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CHAPTER -1
UNIT – 1
ADDITIONAL PROBLEMS ON “PLYING WITH NUMBERS”
I.
Choose the correct option
1. The general from of 456 is
a) ( 4 x 100) + ( 5 x 10 ) + ( 6 x 1 )
b) ( 4 x 100) + ( 6 x 10 ) + ( 5 x 1 )
c) (5 x 100 ) + ( 4 x 10 ) + (6 x 1 )
d) ( 6 x 100 ) + ( 5 x 10 ) + ( 4 x 1)
[a]
2. Computers use
a) Decimal system
b) binary system
c)base 5 system
d) base 6 system [b]
3. If abc is 3 digit number , then the number n = abc + acb + bac + cab + cba is
always divisible by
a) 8
b) 7
c) 6
d) 5
[c]
4. If abc is 3 – digit number, then n = abc - acb - bac - bca - cab - cba is always
divisible by
a) 12
b) 15
c) 18
d) 21
[c]
5. If 1 K x K1 = K 2 K , the letter K stands for the digit
a) 1
b) 2
c) 3
d) 4
[a]
6. The number 345111 is divisible by
a) 15
b) 12
c) 9
d) 3
[d]
7. The number of integer of the form 3AB4 , where A, B denote some digits, which
divisible by 11 is
a)
0
b) 4
c) 7
d) 9
[d]
2.What is the smallest 5 – digit number divisible by 11 and containing each of the
digits 2, 3, 4, 5, 6?
The smallest 5- digit number divisible by 11 which contain the digits 2, 3, 4, 5, 6 is 24365
[ ∵ 2 – 43 – 6 + 5 = 0 ]
3.How many 5 – digit number divisible by 11 are there containing each of the
digits 2,3, 4, 5, 6 ?
24365 ,
26245,
24563,
26543
36542,
34562,
34265,
36245
54263,
56243,
54362,
56342
There are 12 5 digit numbers
4.If 49A and A49 where A > 0 , have a common factor , find all possible values of A .
If A = 2, 5, 7, 8 then the numbers formed by 49 A and have common factors.
Like
1. 492 and 249 common factor 3
2. 495 and 549 common factor 3
3. 497 and 749 common factor 7
4. 498 and 849 common factor 3
But if A = 1, 3, 4, 6 , 9 donot have common factor.
5. Write 1 to 10 using 3 and 5, eah al last once , and using addition and subtraction,
(For example , 7 = 5 + 5 – 3 )
1= 3 + 3– 5
2= 5– 3
3= 5+ 3 – 5
4= 3+ 3 + 3– 5
5 = 3 -3 + 5
6= 3+ 3 + 5– 5
7= 5+ 5 – 3
8= 5+ 3
9= 5+ 5 + 5– 3 –3
10 = 3 – 3 + 5 + 5
6. Find all 2 – digit numbers each of which is divisible by the sum of digits.
2 – digit numbers which are divisible by the sum of its digits are
10, 12, 18 , 20 , 21 , 24 , 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80,
81, 84 , 90.
7. The page numbers of a book written in a row gives a 216 digit number. How many
pages are there in the book.
From Page No 1 to 9 there are
9 digits
From page No 10 to 99 there are 180 digits
From Page No 100 to 108 there are 27 digits
216
∴ Total No of pages 108 in a book when has 216 digit numbers.
8. Look at the following patters:
1
1
1
1
1
1
2
3
4
1
3
6
1
4
1
1
1
1
1
1
1
1
1
5
7
8
3
4
6
6
10
15
21
20
1
3
1
1
2
4
10
20
35
56
1
5
15
35
70
1
1
6
21
56
1
7
28
1
8
1
9. This is called Pascal’s triangle. What is the middle number in the 9-th row?
The Middle number is the 9th Row is 70.
10. Complete the adjoining magic square. [ Hint: in a 3 x 3 magic square , ……., Magic
some is three times the central number?
Ans: The Number is 108
[ ∵ 108 = 1 + 0 + 8 = 9]
9 x 12 = 108
11. Find all 3- digit natural numbers which are 12 times as large as the ……. m of
their digits.
Find all digits x, y such that 34 x 5y is divisible by 36.
The no.s are divisible by 36 only when they are divisible by 4 and 9.
According to the divisibility condition of 4 and 9, then the numer is divisible by
4. If the sum of all the digits of a number is divisible 9, then the number s
divisible by 9.
