adj A = A11A21A31A12A22A32A13A23A33,where Aij is the co-factor of aij.
• If A is a square matrix, then A (AdjA) = (AdjA) A = A I
• A square matrix A is said to be singular, if A=0
• A square matrix A is said to be non-singular, if A0
• If A and B are non-singular matrices of same order, then AB and BA are also non-singular matrices of same order.
• If A and B are square matrices of same order, then AB=AB
• If A is a non-singular matrix of order n, then Adj A=An-1
• A square matrix A is invertible, if and only if A is non-singular and inverse of A is given by the formula:
A-1=1AAdj A
 
Application of determinants and matrices
The system of following linear equations
a1x+b1y+c1z=d1a2x+b2y+c2z=d2a3x+b3y+c3z=d3
can be written as AX = B, where
A=a1b1c1a2b2c2a3b3c3, X=xyz, B=d1d2d3
Consistent system: A system of linear equations is said to be consistent, if its solution (one or more) exists.
Inconsistent system: A system of linear equations is said to be inconsistent, if its solution does not exist.
• Unique solution of equation AX = B is given by X = A–1 B, where A0
• For a square matrix A in equation AX = B, if
a) A0, then there exists unique solution
b) A=0 and (adj A) B 0, then no solution exists
c) A=0 and (adj A) B = 0, then the system may or may not be consistent
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