Areas of Parallelograms and Triangles
Concept 1: Figures on the same base and between the same parallels
Example: Consider two parallelograms on the same base and between the same parallels.
Shown are two parallelograms ABCD and CDEF on the same base CD and between the same parallels CD and AF. We know the area of a parallelogram is base*height and that of triangle is ½*base*height.
In the above case, triangles AEC and BFD are congruent.
The areas of congruent figures are equal.
So, area of ABCD and CDEF are equal since ERCD is common in both parallelograms. So,
Parallelograms on the same base and between the same parallels are equal in area.
Similarly, if two triangles are on the same base, and between the same parallels, their height is also the same since it is the distance between the parallel lines which never changes. Therefore,
Two triangles on the same base and between the same parallels are equal in area.
Conversely, two triangles having same or equal base, and equal areas, lie between the same parallels.
Example: Prove that the area of the quadrilateral PQRS is equal to the triangle PQX if PR||SX.
Since PR||SX, the areas of triangles PRS and RPX are equal by the above theorem.
Since triangle PQR is common between PQRS and triangle PQX,
and the total area of PQRS equals ar.(PQR)+ar.(PRS) and
area of triangle PQX equals ar.(PQR)+ar.(RPX).
So, the total area of PQRS and PQX will be equal.
Circles
Concept 1: Basic Terminology
A circle is a collection of all points in a plane, which are equidistant from a fixed point i.e. the center of the circle.
The perimeter of the circle is called circumference.
A chord is formed when any two points on the circumference of the circle are joined. The chord passing through the center of the circle is the diameter.
Diameter is the longest chord and its length is twice the length of the radius, which is the distance of any point on the circumference from the center of the circle.
The area between a chord and an arc is called a segment.
The area between two radii joining two points on a circle to the center, and the related arc is known as a sector.
Segments and sectors are of two types as depicted in the figures.
XY is a chord and OA and OB are radii.
Z ,V,Q and P are points on the circumference of their respective circles.
XZY is the major arc and XVY is the minor arc. Similarly, APB is major arc, AQB is minor arc.
As shown, XZY is the major segment and APB is the major sector and vice versa.
In the above figures, if AB is joined to make a chord, we notice that it subtends an angle AOB at the center. This angle is called angle subtended by a chord at the center. If AP and BP are joined, then the chord AB will subtend an angle at point P. If there are more than one chords in a single circle,
Equal chords subtend equal angles at the center.
Conversely, if the angles subtended at the center by two chords are equal, the chords are equal.
If a perpendicular is drawn to a chord from the center of the circle, the following theorems exist.
The perpendicular from the center of a circle, bisects the chord.
Conversely, a bisector drawn from the center of a circle to a chord, is the perpendicular to the chord.
Also, equal chords in a circle, are at equal distances from the center.
Conversely, chords equidistant from the center of a circle, are equal.
Also, only one circle exists which can pass through a set of three non-collinear points.
Example1: If an arc is given, how can you form the complete circle?
Solution: Join the end points of the arc to any other point on the inside of the arc, making two chords. Construct the perpendicular bisectors of these chords. According to the theorems, the point where these bisectors will intersect will be the center of the circle. The rest of the circle can be made using the center and a point in the circumference of the arc.
Example2: If two equal chords intersect within a circle, prove that their corresponding segments are equal.
Solution:
Let H be the center of the circle and AB be a diameter as shown. Let XY and PQ be the intersecting chords. Prove that XE=PF, EO=FO and OY=OQ by using congruency of triangles OEH and OFH and it is proved that all the corresponding segments will be equal.
Constructions
Concept 1: Reasoning of constructions and Construction of Triangles
The explanation of the construction processes followed for construction of angle bisectors, perpendicular bisectors and angles of various measurements are all done by using the congruency of triangles as the proof.
These constructions have been in use since previous classes.
Example: Construction of bisector of given angle.
Let the given angle be ABC as shown in figure. With the center at
B, with any radius cut an arc through BA and BC at E and D
respectively. Cut arcs of same radius with centers at E and D.
Join the point of intersection of these arcs (F) with B. This is the required angle bisector.
Proof: BE=BD=EF=DF. Hence, triangles BEF and BDF are congruent. So, LEBF=LDBF (CPCT).
In similar ways, other constructions can be explained.
Construction of Triangles
To construct a triangle(ABC), given its base(BC), a base angle(LB) and sum of the
other two sides(AB+AC).
Method: Draw BC and LXBC. On BX, make BD=AB+AC. Join CD.
Make LDCA=LBDC. Mark A on BD. ABC is the required triangle.
Proof: BD=AB+AC. AC=AD(because their opposite angles are equal). Hence, BD=AB+AC.
Alternate Method: Construct perpendicular bisector(PQ) of DC. The point where it meets BD is A.
Again, AC=AD because A is on the perpendicular bisector of CD.
To construct a triangle given its base(BC), a base angle(LB) and the difference of the other sides.
Method: Make BC and LXBC.
1. Case1: AB>AC, the difference of other two sides is AB-AC. Make BD=AB-AC on BX.
OR
1. Case2: AC>AB, the difference of other two sides is AC-AB.
Make BD=AC-AB on extended part of BX on the opposite side of X.
2. Join DC and draw perpendicular bisector(PQ) of DC.
3. PQ intersects BX at A. Join AC. ABC is the required triangle.
Proof: AD=AC. Use this to proof BD = (given) Difference of other two sides.
To construct a triangle(ABC), given its perimeter(AB+BC+CA) and its two base angles(LB and LC).
Method: Draw a line XY=AB+BC+CA and make LLXY=LB and LMYX=LC.
Bisect LLXY and LMYX and let the bisectors meet at A.
Make perpendicular bisectors of XA and YA to meet XY at
B and C respectively. ABC is the required triangle.
Proof: XB=AB and YC=AC because B and C exist on
the perpendicular bisectors of XA and YA respectively.
Also, LABC=2LAXY (exterior angle property of triangle ABX).