These condition one satisfied only
When x = 4, y = 2
Or
x = 0, y = 6
Or
x =9,y=6
12.Can you divide the numbers 1,2,3,4,5,6,7,8,9,10 into two groups such that
the product of numbers in one group divides the product of numbers in the
other group and the quotient is minimum?
Group 1 1 x 2 x 3 x 4 x 5 x 6 7
= 5040
Group 2 8 x 9 x 10
= 720
5040 = 7
720
Minimum quotient i s 7
13.Find all 8 digit numbers 273A49B5 which are divisible by 11 as well as 25.
The number is divisible by 25
When the number ends with 00, 25 , 50 or 75
∴ In the number ‘B’ must be either 2 to 7.
The number is divisible by 11 if and only if ( Sum of odd placed digits) – (
Sum of even placed digits) is divisible by 11.
It is possible only when
i)
If A = 1 and B = 2
Then the number is = 27314925
Or
If A = 6 and B = 7
Then the number is 27364975
14.Suppose a, b are integers such that 2 + a and 25 –b are divisible by 11. Prove that
a + b is divisible by 11.
It two numbers are divisible by 11 then their sum and difference is also divisible by 11
Let us take the difference of the no. s
Then ( 2 + a ) – ( 35 – b ) is divisible by
= 2 + a – 35 + b
= 2 – 35 + a + b
= - 33 + a + b divisible by 11
Since – 33 is divisible by 11
a + b also must be divisible by 11
15.A list of numbers and corresponding codes are given in the adjacent table. Find the
Number L.
As per the given of numbers and their codes
2 corresponds to 8
0 corresponds to 0
1 corresponds to 6
∴ the number L = 8066
2011
16. In the multiplication table A 8 x 3B = 2730 , A and B represent distinct digits different
from 0. Find A + B.
In A8 x 3B = 2740
To get 0 in unit place
8 is to be multiplied by 5
Ie is 8 x 5 = 40
A8 = 2730 = 78
35
AB = 78
⇒A=7
17. Find the least natural number which leaves the remainders 6 and when divided by 7 and 9
respectively.
LCM of 7 and 9 is 63
Least number which leaves the remainder 6 and 8
Go on finding the remainder by subtract in natural no. from 63
Ie 63 – 1 = 7 =
8
7) 62
56
6
II ly 63 – 1 =
6
9) 62
54
08
There is no other no. which gives 6 and 8 when divided by 7 and 9
Least number is 62.
18.Prove that the sum of cubes of three consecutive natural numbers always
divisible by 3.
Let the 3 consecutive nature no.s be
X, x + 1 and x + 2
x³ + ( x + 1) ³ + ( x + 2 ) ³
x ³ + x ³+ 1 + 3x ( X + 1) + x ³ + 6 + 6x ( x + 2)
x³ + y³ + 1+ 3x³ + 3x + x³ + 6 + 6x + 12x
3x³ + 9x² + 15x + 9
3 ( x² + 3x + 5x + 3)
⇒ Sum of cubes of three consecutive natural numbers is always divisible by 3.
CHAPTER – 2
UNIT – 1
ADDITIONAL PROBLES ON ALGEBRIC EPXPRESSIONS
I.
Choose the correct answer:
a. Terms having the same literal factors with same exponents are called
a) Exponents
b) like terms
c)factors
d) unlike terms
[b]
b. The coefficient of ab in 2ab is:
a)ab
b) 2
c) 2a
d)2b
[b]
c. The exponential from of a x a x a is :
a) 3a
b) 3+a
c) a³
d) 3-a
[c]
d. Sum of two negative integer is:
a) Negative
b) positive
C)zero
d) infinite
[a]
e. What should be added to a² + 2ab to make it a complete square?
a)b²
b) 2ab
c) ab
d)2a
[a]
f. What is the product (x + 2) ( x – 3 )?
a) 2x – 6
b) 3x – 2
C)x² -x – 6
d) x² - 6x
[c]
g. The value of (7.2)² is (use an identity to expand):
a) 49.4
b) 14.4
c) 51.84
d)49.04
[c]
h. The expansion of (2x – 3y)² is:
a) 2x² + 3y² + 6xy
b)4x² + 9y² - 12 xy
c)2x² + 3y² - 6xy
d) 4x² + 9y² + 12xy
[b]
i. The product 57 x 62 is (use an identity):
a) 4596
b) 2596
c) 3596
d) 6596
2.Take away 8x – 8p + 10q from 10x + 10y – 7p + 9q.
