# Study Material XM aths

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```					KENDRIYA VIDYALAYA SANGATHAN- LUCKNOW REGION

STUDY MATERIAL 2009-2010

MATHEMATICS CLASS X

TEAM MEMBERS:

PATRON:

SHRI M.S.CHAUHAN, ASSTT. COMMISSIONER, KVS (LR)

CO-ORDINATOR:

SHRI I.U Haque, PRINCIPAL, KV Aliganj , Lucknow

RESOURCE PERSONS:
1. MR.V.K.SABHARWAL          TGT(MATHS)       KV    CANTT
KANPUR
2. MRS S.K. SHARMA TGT(MATHS)KV NO.2 AFS
CHAKERI KANPUR
3. MR. R.K.KATIYAR TGT(MATHS) KV IIT KANPUR
4. MR.NAND       LALYADAV       TGT(MATHS)       KV    NO.1
ARMAPUR KANPUR
5. MR.A.K.DWIVEDI TGT(MATHS) KV NO.2 ARMAPUR
KANPUR
6. Mr. P. K. Shrivastava T.G.T.(Maths) K. V. 01 Jhansi
7. Mr. D. S. Sharma T.G.T.(Maths) K.V. M.R. N. Mathura
Lesson             1
REAL NUMBER
A Euclid’s division lemma- given positive integers a and b there exists a unique pair of inters
q and r such that≠≠≠≠
a=bq+r , 0≤r<b

Euclid’s division algorithm- according to this algorithm the H.C.F of two positive integer a and
b, with a >b can be obtained as follows.

Step1. Apply the Euclid’s division lemma to a and b to find the whole numbers q and r
such that a=bq+r, 0≤r<b

Step2. If r=0 ,the H.C.F of a and b is b If r ≠0, apply Euclid’s lemma on b and r again .

Step3. we continue the division process till the remainder becomes zero. the divisor at the last
stage will be H.C.F of a and b.

Questions-

1 Using Euclid’s division algorithm find H.C.F of following pairs-
(A) 867and 225 (B) 12576 and 4052 (C) 225 and 135
[Ans A. 3         B. 4              C. 45

Q2. show that one and only one out of n, n +4 ,n +8 ,n +12, n +16 is divisible by 5 where n is
any positive integer

[H int Consider n= 5q, 5q+1, 5q+2, 5q+3 or 5q+4]

Q3. Show that any positive odd integer is of the 4q+1 or 4q+3 where q is some integer.

Concept-
Fundamental theorem of arithmetic -every composite number can be expressed as a product of
primes and this factorization is unique apart from the order in which the prime factor occur.

Question-
1. find the prime factorization of given numbers .

(A) 1125 (B) 840 (C) 16755
Answer (A) 32 x 53 (B) 23 .3.5.7 (C) 32 .1117
Concept- with the help of prime factorization

A. working rule for determining H.C.F-

(I) factories the number in to the product of primes expressed in exponential form.

(II) Select the lowest of the powers of common prime.

(II) The product of primes with lowest powers is HCF

B. working rule for determine L.C.M-
(I) factories into product of primes expressed in exponential form.
(II) Select the highest power of a prime present in all or some of the numbers.

(III) Product of primes with highest power is L.C.M.
Question -
1. Find GCD or HCF and LCM of following using prime factorization method.

(A)252 and 1650    (B)225, 336, 360
[Ans. A.LCM= 69300,HCF=6     B. LCM=25200, HCF=3]

2. Check whether 8n can end with the digit zero(0) for any natural number n .

Concept- for any two positive integer a and b.

HCF (a, b) X LCM (a, b)=a X b.

Question-

Q1. HCF(867,255)=51 find LCM(867,255)
[LCM= 4335]

Q2. Find LCM and HCF of 306 and 657 also verify that -LCM X HCF=product of        numbers
.                                             Ans. LCM=22338
HCF= 9
Q3. If LCM and HCF of the two numbers are 180 and 12 If the one number is 36 then
find the other number .
[Ans.60]

Concept –

(A) Irrational Numbers- an irrational number is non terminating and recurring decimal and
can not be put in the form P/Q where P and Q are integers and Q≠0

(B)Let X=P/Q be a rational number, such that the prime factorization of Q is of the form
where 2n 5m ,where m, n are non negative integer then X has a terminating decimal expansion
example 7/20, 11/8, 51/50 etc-

(C) Let X=P/Q is a rational number, such that prime factorization of Q is not of the form of
where m and n are non negative integers. then X has a non terminating repeating expansion
example -2/9, 29/343

Question-
Q1.Prove that 5+√2     irrational number

Q2. check whether(7+√5) (7-√5) irrational.
Q3.Prove that sum of two irrational numbers may be a rational number .
Q4. with out actual division state whether the following rational number will have a
terminating decimal expansion or non terminating repeating decimal expansion

(A) 311/312     (B) 23/2352   (C) 77/210    (D) 64/455

[A.non terminating B. terminating C.non terminating       D.non terminating]

Q5. Show that √2+√5 is irrational number.
Q6. Prove that 3- 2√7 is an irrational number.
Chapter-2 (Polynomial)

1. Look at the graph of y=p(x), p(x) is polynomial for each of the graph
find number of zero.

Hint: check of intersection on X Axis.

Ans. (1)1       (2) 2 (3) 3 (4) 1 (5) 1 (6) 4

2. Find the zeroes of quadratic polynomial x 2 +7x +10
Ans = -2, -5
3. Find the polynomial, the Sum and product of its zeroes are -3 and 2
respectively.
Ans = x 2 + 3x+2
4. If I is a zero of polynomial 3t3 – t2 -3t +1. Find the other zeroes.