So, LABC=LLXY(since LBXA=LBAX). Similarly, LACB=LMYX.
Example: Construct a triangle PQR with perimeter 36cm and base angles LP=30° and LQ=90°.
Solution:
Steps of construction:
Make XY=36cm, LLXY=30° and LMYX=90°.
Construct angle bisectors of LLXY and LMYX and let them meet each other at R.
Construct perpendicular bisectors of XR and YR and let them meet XY at P and Q respectively.
Join RP and RQ. PQR is the required triangle.
Coordinate Geometry
Concept 1: Cartesian System
Coordinate geometry is required to locate the position of a point in a 2-dimensional plane.
We know how to mark the value of a point on a number line. In cartesian system, we have two perpendicular lines and we mark points in the reference of both these lines.
We draw 2 number lines perpendicular to each other, called coordinate axes, intersecting at origin or 0.
The horizontal line is called the x-axis and the vertical line is called the y-axis.
Every point is marked as the distance from the y-axis in the x direction or horizontal direction, and the distance from x-axis in the y or vertical direction.
The values of these distances from the x and y axes respectively, are called coordinates.
These coordinates can be negative or positive, denoting the direction of the distance from any axis.
The coordinates are written in a specific order (x,y), x comes first, then y.
The value of x coordinate is known as the abscissa and that of y coordinate is known as ordinate.
NOTE: If x≠y, then the points (x,y) and (y,x) will always be different in the coordinate system.
The cartesian system is divided in 4 quadrants as shown above and the form of coordinates as per their sign in the quadrants is also shown.
The coordinates of the points on the x-axis are of the form (x,0), because their distance from the x axis is zero. Similarly, the points on y-axis have coordinates of the form (0,y).
The point of intersection of x and y axes is the origin and its coordinates are (0,0).
Example: Identify the quadrant by marking the following points on the cartesian plane: (1,3),(4,-5),(0,2).
(1,3)- I Quadrant, (4,-5)- IV Quadrant, (0,2)- It is on the y-axis, hence it does not exist in any quadrant.
Heron’s Formula
Concept 1: Heron’s Formula and its Applications
We know the area of triangle is ½*base*height.
But this formula is valid for right triangles and for the triangles whose height is given or can be found.
For example, the height of isosceles and equilateral triangles can be found given the dimensions of their sides.
For scalene triangles, finding of area is not possible without knowing its height and the associated base.
For this purpose, Heron, a mathematician, gave a formula to calculate the area of scalene triangles.
where, A = area, s = semi perimeter of the triangle and a, b, c are its sides.
So, s = (a + b + c)/2
Using this formula, area of scalene triangles can be calculated by using the dimensions of its sides alone,
without knowledge of its height.
Example: From the figure shown, two triangles share the same base. ABC is a triangle with perimeter = 36cm.
Its sides are in the ratio of 2:3:4. PB=12cm and PC=11cm. Calculate the area of ABPC.
Given, the perimeter of ABC=36cm, let the common multiple of its sides be ‘x’.
Then, 2x+3x+4x=36; 9x=36; x=4.
AB=8cm, BC=12cm, AC=16cm. (since the sum of two sides is greater than the third side, this triangle exists.)
Now, using heron’s formula in triangle ABC, s = 36/2 = 18cm.
So, A1=[18(18-8)(18-12)(18-16)](1/2), A1=[18*10*6*2](1/2)
A1=[2160](1/2), A1=46.475cm2.
Now, in triangle PBC, s=(PB+BC+CP)/2; s = (12+12+11) / 2; s = 35/2; s = 17.5cm.
So, A2=[17.5(17.5-12)(17.5-12)(17.5-11)](1/2), A2=[17.5*5.5*5.5*6.5](1/2)
A2=[3440.9375](1/2), A2=58.659cm2.
Total area of ABPC = A1 + A2
A = 46.475 + 58.659
A = 105.134cm2.
ntroduction to Euclid’s Geometry
Concept 1: Axioms and postulates
Euclid was a Greek mathematician who wrote a book called ‘elements’ containing the axioms and postulates given by him. These are used till date with slight modifications.
Axioms and postulates are both considered obvious universal truths by Euclid and distinguished as follows:
Axiom: Common notions or the things commonly accepted by people are known as axioms.
Postulates: Assumptions related to geometry.
NOTE: These are not the exact meanings of these words as used today.
Some of Euclid’s axioms were:
Things which are equal to the same thing are equal to one another.
If equals are added to equals, the wholes are equal.
If equals are subtracted from equals, the remainders are equal.
Things which coincide with one another are equal to one another.
The whole is greater than the part.
Things which are double of the same things are equal to one another.
Things which are halves of the same things are equal to one another.
Euclid’s postulates were:
Postulate 1: A straight line may be drawn from any one point to any other point.
It is also assumed along with the postulate that for 2 distinct points, there is a unique line that passes through them.
Postulate 2: A terminated line can be produced indefinitely.
Postulate 3: A circle can be drawn with any center and any radius.
Postulate 4: All right angles are equal to one another.
Postulate 5: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the
two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.
Above figure is to prove the 5th postulate. Since ‘n’ is a straight line falling on two straight lines ‘m’ and ‘l’, the sum of angles ‘a’ and ‘b’ is less than 180° because the lines are not parallel. Hence, these lines will intersect on this side where angles ‘a’ and ‘b’ are.
Using his axioms and postulates, Euclid proved other results which were called propositions or theorems.
A theorem deduced by Euclid based on the above axioms and postulates is:
Two distinct lines cannot have more than one point in common.
A system of axioms is called consistent, if any of the axioms of the system does not contradict any other axiom or previously proved statement.
Example 1: If P,Q and R are points on a line and Q lies in the middle of P and R, prove that PQ+QR=PR.
Solution: From the figure, PR is coincident with PQ+QR. Also, by Euclid’s axiom 4, PR=PQ+QR.
From the 5th postulate of Euclid, two equivalent versions are also made.