10x + 10y - 7p + 9 q
8x - 7 y – 8 p + 10 q
( - ) (+) (+ ) ( - )
+2x + 17 y + p - q
[c]
3.Expand:
(i) (4x + 3)²
ii) (x + 2y)²
(iii) ( x + 1/x)²
iv) ( x – 1 / x)²
i)
(4x + 3) ²
this is in the form
( a + b )²
= a² + 2ab + b²
(4x + 3) ²
Here a = 4x
= (4x)² + 2 (4x) (3) + 3²
b= 3
= 16x² + 24x + 9
ii)
(x + 2y )²
(a + b) ²
this is in the form
= a² + 2ab + b²
Here a= x
= x² + 2 (x) (2y) + ( 2y)²
b = 2y
= x ²+ 4xy + 4y²
iii)
( x + 1 )²
x
This is in the form
( a+ b) = a + 2ab + b
= x² + 2 . x . 1 + ( 1 )²
X
x
= x² + 2 + 1
x²
Here a = x
b=1
x
iv)
( x- 1 ) ² This is in the form
X
( a+ b) ² = a² – 2ab + b ²
Here a = x
= x² + 2 , x , 1 + (1 ) ²
x
x
b=1
x
= x² + 2 + 1
x²
3. Expand:
i)
(2t + 5) ( 2t – 5)
ii) ( xy + 8 ) ( xy – 8 )
iii)(2x + 3y ) ( 2x – 3y)
i)
( 2t + 5) ( 2t – 5) This is in the form
= ( a+ b) ( a- b) = a² – b²
= ( 2t + 5 ) ( 2t – 5) = (2t) ² – 5²
= 4t² - 25
ii)
(xy + 8 ) ( cy – 8) This is in the form of
= ( a + b) ( a- b) = a² – b²
= ( xy + 8 ) ( xy – 8) = (xy) ² – 9²
= x² y² - 64
iii)
( 2x + 3y ) ( 2x – 3y ) This is in the form of
= (a + b) ( a- b) = a² – b²
= ( 2x + 3y ) ( 2x – 3y ) = (2x) ² - (3y) ²
= 4x² – 9y²
5.Expand
i) ( n-1 ) ( n + 1) ( n² + 1)
ii) ( n – 1/n) ( n + 1 /n) ( n² +1/n²)
iii)(x -1 ) ( x + 1) ( x² + 1) ( x⁴ + 1)
iv)(2x – y ) ( 2x + y) ( 4x² + y²)
i)
( n – 1) ( n²+ 1) ( n -1) ( n + 1 ) = n² -1²
= ( n²- 1² ) ( n²+ 1²) This is in the form
= ( a- b) ( a + b) = a² - b²
= ( n² – 1²) ( n² + 1²) = (n²)² – ( 1²)²
= n⁴ - 1
ii)
iii)
( n – 1 ) ( n + 1 ) ( n² + 1 )
n
n
n²
= ( n² – 1 ) ( n² + 1 )
n²
n²
= n⁴ – 1
n⁴
( x – 1) ( x + 1) ( x ² + 1) ( x⁴ + 1)
= (x² – 1² ) ( x² + 1²) ( x⁴ + 1)
=[ ( x² ) ² ) – ( 1²)²) ] [ x⁴ + 1⁴ )]
=(x⁴ – 1⁴ ) ( x⁴ + 1⁴ )
=( x⁴ )² – ( 1⁴ )²
= x ⁸- 1
iv)
( 2x – y) (2x + y) ( 4x² + y² )
=[(2x) ² – (y) ² ] ( 4x² + y² )
= ( 4x² + y² ) ( 4x² + y²)
= (4x²)² – ( y²)²
=16x⁴ - y⁴
6. Use appropriate formula and compute:
i) ( 103) ²
ii) (96) ²
iv)1008 x 992
i)
iii) 107 x 93
v) 185² - 115²
( 103) ² = (100 + 3) ² This is in the form
( a + b) ² = a² + 2ab + b²
( 100 + 3 ) ² = ( 100 ) ² + 2 ( 100) (3) + 3²
= 10000 + 600 + 9
( 103 ) ²
ii)
(96) ²
= 10609
= ( 100 – 4) ² This is in the form
( a – b) ² = a ² – 2ab + b²
(100 – 4) ² = ( 100) ² + 2 ( 100) (4) + 4²
= 10000 – 800 + 16
= 10016 – 800
( 96) ² = 9216
iii)
107 x 93
= ( 100 + 7) ( 100 – 7) This is the form of
( a + b) ( a- b) = a² – b ²
( 100 + 7) ( 100 – 7) = ( 100) ² – 7²
= 1000 – 49
= 9951
iv)
1008 x 992
( 1000 + 8 ) ( 1000 – 8) This is in the form
( a+ b) ( a- b) = a² – b²
( 1000 + 8) ( 1000 – 8) = ( 1000) ² – 8²
= 100000 – 64
= 999936
v)
185² – 115² This is in the form
a² – b² = ( a + b) ( a- b)
( 185) ² – (115) ² = ( 185 + 115) ( 185 – 115)
= ( 300) (70)
(185) ² – (115) ² = (300) (70)
(185) ² – ( 115) ² = 21000
7.If x + y and xy = 12 find x² + y².