Ans = -1, 1/3

5. If (x+1) is a factor of x2 -3ax +3a-7, find the value of a
Ans . a=1

6. Find the zeroes of the polynomials x 2 -3 and verify the relation
between zeroes and coefficients.              Ans. √3,-√3

7. Divide 3x 2 –x 3 -3x+5 by x-1-x 2 and verify the division algorithm.
Hint: Divided = Division X Quotient + Remainder

8. Find all the zeroes of 2x 4 -3x 3 - 3x 2 +6x-2. If two of its zeroes are √2
and -√2
Ans . √2, -√2, ½ and 1

9. On dividing x 3-3x 2+x+2 by a polynomial g(x) the quotient and
remainder are x-2 and -2x+4 respectively . find g(x)

Ans g(x) = x 2 –x+1

10.Form a quadratic polynomial whose zeroes are(2+√5) and (2-√5).
Ans. X2-4X-1

SUMMARY

1. Quadratic polynomial ax2+bx+c has two zeroes. If α and β are zeros
then
α+β = -b/a and αβ =c/a

2. If α and β are zeroes of poly then it can be written as

(x - α) (x - β)

OR x2-(α + β) x+ αβ
CHAPTER 3
PAIR OF LINEAR EQUATIONS
IN TWO
VARIABLES
A- To find the value of K when linear equations in two variables having
unique solution or infinitely many solutions or no solution.
Q. Find the value of K for which the following pair of linear
equations have infinitely many solutions –

a) 2x – 3y = 7                      (i)
Ans – K = 5
(K + 1) x + (1-2 K) y = 5K – 4    (ii)

b) 4 x + K y = 8                    (i)
2x – 5 y = 4                     (ii)           Ans – K = -10

c) K x + 3 y = k – 3                  (i)
12 x + K y = K                    (ii)            Ans – K = 6

B-Solving a pair of linear equations by
Elimination Method
Solve the following system of equation by elimination method

4
1-      3 y 14                            (i)               Ans. - x = 1
X                                                                   5

3                                                               Y = -2
 4 y  23                           (ii)
x

6
2-   x     6                              (i)               Ans. x = 3
y
y=2
8
3x   5                               (ii)
y
X=1
3- a x + b y = a – b                         (i)              Ans.-
y = -1
bx–ay=a+b                               (ii)

C-                  Solve the following system of equations
by cross multiplication method

2x y
1-        2                                (i)
a b                                                              X=2a
Ans.-
x y                                                               y = -2b
 4                                  (ii)
a b

X=a
2- x + y = a – b                             (i)              Ans.-
y = -b
a x  by  a  b
2       2
(ii)

3- 2 x + y = 35                                 (i)                    X = 15
Ans.-
y=5
3 x + 4 y = 65                           (ii)

D-                                  Graphical Method

1-          Solve graphically the following system of Equations.

i) x + y = 3                                  (i)                 X=2
Ans.-
3x–2y=4                                   (ii)                y=1

ii) x + 2 y = 5                               (i)                 x=1
Ans.-
y=2
2x–3y=-4                                   (ii)
2-        Solve the system of the equations 4 x – y = 4 and 3 x + 2 y = 14
graphically and Shade the region between the lines and y axis.
X=2
Ans.-
y=4

Word Problems
1
1. A fraction becomes      when 1 is subtracted from the numerator and 2
2
1
is added to the denominator. It becomes when 7 is subtracted from
3
the numerator and 2 is subtracted from the denominator. Find the
fraction.

Ans. - 15
26

2. A fraction is such that if the numerator is multiplied by 3 and the
18
denominator is reduced by 3, we get     . But if the numerator is
11
2
increased by 8 and the denominator is doubled we get . Find the
5
fraction.
Ans. - 12
25

3. A man travels 600 km. partly by train and partly by car. If he covers
400 km by train and the rest by car, it takes him 6 hours 30 minutes.
But if he travels 200 km. by train and the rest by car, he takes half an
hour longer. Find the speed of train and that of car.

Ans. - Speed of train = 100 km. / h

Speed of car = 80 km. /h

4. A boat can go 20 km. upstream and 30 km. downs stream in 3 hours.
2
It can go 20 km. downstream and 10 km. upstream in 1 hrs. Find the
3
speed of the boat in still water and speed of the stream.

Ans. - 20 km. /h, 10 km. /h
5. A and B are two points 150 km apart on a highway. A car starts from
A and another from B at the same time. If they move in the same
direction, they meet in 15 hours, but if they move in opposite
directions, they meet in one hour. Find their speeds.

6. The sum of the digits of a two-digit number is 9. Also,9 times this
number is twice the number obtained by reversing the order of the
digits. Find the number.
Ans. 18
7. The area of a rectangle gets reduced by 9 sq.units, if its length is
reduced by 5 units and breath is increased by 3 units. If we increase
the length by 3 units and breath by 2 units, the area increased by 67
sq. units. Find the dimensions of the rectangle.
Ans. 17units, 9units
8. A part of monthly hostel charges is fixed and the remaining depends
on the number of days one has taken food in the mess.
When a student A takes food for 20 days she has to pay Rs.1000
as hostel charges,whereas a student ,who takes food for 26 days,
is Rs.1180 as hostel charges. Find the fixed charges and the cost of
food per day.
Ans. Rs. 400, Rs. 30
9. Five year hence the age of Ramesh will be three times that of
this daughter. Five years ago,the age of Ramesh was seven times
that of his daughter. What are their present ages.
Ans. 40years, 10years
CHAPTER 4

Solution of quadratic equation by –

A) Factorization Method –

Solve the following quadratic by factorization Method –

2 1
i) 6x 2 – x – 2 = 0                                Ans. -   x       ,
3 2

ii) 5x2 – 3 x – 2 = 0                              Ans.- x    2
5    ,1

x 1 x  3    1                                                   5
iii)             3                               Ans.- x = 5,
x2 x4       3                                                   2

1   1 1 1
iv)                                      Ans. - x = - a, - b
abc a b x

B)          By Completing Square Method
Solve the following equations by the Method of completing the square.
1
i) 2x2 – 3x + 1 = 0                             Ans. x = 1,
2

3  19
ii) 5x2 – 6x – 2 = 0                            Ans. x 
5

iii) x2 + 2 bx – (a2 – b2) = 0                 Ans. x = - b  a

Solve the following equations by using quadratic formula

 3  3
i) 4 x 2  4 3x  3  0                         Ans. x       ,
2   2

ii) y2 – 6y + 2 = 0                             Ans. y  3  7

a b
iii) 9x2 - 3 (a + b) x + a b = 0                Ans. x  ,
3 3

1     2   6                                           4
iv)                                           Ans.x = 3,
x  2 x 1 x                                            3

D)      Find the value of K for which the equations have
real and equal roots
4
i) 4x2 – 3 Kx + 1 = 0                             Ans. K  
3

ii) x2 – 2 Kx + 7K – 12 = 0                       Ans. K = 4, K = 3

iii) Kx2 + Kx + 1 = -4x2 – x                      Ans. K = 5, -3

Word Problems
1. The sum of the two numbers is 15 and the sum of their reciprocals
3
is      . Find the numbers.
10
Ans. 5, 10

2. Two pipes together can fill a reservoir in 12 hours. If one pipe can fill
the reservoir 10 hours faster than the other, how many hours will the
second pipe take to fill the reservoir?
Ans. 30 hours

3. In a flight of 2800 km, an aircraft was slowed down due to bad
weather. Its average speed of the trip was reduced by 100 Km/hr. and
time increased by 30 minutes. Find the original duration of flight.