Playfair’s Axiom: For every line ‘l’ and for every point ‘P’ not lying on ‘l’, there exists a unique line ‘m’ passing through ‘P’ and parallel to ‘l’.
This can also be stated as ‘Two distinct intersecting lines cannot be parallel to the same line’.
Linear Equations in two variables
Concept 1: Basics
An equation written in the form ax+by+c=0, where ‘a’, ‘b’ and ‘c’ are constants and at least one of ‘a’ and ‘b’ is non-zero, is called a linear equation in two variables, ‘x’ and ‘y’.
Example1: 3x-6y=37, x+9z=29, x=2 etc.
x=2 can be rewritten as x+(0.y)-2=0, in the standard form.
In the above examples, in 3x-6y=37 and x+9z=29, substitute x=0,1,2,3,4, -1, -2, -5, -7 etc. to get many values of the other variable.
In x=2, all the values of y will satisfy this equation for x=2.
The solution of linear pair of equations are determined by substituting some values for a variable, to find the corresponding value of the other variable which will satisfy the equation.
The solutions to such equations are written in the form of (x,y), where ‘x’ and ‘y’ are the variables in the equation. This format of obtaining solutions is used in plotting the graphs of such equations. Every point on the graph of a linear equation is a solution to it.
Example2: Find the 5 solutions of 2x-6y=20.
At x=0, y= (-20/6) = (-10/3) = (-3.34)
At x=1, y= (-18/6) = (-3)
At x=2, y= (-16/6) = (-8/3) = (-2.67)
At x= (-1), y= (-22/6) = (-11/3) = (-3.67)
At x= (-2), y= (-24/6) = (-4)
The solutions are: (0,-3.34), (1,-3), (2,-2.67), (-1,-3.67), (-2,-4)
Other solutions can be found by substituting ‘y’ with any values and finding corresponding values of ‘x’, or simply substituting more values of ‘x’ to find the associated value of ‘y’. For example, at y=0, x=10.
A linear equation in two variables has infinite solutions.
The graphs of all linear equations in two variables are straight lines.
For plotting a linear equation in two variables on a Cartesian Plane, we find at least two solutions of the equation in the form of (x,y), plot them as points on the graph and join them to obtain a line on the graph. In general, if you find the values of (x,0) and (0,y), the line can be easily plotted.
The graph of an equation x=a for any value of ‘a’, is a line parallel to y-axis at x=a, because all the values of ‘y’ satisfy the equation at x=a (represented by blue line in the graph below).
Similarly, y=b for any value of ‘b’ is a line parallel to x-axis at y=b (represented by yellow line the graph below).
By this theory, x=0 is the y-axis and y=0 is the x-axis.
NOTE: The representation of x=a or y=b for an equation in one variable is done on the number line as a single point. For equations in two variables, cartesian plane is used.
Example3: Plot the graph of the equation 2x-6y=20 using the solutions obtained in example2.
The solutions are: (0,-3.34), (1,-3), (2,-2.67), (-1,-3.67), (-2,-4), (10,0).
Lines and Angles
Concept 1: Basic Terminology
Lines can be extended indefinitely in both the directions. It’s a line segment which has a fixed length and its end points are fixed.
Intersecting lines are those which meet each other at one point. If it is not evident that the lines are intersecting, they may be produced or extended further in any direction to check the intersection.
Non-intersecting lines are parallel lines and never meet each other at any point in space.
We know that the angle formed by a straight line is equal to 180°. So, if a ray has its end point on a straight line, the angles formed by the ray with the line will sum up to 180°. Such pairs of angles are called linear pair of angles.
Conversely, if the sum of two adjacent angles is 180°, they are said to be present on a line or the non-common arms of the angle are on a line.
If two lines intersect at a point, the vertically opposite angles are always equal. It can be proven easily by the following figure:
By linear pair of angles property, the sum of angles AOP and POQ is 180°. Similarly, the sum of angles AOP and AOB is 180°. So, LAOP+LPOQ=LAOP+LAOB. So, LPOQ=LAOB. These are vertically opposite angles. Similarly, LAOP=LBOQ. Hence, it is proved that when two lines intersect, the vertically opposite angles are always equal.
Example: Find the value of angle POA from the given figure if OB is angle bisector of AOQ and AOB=90°.
To solve this, we know that angle AOB=90°, and angle AOB=angle QOB=90°.
Since vertically opposite angles are equal, angle POA=angle QOB=90°.
Alternate solution: the sum of a linear pair of angles is 180°. So, the sum of the angles AOP and BOA=180°.
Since, angle BOA or angle AOB = 90°, angle AOP comes out to be 90°.
In this way, the above explained properties of angles can be applied to a variety of problems.
Chapter 6: Lines and Angles
Parallel lines and a Transversal
Shown here are two parallel lines ‘l’ and ‘m’ and a transversal ‘n’.
A transversal is a line which intersects with two or more lines at distinct points.
Since the lines ‘l’ and ‘m’ are parallel, the angles 1,2,3,4,5,6,7 and 8 have some properties and names. They are as follows:
Corresponding angles: 1 and 5, 2 and 6, 4 and 8, 3 and 7. These pairs are corresponding angles and such pairs are always equal.
Alternate interior angles: 4 and 6, 3 and 5. These are the pairs of alternate interior angles and such pairs are always equal.
Alternate exterior angles: 1 and 7, 2 and 8. These are the pairs of alternate exterior angles and such pairs are always equal.
Interior angles on the same side of the transversal: 4 and 5, 3 and 6. These are the pairs of interior angles on the same side of the transversal. Such pairs are always supplementary.
The converse of these conditions is also true.
For example: Any pair of lines intersected by a transversal having equal corresponding angles, is said to be parallel. Similarly, the converse of other conditions is also true.
NOTE: These conditions are all true only for a transversal intersecting a pair of parallel lines.
Lines parallel to the same line are parallel to each other.
Example: Find the value of angle ‘a’ if the shown lines are parallel with a transversal.