( x + y ) ² = x² + y² + 2 xy
(x+y) ² =
( x + y ) ² = 2xy = x² + y ²
x² + y² – 2xy
7² = 2 ( 12) = x² + y²
∴ x² + y² =
49 – 24 = x ² + y²
25 = x² + y²
( x + y) – 2xy
X² + y² = 25
8. If x + y = 12 and xy = 32 , find x² + y².
( x + y) ² = x² + y² + 2xy
( x + y ) ² – 2 xy = x² + y²
(12) ² – 2 ( 32)= x² + y²
144 – 64 = x² + y²
X² + y² = 144 – 64
X² + y² = 80
9. If 4x² + y² = 40 and xy = 10 , find x ² + y² .
4x² + y² = 40
= 2² x² + y² = 40
= ( 2x) ² + y² = 40
( 2x + y) ² = ( 2x )+ y² + 2 (2x) (y)
(2x + y) ² = 4x² + y² + 4 ( xy)
= 40 + 4 (6)
= 40 + 24
( 2x + y) ² = 64
2x + y
=√ 64
2x + y =8
10. if xy = 3 and xy = 10 , find x² + y²
( x-y) ² = x² + y² – 2xy
3² = x² + y² – 2 (10)
9 = x² + y² – 20
X² + y² = 9 + 20
X² + y² = 29
11. If x =( 1 )= 3, find x² + (1 ) and x³ + ( 1 )
x
x²
x³
case – 1 ( x + 1 ) = 3
x
Squaring on both side
( x + 1 )² = 3²
X
X² + 1 + 2 . x . 1 = 9
X²
x
X² + 1 = 9 – 2
x²
x² + 1 = 7
x²
Case – 2 x + 1 = 3
x
Cubing on both side
(x + 1)³ = 3³
x
x³ + 1 + 3 x . 1 ( x + 1 ) = 27
x³
x
x
x³ + 1 + 3 ( 3) = 27
x³
x³ + 1 + 9 = 27
x³
x³ + 1 = 27
x³
x³ + 1 = 18
x³
12.If x + 1 = 6 , find x ²+ 1 and x⁴ + 1
X
x²
x⁴
Case – 1 x + 1 = 6
x
Squaring on both side
( x + 1 )² = 6²
X
X² + 1 + 2 . x. 1 = 36
X²
x
X² + 1 + 2 = 36 - 2
X²
X² + 1 = 34
X²
Case – 2
Consider X² + 1 = 34
X²
Squaring on both side
( x + 1 )³ = 3³
X
( x² + 1 )² = ( 34)²
X²
X⁴ + 1 + 2 . x² 1 = 1156
X⁴
x²
34² = 1156
X⁴ + 1 + 2 = 1156
X⁴
X⁴ + 1 = 1156 – 2
x⁴
∴
X⁴ + 1 = 1154
x⁴
13.Simplify : (i) ( x + y)² + ( x – y)²
ii) ( x + y )² x ( x – y )²
i)
( x + y ) ² + ( x – y) ²
= [x² + y² + 2 xy ] + [x² + y ²– 2xy]
=x² + y ²+ 2xy + x² + y² – 2xy
= 2x² + 2y ²
= 2 ( x² + y² )
ii)
(x + y) ² x
( x- y) ²
= [ ( x + y ) ( x – y) ] ²
( a + b) ( a- b) = a² - b²
= [ x² – y ²]²
This is in the form
= ( a – b) ² = a² + b² – 2ab ( a- b) ² = a² + b² – 2ab
= ( x² – y² ) ² = x⁴ + y⁴ – 2 x² y² ( or)
(x² – y² ) = x⁴ – 2x²y ² + y⁴
14.Express the following as difference of two squares:
i) ( x + 2z) ( 2x + z)
ii) 4 (x + 2y) ( 2x + y)
iii)
(x + 98) ( x + 102)
iv)