1
Ans. 3 hours
2

4. Rs. 9000 were divided equally among a certain number of persons.
Had there been 20 more persons each would have got Rs. 160 less.
Find the original number of persons.

Ans. 25

5. A fisherman can row a boat 8 km down stream and return in 100
Minutes. If the speed of the stream is 2 km/hour. Find the speed of the
boat in still water.
Ans. 10 km/hour
6. By selling a pen for Rs. 24 a man earns the profit percent as much as
the cost price of the pen.Find the cost price of the pen.
Ans. Rs. 20

7. Seven years ago Varun’s age was five times the square of Swati’s age.
Three years hence Swati’s age will be two-fifth of Varun’s age. Find
their present ages.

Ans. 27, 9

8. B takes 16 days less than A to do a piece of work. If both working
together can do it in 15 days, in how many days will B alone complete
the work?

Ans. 24 days
CHAPTER 5
Arithmetic Progressions

A) an = a + (n – 1) d

(i)     Find the common difference of an A.P if nth term an = 4-2n

Ans. d = -2

(ii)    Find the A.P. whose third term is 16 and the difference of the
9th term from 11th term is 12.

Ans. 4, 10, 16……

(iii)   Write the next term of the A.P 8, 18, 32,........

Ans. 50

(iv)    For what value of K, are the numbers x, 2x + K and 3x + 6
three consecutive terms of an A.P.?

Ans. 3

(v)     Which term of the arithmetic progression 3, 10, 17, ………
will be 84 more than its 13th term?

Ans. 25th term

(vi)    Find the 10th term from end of the A.P. –
4, 9, 14, ……….., 254.

Ans. 209

(vii) The 8th term of an arithmetic progression is Zero. Prove that
its 38 th term is triple of its 18th term.

(viii) The 51st, 11th and the last terms of an A.P. are 0, 8 and
1
     respectively. Find the common difference and the number
5
of terms.
Ans. -1/5, 52
B)
Sn 
n
2a  n  1d 
2

Sn 
n
a  l 
2

1. Find the sum of the first 25 terms of an A.P. whose nth term is
given by tn = 2 – 3n.

Ans. -925

2. If the sum of the first n terms of an A.P. is given by Sn = 3n 2 +2n .
Find the n th term of the A.P.

Ans. 6n – 1

3. How many terms of the A.P. 3, 5, 7,………… must be taken so that
the sum is 120 ?

Ans. 10

4. Find the sum of all two digit natural numbers which are divisible
by 4.

Ans. 1188

5. The 5 th and 15th terms of an A.P. are 13 and – 17 respectively. Find
the sum of first 21 terms of the A.P.

Ans. -105

6. Find the sum of all 3 digit numbers which when divided by 16 leave
remainder 7.

Ans. 31407
7. If the pth terms of an A.P. is q and the qth term is p, then find
the the sum of (p + q)th term .
Ans. (p+q)/2 {p+q-1}

8. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of its
6th and 10th terms is 44. Find the first 3 terms of A.P.

Ans. -13, -8, -3

9. If m times of mth term of an AP is equal to n times of its nth term ,find
its (m+n)th term .

Ans. a(m+n)=0
Lesson 6
SIMILAR TRIANGLES
BASIC CONCEPTS
Two figures having the same shape but not necessarily the same size are called similar figures.
1. Two polygons of the same number of sides are similar if:

(a) Their corresponding angles are equal

(b) Their corresponding sides are in the same ratio.

Similar Triangles
Two triangles are said to be similar if their corresponding angles are equal and their
corresponding sides are proportional.
Basic Proportionality Theorem
(Proof can be asked in the exam)
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct
points the other two sides are divided in the same ratio.

I.                                                                        A
DB EC

II.                                                              D             E
AB AC

AB AC
III.                                                          B        B            C
DB EC

Problems related to Basic Proportionality Theorem

1. In a triangle ABC, DE II BC. If AD    (4x – 3 )cm, DB        (3x – 1 )cm, AE     (8x – 7 )cm and EC
(5x – 3 )cm. Find the value of x.                          (Ans x = 1)

2. In a triangle ABC, DE II BC and            . If AC    4.8cm, find AE.

(Ans AE=1.8 cm)

3. Prove that the diagonals of a trapezium intersect each other in the same ratio.

4. In a triangle ABC, DE II BC. If AD     x cm, DB      (x – 2 )cm, AE       (x + 2 )cm and

EC     (x – 1 )cm. Find the value of x.                      (Ans x=4)

Criterion for similarity.
1. AAA Similarity:

2. ∆ABC       ∆DEF      when    A = D,      B=    E,        C=       F

3. AA Similarity

∆ABC      ∆DEF      when    A=     D,    B=        E,       C=       F

4. SAS Similarity

∆ABC      ∆DEF     when             and      A=        D

5. SSS similarity

∆ABC      ∆DEF      when

Problems :
1. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ
and QR and the median PM of triangle PQR. Show that ∆ABC          PQR

2. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show
that ∆ABE       ∆CFB

3. If AD and PM are the medians of ∆ABC and ∆ PQR respectively where ∆ABC                 ∆PQR . Prove
that

4. D is a point on the side BC of a triangle ABC such that           ADC = BAC. Show that CA2 = CB× CD

5. In a triangle ABC , AB = AC and D is a point on side AC such that BC2 = AC × CD. Prove that

BD = BC

Theorem:
The ratio of the areas of two similar triangles is equal to the ratio of the squares of their
corresponding sides. ( Can be asked in the examination)
A                                      P

B                         C              Q                             R

If ∆ABC        ∆PQR then

=        =
a) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their
corresponding altitudes.
b) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their
corresponding medians.