Angle a and the angle shown to be 55° are corresponding angles. Since, for a pair of parallel lines these are
equal, so a=55°.
Angle sum property of a triangle: it is known that the sum of the angles of a triangle is 180°.
This property can be proved by making a line parallel to one side of a triangle at the vertex of the opposite side.
Then by using alternate interior angles and the property of angle in a line, this can be proved.
Using this property, another theorem is made as shown.
According to this theorem for the triangle shown, if a side is produced, then the exterior angle so obtained
Has its value equal to the sum of the other two interior angles.
Proof: La+Lb+Lc=180°, and Lc+LACD=180° (linear pair of angles).
So, La+Lb=LACD.
The angle ACD is called an exterior angle of the triangle.
Polynomials
Concept 1: Polynomials and their Zeroes
A polynomial is an algebraic expression in variables where the powers of the variables are whole numbers.
In every term of a polynomial, there is a coefficient. If there is no coefficient written, it is one. If the term does not exist, then the coefficient is zero.
Example: x2-3x+1; the coefficient of x2 is 1, of x is -3 and of x0(constant term) is 1. Since there is no term of x3, its coefficient is zero.
Degree: the highest power of the variable in a polynomial.
For example: x2-5x+1; the degree is 2, x5-x6+54x9-9x; the degree is 9, x5/7+x16/3-x2; the degree would have been 16/3, but since it is not a whole number, this expression is not a polynomial.
The polynomials with degree= 1, are known as linear polynomials, degree=2 are called quadratic and those with degree=3 are called cubic polynomials.
For example: linear: x+1, 3x-7, 1/2x, etc.
Quadratic: x2-4x+2, x2+6x-8, 5x2+3x-8,1/2x2+5 etc.
Cubic: x3+2x2+6x-1, 1/2x3+6x2-2 etc.
A constant is a polynomial with zero degree. Example: 5x0, 6, 89x0, 1/2 etc.
Polynomials are represented as p(x), q(x), r(x), s(y) etc. , where the alphabet p,q,r,s denote the polynomial and the variable inside bracket denotes the variable of the polynomial.
A polynomial in one variable x of degree n is an expression of the form:
anxn + an–1xn–1 + . . . + a1x + a0
where a0 , a1 , a2 , . . ., an are constants and an ≠ 0.
If all constants of a polynomial become zero, it is called a zero-polynomial denoted by 0. The degree of such a polynomial is not defined.
A zero of a polynomial, is that value of the variable, for which the value of the polynomial=0.
For example: p(x)=x2-4, f(x)=5x-2, p(y)=y3+y2-2.
If we put p(x)=0, we find that x=2 and x=-2 are the zeroes.
Put f(x)=0, to find that the value x=-2/5 makes the value of f(x) = 0.
Put y=1 in p(y) to obtain p(y)=0.
These values of x and y are the zeroes of the respective polynomials.
Polynomials
Concept 2: Remainder and Factor Theorem
Division of two polynomials – Example1: Divide x2-24-2x by 4+x.
x-6
x+4 x2 – 2x – 24
– [x2 + 4x]
-6x + 24
– [-6x + 24]
0
Arrange the polynomials in the standard order of decreasing degrees.
Divide the first term of the dividend by first term of divisor, to obtain first term of quotient: here, x2/x=x.
If the consequent term is not in the present dividend, just add the term in the next dividend.
Multiply other terms of the divisor by the quotient obtained, and subtract the expression obtained from the dividend, to obtain a new expression which is now considered the dividend.
Follow the above 2 steps till you obtain 0 or a remainder which cannot be divided further.
On successful division, we have obtained a quotient (x-6) and remainder is zero. We know that: Dividend=(Divisor*Quotient) + Remainder
So, x2-2x-24 = (x-4)*(x-6) + 0.
Remainder theorem: If any polynomial p(x) having its degree > 1, is divided by a linear polynomial (x-a), then the remainder is p(a).
If p(a)=0, (x-a) is a factor of p(x).
To explain it better, the degree of the remainder is always less than that of the divisor.
Factor theorem: If (x-a) is a factor of p(x), then p(a)=0 and conversely, if p(a)=0, then (x-a) is a factor of p(x).
Applying to example1, (x-4) and (x-6) are the factors of p(x), so 6 and -4 will be the zeroes of p(x) or p(-4) and p(6) will be equal to zero. This can be proved by substituting -4 and 6 in p(x).
Factorisation: This is the process of writing a polynomial as product of its factors.
For a quadratic polynomial ax2+bx+c, factorisation is done by splitting the middle term (i.e. the term having x) into two terms. Let the new terms be px and qx, they must satisfy:
p+q=b and pq=ac
In example1, ac = 1*(-24)=(-24); b=(-2), so p and q are to be found such that their product is -24 and their sum is -2. Such a pair is -6 and 4. So, x2-6x+4x-24 = x(x-6) + 4(x-6)= (x-6)(x-4).
For a cubic polynomial, one factor is found by hit and trial, and then the polynomial is divided by that factor to obtain the other factors or the factor is taken common from the polynomial.
Example2: Factorise p(x) = x3+8x2-5x-4.
By hit and trial, we find that p(1)=0. So, (x-1) is a factor of p(x).
x2 + 9x – 4
x – 1 x3 + 8x2 – 5x – 4
– [x3 – x2]
9x2 – 5x – 4
– [9x2 – 9x]
-4x – 4
– [-4x – 4]
0
x3+8x2-5x-4=(x-1)(x2+9x-4).
If such quadratic factors are obtained, they can be factorised further.
Polynomials
Concept 3: Algebraic Identities
We know the following identities:
(x+y)2=x2+2xy+y2
(x-y)2=x2-2xy+y2
x2-y2=(x+y)(x-y)
(x+a)(x+b)=x2+(a+b)x+ab
Example: Solve 105*109 using identities.
105*109=(100+5)(100+9) . Here, x=100, a=5 and b=9.