505 x 495 [Hint: 4ab = (a + b) ² – ( a- b) ²
i)
( x + 2z ) ( 2x + z)
= 2x² + xz + 4 xz + 2 z²
= 2x² + 2z ²+ 4 xz + xz
= 2( x² + z² + 2xz ) + xz
= 2 ( x + z ) ² + xz
Multiply and divide by 4
= 8 ( x + z ) ² + 4xz
4
4
= 8 ( x + z ) ² + 4 xz
4
4
( x + z ) ² – (x – z) ²
[x² + z² + 2xy ] – [x² + z² – 2zy]
= 8 ( x² + z ) ² + ( x + z) ² – ( x- z) ²
4
=9(x+z) ² - (x- z) ²
4
4
3(x – z) ²
4
ii)
-
x–z ²
2
4 (x + 2y) ( 2x + y)
4 [2x² + xy + 4xy + 2y² ]
4[2x² + 2y² + 4xy + xy ]
4[ 2 ( x² +y² + 2xy ) + xy]
4[2 (x +y) ² + xy]
8(x+ y ) 2 + 4xy
8 (x + y ) ² + ( x + y) ² – (x – y) ²
=9 ( x + y) ² – ( x- y) ²
=[3 (x + y)] ² – ( x- y ) ²
iii)
( X + 98) ( x + 102)
= x² + 102 x + 98 x + 9996
= x ²+ 200x + ( 10000 – 4)
= x ²+ 200x + 10000 – 4
= x² + 200 x + (100) ² – 2²
=(x + 100) ² - 2²
iv)
505 x 495
= ( 500 + 5) ( 500 – 5)
= ( 500) ² – 5²
15. If a = 3x – 5y , b = 6x + 3y and c= 2y – 4x, find
i) a + b – c
ii) 2a – 3b + 4c
i) a + b + c
= ( 3x – 5y) + ( 6x + 3y) – (2 y- 4x)
= 3x – 5y + 6x + 3y – 2y + 4x
= 13x – 4y
ii)
2a – 3b + 4c
=2(3x – 5y) – 3 ( 6x + 3y) + 4 ( 2y – 4x)
=6x – 10y -18x – 9y +8y -16 x
= 6x – 13x – 16x -10y- 9y + 8y
= 28x – 11y
16. The perimeter of a triangle is 15x² – 23 x + 9 and two of its side
are 8x² + 8x – 1 and 6x² – 9x + 4 . Find the third side.
Perimeter of a triangle = sum of all the three sides of a triangle
Let the sides of the triangle be a, b and c then P = a + b + c
Let a = 5x² + 8x – 1
b = 6x² - 9x + 4 , and
P = 15x² – 23 x + 9
C=?
C = P – ( a + b)
= ( 15x² – 23 + 9 ) – [( 5x² + 8x – 1 ) + (6x² – 9x + 4)
= ( 15x² + 23x + 9 ) – [11x² – x + 3]
= (15x² – 23x + 9)- 11x² + x -3
= 4x² – 22 x 6
17. The two adjacent sides of a rectangle are 2x² – 5xy + 3x² and
4xy – x² – x² . Find its perimeter.
Let a and b are the sides of the Rectangle
ℓ = 2x² - 5xy + 3x² b = 4xy - x² - x²
Then its perimeter P = 2 ( ℓ + b)
P = 2 [2x² - 5xy + 3z² + 4xy - x² - z² ]
P= 2 [ x² - xy + 2z²]
P = 2x² - 52 xy + 4z²
18.The base and altitude of a triangle are ( 3x – 4y) and ( 6x + 5y)
respectively. Find its are.