1. If ∆ABC       ∆DEF, BC = 3cm, EF = 4cm and the area of ∆ABC = 54cm2. Find the area of
∆DEF.
(Ans = 96 square cm)
2. ∆ABC       ∆DEF. If AB = 2DE and area of ∆ABC = 56cm2. Find the area of ∆DEF.
( Ans = 14 cm2 )
3. The area of two similar triangles are 100cm2 and 40cm2 respectively. If the altitude of the
bigger triangle is 5cm, find the corresponding altitude of the other triangle.
( Ans = 3.16cm)
4. ∆ABC is right angled at A and AD is perpendicular to BC. If BC = 13cm and AC =
5cm. Find the ratio of the areas of ∆ABC and ∆ADC.
( Ans = 169:25)
5. In ∆ABC, D is a point on AB and E is a point on BC such that DE II AC and the area of

∆DBE =     area of ∆ABC. Find         .

( Ans =        )
6. Two right triangles ABC and DBC are drawn on the same base BC and on the same
side of BC. If AC and BD intersect at P . Prove that AP × PC = BP × PD
7. D, E and F are the mid points of the sides AB, BC and CA of a triangle ABC. Find the
ratio of the areas of the triangles DEF and ABC.                          (Ans 1:4)
8. ABC is an isosceles triangle right angled at B. Two equilaterals triangles are
constructed with the side BC and AC. Prove that ar ∆ BCD = ½ ar ∆ ACE

Pythagoras Theorem:
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other
two sides.         A

B                 C
In a right ∆ABC right angled at B
AC2 = AB2 + BC2
Converse of Pythagoras Theorem (Can be asked in the examination) :               In a triangle , If
the square of one side is equal to the sum of the squares of the other two sides, then the angle
opposite to the first side is a right angle.
Problems related to Pythagoras Theorem:
1. ABC is a right angle triangle , right angle at B. If D and E are any points on AB and BC
respectively. Prove that AE 2 + CD2 = AC2 + DE2

2. Prove that the area of the equilateral triangle described on the side of a square is half the area
of the equilateral triangle described on its diagonal.

3. In an equilateral triangle ABC , the side BC is trisected at D. Prove that 9AD2 = 7 AB2

4. PQR is a right angle triangle right angled at Q. If S is any point on QR such that QS = SR .Show
that PR2 = 4 PS2 - 3PQ2

5. ABC is a right triangle right angled at B. The side BC is trisected at the points D and E. Prove
that 8AE2 = 3AC2 + 5AD2

6. BL and CM are the medians of ∆ ABC right angled at A . Prove that 4 ( BL2 + CM2 ) =
5BC2
7. Prove that the sum of the squares of the side of a rhombus is equal to the sum of the squares
of its diagonals.

8.     B C is a right triangle right angled at C. If p is the length of the perpendicular from C to AB
and a,b and c are the lengths of the sides BC, CA and AB. Prove that

1     1  1
(i)     pc = ab         (ii)         2
 2 2
p    a  b

A
D
b             c
C                     B
a
LESSON 7
CO-ORDINATE GEOMETRY

1) Distance between two point p(x1,y1) and Q(x2,y2)
is given by.

2)The coordinates of the mid point of the line segment joining the
point A(X1,Y2) and B(x1,y2)is given by .
x = x1+x2
2
y = y1+y2
2

Problems
1) prove that the points(-3,0),(1,-3) and (4,1) are the vertices of
an in an isosceles right- angled triangle .
2) find a point an X-axis which is equidistance from A (5,4)
and B(-2,3).
Ans.[2,0]

3) find the values of K for which the distance between (9,2) and
(3,k) is 10 units.
Ans.[-6,10]
4) using distance formula ,show that the points (1,2),(2,0)and (3,-
2)are collinear.
5) Show that the point A (1,0), B(5,3) ,C(2,7) & D(-2,4) are the
vertices of a parallelogram.

6) Find the lengths of the medians of     ABC whose vertices are
A(7,-3) B(5,3) & C(3,-1)

7)Two vertices of a triangle are (1,2)and (3,5).If the centroid of
the triangle is at the origin.Find the co-oridnates of the third
vertex.
Ans. (-4,-7)

8) Let the opposite angular point of a square be (2,0)and
(5,1).find the coordinates of the remaining angular points.

Section formula

the coordinates of the point A(x1,y1) which divides the lines
segment joining the point a (x,y) and B(x2,y2) internally in the
ratio m1:m2 are given by

Figure 1

Problems

1) Find the ratio in which the line segment joining the points
(6,4) and (1,-7) is divided by x-axis [(1,7)]
2) if A and B are two points having coordinates (1,4 )and
(5,2).respectively
,find the coordinates of p such that AP/PB=3/4.

3) If A(-2,-1), B(a,0), C(4,b) and D(1,2) are the vertices of a
parallelogram,
find the values of a and b
.[a=1,b=3]
4)Determine the ratio in which the line 2x+y-4=0.divides the
line segment
joining the points A(2,-2) and B(3,7).
[2:9]
5) Find the coordinate of point which trisect the line segment
joining (1,-2) and (-3,4)[(-1/3,0) and (-5/3,2)]

6) Prove that the coordinates of the centroid of the triangle
whose vertices are
(x1,y1),(x2,y2) and (x3,y3) are

7) In what ratio does the y-axis divide the line segment joining
the point
P(-4,5) and Q (3,-7) ? Also find the coordinate of the point of
intersection
[ 4:3,(0,-13/7)].

AREA OF A TRIANGLE

The area of a triangle, the coordinate of whose vertices are (X1,Y1)
(X2,Y2) and (X3,Y3) is
½ |X1 (Y2-Y3)+X2 (Y3-Y1)+X3 (Y1-Y2)|

PROBLEMS

1) Find the value of K . if the point (k,3),(6,-2) and (-3,4) are
collinear.
[k=-3/2]
2) Find the area of the quadrilateral, the coordinate of whose
vertices are
(-3,2),(5,4),(7,-6) and (-5,-4)
[ 80 sq.units].
3) The vertices of triangle ABC are (-2,1),(5,4) and (2,-3)
respectively. Find the area of the triangle and the length of the attitude
through A.