Using identity, 105*109 = 1002+(5+9)100+5*9= 10000+1400+45 = 11445.
Using the above identities, some other identities are derived. They are:
(x+y+z)2=x2+y2+z2+2xy+2yz+2zx
(x+y)3=x3+y3+3xy(x+y)
(x-y)3=x3-y3-3xy(x-y)
x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)
Example: Factorise 343x3+8y3-64z3+168xyz.
Solution: 343x3+8y3-64z3+168xyz = (7x)3+(2y)3+(-4z)3-3*(7x*2y*-4z)
Using identity; 343x3+8y3-64z3+168xyz = (7x+2y-4z)(49x2+4y2+16z2-14xy+8yz+28zx)
This is the required factorisation of the given expression.
Probability
Concept: Probability
Probability is a value of describing the chances of an event, whether it will happen or not. Its formulated as:
P(E)= (number of favorable outcomes) / (total number of outcomes)
P(E) represents the probability of the event E. Also, the collective probability of all the experiments will always be 1 i.e. if any experiment is conducted, some outcome will always be there.
An event which can never be the outcome of an experiment has the probability 0 and is known as an Impossible Event.
For example: A single dice having numbers 1-6 cannot give the outcome 7. Hence, its probability is 0.
Similarly, an event which will always occur whenever the experiment is performed is called as a sure event or a Certain Event. It has a probability of 1.
For example: The probability of sun rising in the east will always be 1, because it’s a universal truth that sun rises in the east.
Example: In 8 bags of rice labelled 10kg, the following weights of rice (in kg) are present:
9.98 9.5 10.1 10.3 9.9 10.01 9.68 10.34
If any bag is chosen at random, find the probability of the bag having more than 10 kg of rice.
Solution: The number of bags having more than 10 kg of weight, is 4.
So, favorable outcome = 4
Total number of bags = 8
Hence, probability = favorable outcomes / total number of outcomes
Probability = 4 / 8 = 1 / 2 = 0.5.
Quadrilaterals
Concept 1: Parallelogram
A closed figure with four sides and four vertices is known as a quadrilateral.
A quadrilateral can be divided into two triangles by a diagonal, hence the sum of all the angles of a quadrilateral is 360°.
Types of quadrilaterals:
A quadrilateral with one pair of opposite sides to be parallel, is a trapezium.
A quadrilateral with both pairs of opposite sides to be parallel is a parallelogram.
A quadrilateral in which two pairs of adjacent sides are equal is a kite.
A parallelogram in which the opposite sides are equal, and an angle is 90°, is a rectangle.
A parallelogram in which all sides are equal, and an angle is 90°, is a square.
A parallelogram in which all sides are equal, angles may or may not be 90°, is a rhombus.
Diagonals of a rectangle bisect each other and are equal and vice-versa.
Diagonals of a rhombus bisect each other at right angles and vice-versa.
Diagonals of a square bisect each other at right angles and are equal, and vice-versa.
Properties of a Parallelogram
A parallelogram is divided into two congruent triangles by its diagonal.
In a parallelogram, opposite sides are equal, and its converse is also true, that if the opposite sides of a quadrilateral are equal, it is a parallelogram.
In a parallelogram, opposite angles are equal. Its converse is also true.
The diagonals of a parallelogram bisect each other. The converse is also true.
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
All these properties can be proved using congruency of triangles and properties of parallel lines.
Example: The diagonals of a parallelogram are equal. Prove that it’s a rectangle.
Solution: A parallelogram ABCD is shown in the figure.
AC and BD are equal diagonals of the parallelogram.
In triangles AOD and BOC
AO=OD(diagonals of a parallelogram bisect each other)
Since AD=BC (given), ½*AD=1/2*BC ………… ①
From ①, AO=OC=OB=OD and AC=BD (opposite sides of a parallelogram are equal)
So, triangles AOD and BOC are isosceles and congruent to each other. This implies that angles OAC, OCA, OBD, ODB are all equal. Let them be equal to ‘x’.
Similarly, triangles AOB and COD are also congruent and angles OAB, OBA, OCD, ODC are all equal to ‘y’.
So, all the angles of the given parallelogram can now be written as (x+y).
Now, angle sum of a parallelogram is 360°. So,
LCAB+LABD+LBDC+LDCA=360°
(x+y)+(x+y)+(x+y)+(x+y)=360°
4(x+y)=360°
(x+y)=90°
This means that all the angles if the parallelogram are 90°, hence it’s a rectangle.
In similar ways, many questions of quadrilaterals are just an application of properties of triangles.
Quadrilaterals
Concept 2: Mid-Point Theorem
According to the mid-point theorem,
If a line segment is joining two sides of a triangle at their midpoints, then it will be parallel to the third side.
Also, if a line segment is drawn from the midpoint of one side of a triangle and is parallel to a second side, then it intersects the third side at its midpoint.
Example: In the figure shown, PQRS is a quadrilateral in which W,X,Y,Z are mid-points of their respective sides. Prove that WXYZ is a parallelogram.
Solution: Construct a diagonal-QS.
By the above-mentioned theorem, since W and Z are midpoints of PQ and PS respectively, WZ||QS in triangle PQS.
Similarly, XY||QS. Hence, WZ||XY.
Similarly, XW||YZ.
Hence, WXYZ is a parallelogram.
NOTE: This can be proved for all types of quadrilaterals that the quadrilateral formed by joining their midpoints is a parallelogram.
Number Systems
Concept 1: Real Numbers
Number line: A line on which all real numbers can be represented.
Natural Numbers: All numbers like 1,2,3…. are called natural numbers and represented by ‘N’. They are represented on the number line at the right of 0, starting from 1.
Whole Numbers: All numbers like 0,1,2,3…. Are called whole numbers and represented by ‘W’. Basically when 0 is included in natural numbers, whole numbers are formed. They are represented on the number line starting from 0, to the right.
Integers: All numbers like ….-3,-2,-1,0,1,2,3…. Are called integers and represented by ‘Z’. When negative numbers are included in whole numbers, integers are formed. They are represented on both sides of 0.