[ 20 sq.unit,           ]
LESSON-8
INTRODUCATION                         TO       TRIGONOMETRY

Basic concept-

Trigonometry is the study relationships between the sides and angles of a triangle.

A. Trigonometry ratios- Ratio of the sides of right triangle are called trigonometric ratio.

C
B

B
A

Side opposite to ∟A
SinA= ----------------------------------- =BC/AC
Hypotenuse

CosA=------------------------------------- =AB/AC
Hypotenuse

Side opposite to ∟A
TanA=------------------------------------- =BC/AB
hypotenuse
CosecA=----------------------------- =AC/BC
Side opposite to ∟A

Hypotenuse
SecA= -----------------------------= AC/AB

cot A   = ----------------------- = AB /BC
Side opposite to ∟A

Also

sinA=1/cosecA,       cosA=1/ secA,       tanA=1/ cotA

cosecA=1/ sinA,     secA=1/ cosA        cotA=1/ tanA

tanA= sinA/ cosA               cotA=cosA/sinA
Question-

1.∆ABC has right angle at A.In each of the following find SinB, CosC,TanC,TanB

(A) AB=AC=1cm
(B) AB=5cm,BC=13cm.

[Ans. A. sinB=1/√2 ,cos C=1/√2, tanB=1,Tan C=1]
[Ans.B. sinB=12/13,cosC=12/13,tanB=12/5 ,tanC=5/12]

Q2.If CosӨ=3/5, Evalute-SinӨ-CotӨ/2TanӨ
[Ans. 3/160]

Q3.If ∟B and ∟Q are acute- angles such that- SinB=SinQ then prove that ∟B=∟Q .

Q4. If TanӨ=4/3 find the value of.

3SineӨ+2CosӨ/3SineӨ-2CosӨ
[Hint divide numerator and denominator by CosӨ then calculate]
[Ans. 3 ]

Concept- B

trigonometric ratios of some specific angle-

A              0                30            450         600          900
SinA             0                 ½            1/√2        √3/2          1
CosA             1                √3/2          1/√2         ½            0
TanA             0                1/√3           1            √3     Not defined
CosecA      Not defined            2             √2          2/√3         1
SecA             1                2/√3           √2           2      Not defined
CotA        Not defined            √3            1           1/√3         0

Question-

Q1 . 4(sin2300 +cos2600) -3(cos2450- tan2450 ) = 7/2

Q.2 . tan 450/cosec 300 +sec 450/cot 450 -5 sin 900/2 cos 00           Ans √2 -2

Concept-C
Trigonometric ratios of complimentary angles-

In ∆ABC right angle at B.

∟A+∟C=900

∟C =900 -∟A
Q3 .Using formula Tan2A=2TanA/1-Tan2A                 find Tan600 Given that Tan 300=1/√3
(Tan600 = √3)
Q4.   prove that-

4(Sin2 300+Cos2600)-3(Cos2 450-Tan2 450)=7/2
Q5.   Ten450 /Cosec300 +Sec450 /Cot450 -5Sine900 /2Cos00

Sin(90-A)=CosA                       Cos(90-A) = SinA

Tan(90-A)=CotA                        Cot(90-A)= TanA

Sec(90-A)=CosecA                      Cosec(90-A)= SecA

For all values of A lying between 00 and 90 0

Question-Evaluate with out using table

1.Evatuate- [(Cos 2 56 0+ Cos 2 34 0)/ (Sin 2 56 0 + Sin 2 34 0) ]+3 Tan 2 56 0 . Tan 2 34 0
Ans. [4]

Q2. Prove that, Tan9 0 Tan27 0 Tan45 0 Tan63 0 Tan81 0 =1

Q3. Evaluate without using trigonometric table-

Sec 2 Ө-Cot 2 (90- Ө)
-------------------------------- + Sin2 40   + Sin250                            Ans. [2]
Cosec2 67- tan223

Q4.   [Cot(90- Ө).Sin(90- Ө)/ Sin Ө]- cot40 0 / Tan50 0                            Ans [0]

(D) CONCEPT-In A right angled triangle. Right angled at B
1. Evaluate- [Cos 2 56 0+ Cos 2 34 0/ Sin 2 56 0 + Sin 2 34 0 ]+3 Tan 2 56 0 . Tan 12 34 0 [4]

(A)   Sin 2 A+Cos 2 A=1
Sin 2 A=1- Cos 2 A
Cos 2 A=1- Sin 2 A

(B)    Sec 2A-tan 2 A=1
Sec 2A=1+ tan 2 A
tan 2 A= sec 2A-1

(C)    cosec 2A-cot 2 A=1
Cosec 2A=1+ Cot 2 A
Cot 2 A= Cosec 2A-1

Question-
Q1. prove that-
tan 2A-tan2 B=sin 2A-sin 2 B/cos 2A.cos 2B

Q2.prove that
[ cosA/(1- tanA)]- [sin 2 A /(cosA-sinA)]=sinA+cosA .

Q3.prove that
sin Ө (1+tan Ө)+cos Ө (1+cot Ө)=sec Ө +cosec Ө
Lessson-9

Application                   of Trigonometry
Concept-height and distance-

Angle   of depression

Point Of Observation                                    Object
Line of sight

Angle of elevata

Point of Observation                          Object

Q1. From a point on the ground the angles of elevation of the bottom and top of the water
tank kept at the top of a 20m high tower are 450 and 600 find the height of water tank.
[Ans 20(√3- 1) ]

Q2. A man on the top of a vertical tower observes a car moving at uniform speed coming
directly to wards it. If it takes 12minutes for the angle of depression o to change from 300 to
450 .how soon after this, will the car reach the tower.
[Ans 16minute 23 second]

Q3. The horizontal distance between two towers is 140m the angle of elevation of the top of
the first tower when seen from the second towered it is 300. If the height of second tower is
60m. find the height of the first tower. ( 140.73)
[

Q4. From a window 60m high above the ground of a house in a street the angles of elevation
and depression of the top and foot of another house on the opposite side of street are 600 and
450 respectively. Show that the height of the opposite house is 60
(1+√3) m.