Rational Numbers: All numbers which can be represented as (p/q), where q≠0, ‘p’ and ‘q’ are integers and ‘p’ and ‘q’ don’t have any common factor or in other words, ‘p’ and ‘q’ are co-prime, are called rational numbers and represented by ‘Q’. Example: 2/3, 5/6, -9/26 etc. All integers, whole numbers and natural numbers are rational numbers as they can be represented as (p/q). They are represented on the number line as decimals between integers.
Example: 25 can be written as 25/1, -95 can be written as -95 and 0 can be written as 0/1.
Example: Find 5 rational numbers between 2 and 3.
Solution: We can find these numbers by hit and trial and give an answer like 47/23, 63/27, 89/35, 5/2, 100/43 etc. or we can add 1 to the quantity of numbers to be found and find its multiples in the numbers as:
5+1=6; 2=12/6 and 3=18/6. Now, we can write exactly 5 fractions between 2 and 3 as
13/6, 14/6, 15/6, 16/6, 17/6.
Irrational Numbers: All numbers which cannot be written in the form (p/q) as in rational numbers, are known as irrational numbers.
Example: π, 2(1/2), 3(1/2), 0.1001100001111001000……
They are represented on the number line using a construction method as shown in the following example.
Example: Represent 2(1/2), 3(1/2) and 4(1/2) on the number line.
Solution:
Consider 0 and 1 shown on number line as points on the triangles. As shown on the number line, triangle 01A is right angled at 1 and length 01=1unit and length 1A=1 unit. By Pythagoras theorem, 0A=2(1/2) units. Keeping the center at 0 and radius = 0A, an arc is cut on the positive side of the number line showing 2(1/2).
Similarly, for 3(1/2), another right triangle is made with 0A as its base and perpendicular AB=1unit. The hypotenuse 0B so formed is 3(1/2) in length. So, from center 0 and radius 0B, an arc is cut on number line to mark it.
4(1/2) is equal to 2, which can be marked by constructing another triangle on top of the shown construction or marked directly as 2 is a rational number which can be directly marked on the number line.
Real numbers: The set of numbers consisting of all rational and irrational numbers is known as real numbers and is represented by ‘R’.
Real Numbers
Concept 2: Decimal Expansions of Real Numbers
Decimal expansions can be found by the method of long division.
Example: Find decimal expansion of 100/9 and 99/8.
Solution:
In a similar manner, divisions of other rational numbers can be performed to find that they have two types of decimal expansions:
The remainder becomes zero: such decimal expansions are called terminating.
The remainder never becomes zero: such decimal expansions are called non-terminating recurring or non-terminating repeating. Such expansions are represented by a bar on their top. For example, the non-terminating recurring expansion of 100/9 is represented as 11. .
Rational numbers have the above two types of decimal expansions and the numbers having any of the above decimal expansions are rational.
Conversely, the numbers which do not have the above two forms of decimal expansions are irrational numbers.
Or
The numbers which have a non-terminating non-recurring decimal expansion, are irrational numbers.
Example 1: Show that 11.111….=11. is rational.
Solution: Let 11.=x 10x=111.
Since 111.=11.+100, 10x=x+100
x=100/9
Since 11. can be written in the form of p/q, it is rational.
Example 2: Find an irrational number between 1/8 and 3/8.
Solution: 1/8 = 0.125 and 3/8=0.375. Since an irrational number has a non-terminating non-recurring decimal expansion, find any such number between 0.125 and 0.375.
Infinite irrational numbers exist between any two numbers.
One such number is 0.2052200055202055……
Real Numbers
Concept 3: Operations on real numbers and their representation
To represent rational numbers on the number line, the process of successive magnification is used.
Example: Represent 2.567 on the number line.
Operations on rational numbers are known already. They satisfy the commutative, associative and distributive laws for addition and multiplication. On adding, multiplying, subtracting or dividing two rational numbers, the result is always a rational number.
Irrational numbers also satisfy these laws for addition and multiplication. But on performing mathematical operations on two irrational numbers, the results are not always irrational.
To mark any positive integer x(1/2) on the number line:
Draw a line AB of length x, from zero (B), on the negative side of the number line.
Draw a line BC=1unit, on the right of zero.
Mark the midpoint of AC at O. With OA as radius and O as centre, draw a semicircle.
Make a perpendicular at B, intersecting with the semicircle at D. Join OD.
Triangle OBD is right angled at B.
Since AC=x+1, CO=AO=(x+1)/2 OB=[(x+1)/2]-1 OB=(x-1)/2.
OD=(x+1)/2. [since it’s the radius of the semicircle]
By Pythagoras, BD = x(1/2).
Some identities related to square roots are:
(p*q)(1/2)=p(1/2)*q(1/2)
(p/q)(1/2)= (p(1/2))/(q(1/2))
(p(1/2)+q(1/2))(p(1/2)-q(1/2))=p-q
(p+q(1/2))(p-q(1/2))=p2-q
(p(1/2)+q(1/2))(r(1/2)+s(1/2))=pr(1/2)+ps(1/2)+qr(1/2)+qs(1/2)
(p(1/2)+q(1/2))=p+2(pq)(1/2)+q
Example 1: Evaluate
[5(1/2)]+2(1/2)
(6*5(1/2))*(7*5(1/2))
(3(1/2))*5*2(1/2)
[9*15(1/2)]/5(1/2)
Solution: 1. This cannot be solved any further because the terms do not have any commonality. It remains irrational.
2. 5(1/2)*5(1/2)*6*7 = 5*6*7= 30*7=210. It is rational.
3. The terms in the under-root can be multiplied. 5*6(1/2). This remains irrational.
4. The given expression can be rewritten as: [9*5(1/2)*3(1/2)]/5(1/2) 9*3(1/2). This also remains irrational.
Example 2: Rationalize the denominator of 1/[3(1/2)+1].