Q5. The shadow of a flag of a staff is three-times as long as the shadow of the flag staff when
the sun rays meet the ground at an angle of 600 finds the angle between the sun rays and the
ground at the time of longer shadow .
[Ans 30 degree ]
Q6.A man sitting at a height of 20m on a tall tree on a small Island in the middle of a river.
from the top of tree he observes the angle of depression of foot of the two poles standing on the
two banks of river and in line with the foot of tree are 600 and 300 respectively find the width
of river.
[Ans      meter]

Q7. The angle of elevation of a cloud from a point 60m above a lake is 300 and angle of
Depression of the reflection of cloud in the lake is 600 find height of cloud.
[Ans 120 meter]

LESSON 10

CIRCLES
Circle: A closed plane figure consisting of all those points of the plane which are at a
constant distance from a fixed point in that plane .
The fixed point is called its centre and the constant distance is called its radius.

O is the centre and OA is the radius of the circle.
Intersection of a line and a circle.
The following three cases arise:
1. A line may not intersect a circle.
2. A line may intersect a circle in two points.

3. A line may intersect a circle at one point.

A line which intersects a circle in two points is called a Secant.
A line which intersects a circle at one point is called a tangent to the circle.
Thus a line which intersects a circle in one and only one point is called tangent to the circle and
the common point is called point of contact.

The tangent at any point on a circle is perpendicular to the radius through the point of contact.

Theorem: The lengths of the two tangents drawn from an external point to a circle are
equal.

1.     We can not draw any tangent to a circle from any point in the interior of the circle.

2. From a point on the circle , a unique tangent can be drawn.

3. From a point in the exterior of the circle we can draw two tangents to a circle and these
tangents are equal in lengths.

Problems:
1.    A point P is 15 cm from the centre of a circle. The radius of the circle is 5 cm. Find the
length of the tangent drawn to the circle from the point P.      (Ans. 10         )

2. A point P is 5 cm from the centre of the circle. The length of the tangent from the point P
to the circle is 4cm . Find the radius of the circle.             ( Ans. 3 cm )
3. A circle is touching the side BC of a triangle ABC at P and is touching AB and AC when
produced at Q and R respectively. Prove that AQ = ½ (Perimeter Of ∆ABC )

4. Two concentric circles are of radii 10 cm and 8 cm . Find the length of the chord of the
larger circle which touches the smaller circle.                ( Ans . 12 cm )

5. ABC is a right angled triangle right angle at B. A circle is inscribed in the triangle. If AB = 8
cm and BC = 6 cm , find the radius of the circle.                          ( Ans. 2 cm )

6. The radius of the incircle of a triangle is 2 cm. and the segments into which one side is
divided by the point of contact are 3cm, and 4cm. Determine the other two sides of the
triangle.                                                   (Ans.7.5cm. , 6.5cm.)

7. ABCD is a quadrilateral in which angle A is right angle, AB = 50cm. and BC = 40cm. A circle
is inscribed in the quadrilateral and it touches the sides AB, BC, CD and DA at P,Q,R and S
respectively. If CR = 18cm. find the radius of the circle.              (Ans. 28cm. )

8. In the figure, l and m are two parallel tangents to the circle with centre ‘F’ at A and B. The
tangent at C makes an intercept DE between l and m. Prove that DFE = 90°

D
A
l
B

F
f                     C
B

m
E
B                          B
B

9. In the figure , XP and XQ are the tangents from X to the circle with centre O. R is a point on
the circle. Prove that

XA + AR = XB + BR
P       A

X
R

B
Q

10. In the figure AB is a chord of length 8cm. of a circle of radius 5cm. The tangents to the
circle at A and B intersect at P. Find the length of AP.

(Ans. 6.67cm.)

A

P

O

B

Expected Questions:-

1) Two tangents TP and TQ are drawn to a circle with centre ‘O’ from an external point
‘T’. Prove that  PTQ = 2  OPQ
P

T

O

Q
Lesson 11
Constructions
1. To construct a triangle similar to a given triangle as per the given scale factors.

Scale factor means the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle. This construction involves two different
situations.

(a) The triangle to be constructed is smaller than the given triangle, here scale factor is less
than 1.

(b) The triangle to be constructed is bigger than the given triangle, here the scale factor is
greater than 1.

1. Construct a ABC with BC=6 cm, CA=4cm, angle BCA=60 .Construct another triangle similar
to ABC such that the ratio of the corresponding sides is 3/5 of the corresponding sides of
ABC .

2. Construct a triangle PQR in which PQ=6 cm, QR=5 cm and angle Q=45 .Construct another
triangle similar to the PQR whose sides are 7/5 of the corresponding sides of PQR.

3. Construct a triangle in which AB=6.5 cm, angle B=60 and BC=5.5 cm. Also construct a triangle
AB’C’ similar to ABC whose each side is 1.5 times the corresponding sides of the ABC.

4. Construct a triangle ABC with base BC=4.2cm, angle A=45 and altitude through A is
2.5cm.Construct another triangle similar to this triangle with scale factor 1/2.

Construction of tangent to a circle from a point outside the circle.
(i)To construct a tangent to a circle from a point outside the circle when the centre of the
circle,

is (1)known (2) not known.

1. Draw a circle of radius 3 cm. From a point P outside the circle at a distance of 5 cm from the
centre of the circle, draw two tangents to the circle. Measure the length of these tangents.

2. Draw a circle of diameter 6 cm from a point 8 cm away from the centre of the circle, draw two
tangents to the circles. Measure their lengths.

3. Construct a pair of tangents to a circle of radius 6 cm which are inclined to each other at an
angle of 60 .

4. Given a circle C whose centre is not known from the point P in the exterior of C. Draw two
tangents to the circle.

5. Let ABC be a right triangle in which AB = 6cm.,BC = 8cm. and B = 90°. BD is the perpendicular
from B on AC. The circle through B , C and D is drawn. Construct the tangents from A to this
circle.
6. Draw a circle of radius 3cm. Take two points P and Q on one of its extended diameter each at
a distance of 7cm. from its centre. Draw tangents to the circle from these two points P and Q.

7. Construct a circle of radius 3.5cm. From a point on the concentric circle of radius 6.5cm.Draw
a tangent to the first circle and measure the length of the tangent drawn.

8. Draw a line segment PQ = 9cm. Taking P as centre ,draw a circle of radius 5cm and taking Q as
centre , draw another circle of radius 3cm. Construct the tangents to each circle from the
centre of the other circle.
Chapter -12 Mansuration Pt-I
Chapter (Area Related to the Circles)
Basic Concepts with Examples

1. Circle

Fig.

Perimeter: =        2πr

Area:        =      πr2

2. Semi Circle

Fig.

Perimeter:          πr + 2r

Area:                     ½ πr2

3. Ring

Fig.

Area:     =               π(R2-r2)

4. Sector of Circles

Fig.
Perimeter: =                 l +2r
 r
Length of an arc l 
180

 r 2
Area of sector : =
360

Nomenclature:                 - Angle of Sector
l - Length of Arc

5. Segments of Circles

Fig.

 r                  
Perimeter: =                  2r sin
180                  2

r 2         1 2
Area:          =         
     r Sin
360           2
Questions
1. If the circumference of a circle is 22 cm, Find the Area (π = 22/7)

Ans. 385 cm2

2. A wire is looped in the form of a circle of radius 28 cm. It is converted into a
square form. Determine the side of square (π = 22/7)

Ans.44 cm

3. The wheels of a car are of diameter 80 cm each. How many complete revolutions
does each wheel make in 10 minutes when the car is travailing at a speed of 66 Km
per hour?

Hint: The distance covered in one revolution = 2πr
Formula used distance = Speed × time
Ans. 4375

4. Find the area of quadrant of a circle whose circumference is 22 cm.

Ans. 77/8cm2
5. A chord of a circle of radius 15 cm subtends an angle of 60º at the centre. Find the
area of the corresponding major and minor segments . (π = 3.14) (√3 = 1.73)

Ans. 20.43 cm2, 686.06 cm2

6. AB and CD are respectively arcs of two concentric circles of radius
21cm and 7cm and center 0. If  AOB = 30º. Find the area of shaded
region.

Ans. 308/3cm2

7. Find the Area of shaded region in given fig. If ABCD is a Square of
Side 14cm and APD and BPC are semicircles.

Ans. 42cm2

8. The length of the minute hand of a clock is 14cm. Find the area swept
by the minute hand in 5 minutes.
Ans. 154/3cm2

9. Find the area of the shaded region. Where a circular arc of radius 6
cm has been drawn with vertex ‘O’ of an equilateral triangle OAB of
side 12cm as centre.

Ans.            
660
     36 3 Cm 2
 7         
10. ABCD is a square inscribed in a circle of radius 10 unit. Find the area
of the circle not included in the square.(Use π = 3.14)

Ans. 114Sq. Unit

Chapter 13
Surface Area and Volume

(1) Basic Concept with Formula:
(1) Cuboid:
(i)   Lateral Surface Area/ curved S. A.
=2( l +b)h
(ii) Total Surface Area
=2( l b + bh +h l )
(iii) Volume = l bh

(2) Cube:

(i)     L.S.A / C.S.A = 4 l 2
(ii)    T.S.A = 6 l 2
(iii)   Volume = l 3

(3) Right Circular Cylinder

(i)     L.S.A. / C.S.A. = 2π r h
(ii)    T.S.A. = 2πr(r+h)
(iii)   Volume = π r2 h
(4) Cone,

(i)     L.S.A. / C.S.A. = π r l
(ii)    T.S.A. = πr( l +r)
(iii)   Volume = 1/3π r2h
l  h2  r 2
l = Slant Height

(5) Sphere

(i)     C.S.A. = 4π r2
(ii)    T.S.A. = 4π r2
(iii)   Volume = 4/3π r3

(6) Hemisphere
(i)   L.S.A. / C.S.A. = 2r2
(ii) Total = 3π r2
(iii) V = 2/3π r3
(7) Spherical Shell

(i)     A = 4 π(R2 – r2)
(ii)    V = 4/3 π(R3 – r3)

(8) Frustum of a cone

(i)     C.S.A. = π(R + r) l
(ii)    T.S.A. = π [(R2 + r2 + l (R + r)]
(iii)   V = 1/3πh[R2 +r2 +R.r]
2        2
l =√(h +(R-r) )

QUESTIONS

(1) A toy is in the form of a cone of a radius 3.5 cm mounted on a hemisphere of same
radius the total height of toy is 15.5 cm. Find the total surface Area of toy.

Ans: 214.5 Cm2

(2) From a solid cylinder where height is 2.4 cm and diameter 1.4 cm a conical cavity of
the same height and same diameter is hollowed out. Find the total surface area of
remaining solid to the nearest centimeter square.
Ans: 18 cm2

(3) A 20 m deep well diameter 7 m is dug and the earth from digging is evenly spread out
to from a platform 22m by 14m. Find the height if platform.
Ans. 2.5m

(4) A solid iron pole consists of a cylinder of height 220 Cm and base diameter 24 Cm,
which is surmounted by another cylinder of height 60 Cm and radius 8 Cm. Find the
Mass of the pole, Given that 1 Cm3 of iron has approximately 8 gm mass. (use π =
3.14)

Ans. : 892.2624 Kg.
(5) The radii of the ends of a frustum of a cone 45 cm high are 28 cm & 7 cm. Find its
22
volume, the curved surface area & total surface area. (use π =      )
7
Ans.: Volume: 48510 cm3
C.S.A. : 5461.5 cm2
T.S.A. : 8079.5 cm2
(6) Water flows at the rate of 10 m per minute through a pipe having diameter 5m.m How
much time will it take to fill a conical vessel whose diameter of base is 40 cm and
depth 23cm.

Ans. 51.2Min

(7) A right triangle, whose sides are 3 cm & 4 cm ( other than hypotenuse) is made to
revolve about its hypotenuse. Find the volume and surface area of the double cone so
formed.                               Ans: 3.14 cm3 ; 52.75 cm3

(8) If the radii of the circular ends of a bucket 45cm height are 28cm and 7 cm. Find the
capacity of bucket.
Ans. 48510cm2
(9) A metallic right circular cone 20 cm high and whose vertical angle is 600 is cut into
two parts at the middle of its height by a plane parallel to its base. If the frustum so
obtained be drawn into a wire of diameter 1/16 cm. find the length of the wire.
Ans- 7964.44 mtr

(10)An open metal bucket is in the shape of a frustum of a cone, mounted on a hallow
cylindrical base made of same metallic sheet. The diameter of the two circular ends of
the bucket are 45 cm and 25 cm. the total vertical height of the bucket is 40 cm and that
of a cylindrical base is 6 cm. Find the area of metallic sheet use to make the bucket,
where we do not take into account the handle of the bucket. Also find the volume of
water the bucket can hold.                        Ans- 4860.9 cm2, 83.62 ltrs approx.)