Solution: Multiply the numerator and denominator by [3(1/2)-1]
= [3(1/2)-1]/[3(2/2)-12]
= [3(1/2)-1]/2.
Real Numbers
Concept 4: Laws of exponents
There are certain laws by which the changes in powers of real numbers are governed. These laws are called laws of exponents for real numbers. They are:
am . an = am+n
(am)n = amn
am/an = am-n
ambm = (ab)m
a0 = 1
am/n = n√am
Example 1: 5(2/3).5(3/2)
Solution: Using the identities, 5(2/3*3/2)
5(1)=5.
Example 2: 2(1/2)*18(1/2).
Solution: Using identities, (2*18)(1/2)
36(1/2) = 6.
Example 3: 5(1/7)/5(1/3).
Solution: Using identities, 5(1/7-1/3)
5(-4/21).
Statistics
Concept 1: Basics of Data and Graphical Representation
Data can be collected in various ways, but the presentation and storage of data is very important. For example, if a student comes across the data of the population in his city in different years, he would not remember the data if he reads it in an article, instead, it would be more comfortable to look at this information in the form of a graph as it would show the trend of increase or decrease of population properly organized as per the years. Another way would be to tabulate the data.
Following are the runs scored by a batsman in 20 cricket matches.
10 56 89 111 0 65 12 0 45 10 89 76 12 5 49 26 34 2 59 10
Here, the number of times the same score occurs is the frequency of the score. In other words, the number of times a data occurs is its frequency. For example, frequency of 10 is 3.
Here, table 1 represents the ungrouped frequency distribution. Note that it is tabulated in ascending order
for the ease of understanding. The maximum value of runs was 111 and minimum was 0, hence
the range of the data is said to be 111-0 = 111.
Table 2 represents the same data as grouped frequency distribution.
These groupings are called ‘classes’ or ‘class intervals’ and their size is called ‘class size’ or ‘class width’.
In each of these classes, the lower number is called the ‘lower limit’ of the class and the higher number is
called the ‘upper limit’. Example: in the class interval 10 – 20, 10 is the lower limit and 20 is the upper limit.
20 – 10 = 10, so 10 is the class size or class width. NOTE: 10 comes in the group 10-20. Similar for other values.
Graphical representation
Bar Graphs: We know how to make a bar graph from previous classes. It is made for ungrouped data.
Histogram: For grouped data, histogram is made. This is a form of representation like bar graph, but it is used
for continuous class intervals. For example, if table 2 from the above data is used to plot a histogram, it will
look as shown.
For the data from 90 to 120, since the grouping is not uniform, the length of the rectangle in such cases is determined as (frequency/class width*proportionate class interval). Here, frequency and class width belong
to the class interval which is different. Proportionate class interval is the common class interval chosen
according to which the histogram is made, which is 10 in this case. So, length of rectangle = 1/30*10 = 0.34.
This can be done for other class intervals also. For 10-20, 4/10*10 = 4.
In some cases where the class intervals don’t start with zero, a kink can be shown on the x-axis.
Frequency Polygon: This is obtained by joining the midpoints of the rectangles in the histogram as shown in
the above example by a dotted line. A frequency polygon begins at the midpoint of the possible previous rectangle as shown. Similarly, it ends at the midpoint of the possible next rectangle.
Frequency polygon can be made without making a histogram by finding the class mark for every class interval
which will be the x-coordinate of the point and the y-coordinate will be its frequency.
Class mark = (Upper limit + Lower limit) / 2
For example: Class mark of class interval 10-20 in the above example will be (20 + 10)/2 = 15.
Statistics
Concept 2: Measures of Central Tendency
Mean, median and mode are called measures of central tendency because they give a nearly central value of the given data in their own different ways.
We know that mean = sum of observations / total number of observations.
If the frequency of the ith value is represented by fi, and the ith observation is represented as xi, then the mean () is written as:
For example: Consider that in a test 50 students of a class score the marks shown in the table out of 50.
So, mean() = 1430/50 = 28.6.
Median: It is that value of the given number of observations, which divides it into exactly two parts.
So, on arranging the data in an ascending (already done in the above table) or descending order, the median is found as:
– When the number of observations (n) is odd, median is the value of [(n+1)/2]th observation.
– When n is even, median is the mean of (n/2)th and [(n/2)+1]th observation.
For example, find the median of the runs scored by a team in 10 matches shown below:
52 14 15 29 48 26 02 48 75 36.
Solution: Arranging in ascending order: 02 14 15 26 29 36 48 48 52 75
Since n=10 i.e. even, n/2th observation is the 5th observation i.e. 29 and the [n/2 + 1]th observation is the 6th observation i.e. 36. The mean of 29 and 36 is 29+36/2 = 32.5.
Mode: It is the observation which occurs the maximum number of times. For example, in the table shown above, the mode of the data will be 29, because its frequency of occurrence is 9 which is the maximum for the given set of data.
Surface areas and Volumes
Concept 1: Cubes and Cuboids
The surface area of any object is the measure of total area of all the surfaces of the object that exist.
Volume is the measure of the capacity of the object. It is measured as (area of base * height), for any object.
A cuboid is basically a 3-dimensional rectangle. It has 6 rectangular faces and 12 edges.
Surface area of a cuboid is
A = 2(lb + bh + hl)
The volume of a cuboid is (area of base = l*b, height = h)
V = lbh
where ‘l’, ‘b’ and ‘h’ are the edges of the cuboid
A cube is a 3-dimensional square. It has 6 square faces and 12 edges.
Surface area of a cube is
A = 6a2
Volume of a cube is (area of base = a2, height = a)
V = a3
where ‘a’ is the side of cube
Example: An embankment, cuboidal in shape, is to be made of dimensions 50m*50cm*50cm using bricks of
dimensions 5cm*5cm*5cm. Find out the number of bricks required.
Solution: Volume of embankment = (50*100)*50*50 = 12500000 cm3
Volume of one brick = 5*5*5=125 cm3.
No. of bricks required = 12500000/125 = 100000.