(11) A right triangle whose sides are 3 cm and 4 cm(other than hypotenuse) is made to
revolve about its hypotenuse. Find the volume and surface area of the double cone so
formed.                                          (Ans- 30.14 cm3, 52.75 cm2)
LESSON 14
STATISTICS
Mean for ungrouped data

Mean ( X ) =
x   i

n

Mean of grouped data
1) Direct Method

Mean ( X ) =
 f i xi
 fi
2) Assume mean method { short cut method}

x a 
fd   i       i

f        i

a= assume mean
di (deviation)= xi-a
3)Step deviation method

x a 
fu i i
h
f   i

a = assume mean
x a
ui  i
h
h = Class - Size

MODE

 f1  f 0 
Mode = l                               h
2 f1  f 0  f 2 
where, l = lower limit of the modal class.
h = size of the class interval
f1 = frequency of the modal classs
fo = frequency of the class preceding the modal class
f2 = frequency of the class succeeding the modal class.

Median
n     
  cf 
Median = l         h
2
f

where , l = lower limit of median class
n = no. of observation
cf = cumulative frequency of class preceding the
median class
f = frequency of median class
h = class size.

STATISTICS

1)   Find the mean of the following distribution by all the three methods.

Class interval                 10-25                  25-40          40-55           55-70            70-85     85-100

No of student                       2                  3                 7                6            6          6

Ans. :   [62,62,62]
2)   The mean of the following frequency table is 50, but the frequencies f1 and f2 in
classes 20-40 and 60-80 respectively are not known. Find these frequencies.
Class                   0-20             20-40          40-60     60-80        80-100

Frequencies             17                       f1                 32               f2                19

Ans [f1 = 28,f2=24]
3)   If the mean of the following data is 18.75 . find the value of P.
x 10 15 20 25 30
f 5 10 P 8 2
Ans.: [p=7]

4) If the median of the distribution given below is 28.5 ,
find the value of x and y

Class interval               0-10              10-20          20-30          30-40            40-50     50-60     Total

No of student                 5                  X             20             15               Y            5         60

Ans.: [x=8,y=7]

5) Calculate the missing frequency from the following distribution if the median is 24.

Age in years                     0-10              10-20           20-30          30-40            40-50

No of Persons                        5              X25                                18            7
Ans.: 25
6) Find the mean , mode and median of the following data.
Class                     0-10             10-20           20-30                 30-40      40-50         50-60   60-70

Frequency                  3         4           7            15                             10              7     4
[mean=37.4,median=37.3,mode=36.2 approx]

7) Using Empirical formula, find the mode of the following distrubtion.

Class                     20-25            25-30           30-35                   35-40             40-45

Frequency                    3              8                   8                      3              2

[ hint : mode = 3 median -2 mean, mode= 29.8 approx]

8) Using the data given below construct the cumulative frequeny
Table and draw the ogive . From the ogive determine the median.

Class            0-10            10-20       20-30        30-40          40-50     50-60     60-70        70-80

Students           3              8             12         14             10           6       5             2
Ans.: 35

9) The frequency distribution of scores obtained by 230 candidates in a medical entrance
test is as follows :
Scores      400-450 450-500 500-550 550-600 600-650 650-700 700-750 750-800

No of       20          35            40             32             24           27         18         24
Students

Draw the cumulative frequency curves by less than and more than method on the same axes.
Find the median from the above graph.
Ans.: 573.43
10) the medical checkup of 35 students of a class, their weights are recorded as follows

Weight(in kg)                      No of students
Less than 38                       0
Less than 40                       3
Less than 42                       5
Less t han 44                      9
Less than 46                       14
Less than 48                       28
Less than 50                       32
Less than 52                       35

Find the median weight and modal weight for the above data.
(ans Median weight = 46.5 kg)
Modal Weight = 46.95 kg)
CHAPTER 15
PROBABILITY

CHAPTER 15
(PROBABILITY)
1. A Bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. Find
the probabilities of getting.
i)      A white or a Green ball
ii)     Neither a Green ball nor a Red ball.
iii)    Not Green?
Ans. I) ¾ ii) 7/20 iii) 3/5

2. One card is drawn from a well shuffled deck of 52 playing cards. Find the probability of getting.
i)     A non-face card
ii)    A black king or a red queen.
Ans. I) 10/13 ii) 1/13

3. Two unbiased coins are tossed simultaneously. Find the probability of getting
Ans. I) ¾ ii) 3/4 iii) ¾ iv)1/4

4. Find the probability that a Leap year selected at random will contain 53 Sundays and also
calculate the probability for 52 Sundays in a leap year.
Ans. I) 2/7 ii) 5/7

5. A dice is thrown twice, what is the probability that
i)       5 will not come up either time
ii)      5 will come up at least once ?
Ans. I) 25/36 ii) 11/36

6. A Bag contains 90 disc which are numbered from 1 to 90 if one disc is drawn at random from
the box, find the probability that it bears.
i)      A two digit number
ii)     A perfect square number
iii)    A number divisible by 5
Ans. I) 9/10 ii) 1/10 iii) 1/5

7. 17 cards numbered 1,2,3.... 17 are put in a box and mixed thoroughly. One person draws a card
from the box. Find the probability that the two number on the card is -
i)      Odd
ii)     A prime
iii)    Divisible by 3
iv)     Divisible by 2 & 3 or both

Ans. I) 9/17 ii) 7/17 iii) 5/17 iv) 2/17

8. Savita & Namida are friends, what is the probability that both will have-
i)      The same birthday
ii)     Different birthdays (Ignoring the leap year)
Ans. I) 1/365 ii) 364/365

9. A jar contains 24 marbles some are green & others are blue. If a marble is drawn at random
from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Ans. 8

10. Two dice are thrown simultaneously. Find the probability of getting-
i)     The sum as Prime number
iv)    A Multiple of 2 on one dice and multiple of 3 on the other ?
Ans. I) 7/18 ii) 11/36

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