Surface areas and Volumes
Concept 2: Cylinders
If the radius of the base of the cylinder is ‘r’ and the height of the cylinder is ‘h’, then
Curved surface area of the cylinder is calculated by opening the curved surface. Assume that the cylinder does not have any base or top and unroll it, to find that it is a rectangle. The length of this rectangle is the circumference of the base circle and breadth is the height of the cylinder. Hence,
Curved surface area (CSA) = 2πr*h
To calculate total surface area, add the area of the base and top circles to CSA.
Total surface area (TSA) = 2 πrh + πr2 + πr2 = 2 πr (r + h)
Similarly, to calculate volume, the area of base is πr2, and the height is ‘h’. So,
Volume = πr2h
NOTE: Take the value of π as 22/7, unless stated otherwise.
Example: A cylindrical container of radius 7m and height 2m, contains cement filled to the brim. Hollow cylindrical rolls are to be made from this cement of outer radius = 15cm, inner radius = 8cm and height=2cm. Find the number of such rolls which can be made.
Solution: Total volume of cement available= volume of its container = π*49*2 = 22/7*49*2= 308m3
Volume of the hollow cylinders = volume of the outer cylinder – volume of the inner cylinder
Volume of hollow cylinders = π*152*2 – π*82*2 = π*2*(152-82) = 22/7*2*161 = 1012 cm3.
Since, the volume of cement is in m3,converting it into cm3= 308 * 106 = 308000000 cm3.
No. of hollow cylinders = volume of cement/volume of hollow cylinders = 308000000/1012 = 304347.82
The number of cylinders must be a whole number. So, on rounding off, 304347 cylinders can be made.
urface areas and Volumes
Concept 3: Cones
Considering the base radius of the cone to be ‘r’, and the height of the cone to be ‘h’ and the slant height is given as ‘l’, then
The curved surface area of the cone is given as
A = ½ * l * 2πr = πrl
The total surface area of the cone is given as
A = πrl + πr2 = πr(l+r)
Volume of the cone is given as
V = 1/3 * π * r2 * h
Example: The slant height of a cone is 29cm and its base radius is 20cm. Find the volume and CSA of the cone.
Solution: Using Pythagoras theorem in a cone, l2 = r2 + h2
292 = 202 + h2
h2 = 441
h = 21cm
Volume = 1/3* πr2h
V = 1/3*22/7*202*21
V = 8800 cm3
CSA = πrl
CSA = 22/7*20*29
CSA = 1822.857 cm2
Surface areas and Volumes
Concept 4: Spheres
A sphere is basically a 3-dimensional circle, as we know it, it’s a ball.
A hemi-sphere is a 3-dimensional semi-circle, or a ball cut in half.
If the radius of a sphere is ‘r’
Surface area = 4πr2
Volume = 4/3*π*r3
If the radius of a hemi-sphere is ‘r’
Curved surface area = 2πr2
Total surface area = 3πr2
Volume = 2/3*π*r3
Example: Find the volume of a sphere of radius 7cm and the hemisphere formed by cutting this sphere in half.
Solution: Using the formula for volume of a sphere,
Volume = 4/3*22/7*73
Volume = 1437.34 cm3
Volume of the hemisphere formed by cutting this sphere in half, will be the half of the volume of this sphere i.e. 718.67cm3.
Triangles
Concept 1: Congruency of triangles
Two triangles are said to be congruent when they are equal in all respects.
All their corresponding angles and the length of the corresponding sides must be equal.
Congruent triangles, if placed over one another, must completely overlap each other.
The sign of congruency is ( ).
Criterion for congruency:
SAS Congruency rule: If two sides and the included angle of two triangles are equal, they are said to be congruent. SAS represents Side-Angle-Side.
ASA Congruency rule: If two angles and the included side of two triangles are equal, the triangles are congruent. ASA stands for Angle-Side-Angle.
AAS Congruency rule: If two angles and a corresponding side of two triangles are equal, they are congruent. AAS stands for Angle-Angle-Side.
The above criterions can be proved by drawing relevant figures and checking the values of other dimensions of the triangles when the above conditions are stated.
An isosceles triangle is a triangle which has two equal sides.
By theorem, if a triangle has two equal sides, the angles opposite to these equal sides are also equal.
Conversely, the sides opposite to equal angles of a triangle, are equal.
Example: In the given figure, PQ=PR and QY=RX. Prove that PX=PY. P
It is given that PQR is an isosceles triangle. So, by its properties, LQ=LR.
Now, in triangles PQY and PRX: PQ=PR Q X Y R
LQ=LR
QY=RX [Note: QY=XR is wrong. The nomenclature is done as per the correspondence of the sides.]
So, by SAS congruency rule, triangle PQY triangle PRX. Hence, PY=PX.
Triangles
Concept 2: Some more criterions of congruency and Inequalities in Triangles
Some more criterions for congruency of triangles:
SSS Congruency Rule: If all the corresponding sides of the two triangles are equal, they are congruent. SSS stands for Side-Side-Side.
RHS Congruency Rule: If the hypotenuse and one more corresponding side of two right triangles are equal, they are congruent. RHS stands for Right Angle-Hypotenuse-Side.
These criterions can also be proved similarly to the other criterions.
Inequalities in Triangles
For two unequal sides of a triangle, the side which is longer subtends a bigger angle opposite to it.
Conversely, the side opposite to the larger angle is longer.
An important property of triangles which is used to determine whether a certain group of lines can be joined to form a triangle is:
The sum of any two sides of a triangle is always greater than the third side.
If this condition is not satisfied, a triangle cannot exist.
Example1: Prove that the altitude on the unequal side in an isosceles triangle, bisects the side.
Solution: Let ABC be an isosceles triangle and AT be the altitude. Now, AB=AC.
In triangles ATB and ATC, LATB=LATC and AT is common.
Hence, ATB and ATC are congruent by RHS Congruency Rule.
So, BT=CT. Hence, AT is the bisector of BC.