Area of a ∆ = 1 b x h
2
Let b and h are the base and altitude of the triangle
Then Area of triangle = 1 x b x h
2
A = 1 ( 3x – 4y ) ( 6x + 5y)
2
= 1 [ 18x + 15 xy - 2 xy – 20 y² ]
2
= 1 [ 18x² - 9xy – 20 y² ]
2
19.The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of
side length x + y is removed . What is the area of the remaining region?
( 2x + 3y)
X+y
X+y
( 3x + 2y)
Area of the remaining region =
(Area of the Rectangle ) – (Area of the square)
(ℓ x b )
- ( side) ²
=[(2x + 3y) (3x + 2y)] – ( x + y) ²
= [ 16² + 4xy + 9xy + 6y² ] – ( x² + y² + 2xy)
= 6x² + 13xy + 6y² - x² - 2xy - y²
= 5x² + 11xy – 5y²
20. a, b, c are rational numbers such that a² + b² + c² -bc – ca = 0.
Prove that a = b = c.
a² + b² + c² - bc – ca = 0
Multiply both side by 2
2(a² + b² + c² - ab – bc –ca ) = 2 x 0
2a² + 2b² + 2c² - 2ab – 2bc – 2ca = 0
a² + a² + b² + c² + c² - 2ab – 2bc – 2ca = 0
a² + b² - 2ab + b² + c² - 2bc + c² +a² - 2ca = 0
( a- b) ² + ( b – c) ² + ( c- a) ² = 0
If the sum of squares of three terms is ‘0’
Then each term is equal to ‘0’
⇒ ( a- b) ² = 0
a- b= 0
(b–c)² =0
b –c=0
( c- a) ² = 0
c–a=0
a=b …. ( 1)
b = c ….(2)
c = a ….(3)
from equation ( 1) ( 2) and ( 3)
a=b=c
CHAPTER – 3
UNIT – 1
ADDITIONAL PROBLEMS ON “AXIOMS, POSTULATES”
1. Choose the correct option:
i)
a= 60 and b = a , then b= 60 by ….
a. Axiom 1
b. Axiom 2
c.Axiom 3
d.Axiom 4
[…..]
ii)
Given a point on the plane , one can draw …….
a. Unique
b . two
c.finite number of
d.infinitely many …… through that point
[d]
iii) Even two points in a plane, the number of lines which can be
drawn to pass through these two points is ….
a. Zero
b. exactly one
c.at most one
d. more than on
[b]
iv) Two angles are supplementary , then their sum is …..
a. 90˚
b. 180˚
c. 270˚
d. 360˚
[b]
v)The measure of an angle which is 5 times its supplement is
a. 30˚
b. 60˚
c.120˚
d.150˚
[d]
2. What is the difference between a pair of supplementary angles and
a pair of complementary angle?
Supplementary angles : If the sum of two angles is 180˚ then they are
supplementary angles.
Complementary angles: If the sum of two angles is 90˚ then they are
complementary angles.
3. What is the least number of on collinear points required to determine a
plane?
3 non collinerar points.
4. When do you say two angles are adjacent?
The angles are said to be adjacent if they have a common side and a
common and point.
A
C
D
B
5. Let AB be a segment with C and D between them such that the order of
points on the segment is A, C , D, B, Suppose AD = BC . Prove that AC= DB.
AB is s line segment C and D are points on it such that AD = BC
AC + CD = BD + CD
AC = BD
[ Axioms 3 ]
6. Let AB and CD be two straight lines intersecting at 0. Let 0X be the
dissector of BOD. Draw 0Y between 0D and 0A such that 0Y ⟘ 0X .
Prove that 0Y bisects DOA.
Y
A
D
Q˚ P
X˚
0
X˚
X
C
B
AB and CD intersecting at ‘0’
OX bisects BOD
Let = BOX
= DOX = X˚
OY ⟘ OX
Let POY = P and YOA = Q
BOX + XOD + DOY + YOA = 180˚
x˚ + x˚ + p + q = 180˚ ……….. ( 1)
x + p = 90˚
…….. ( 2) given
∴ x + q = 90
……… (3)
( Remaining angle of 180˚ )
From (2) + ( 3)
X + p = x + q = 90˚
P = q = 90˚
⇒ OY bisect DOA
7.Let AB and CD be two parallel lines and PQ be a transversal. Let PQ
intersect AB in L. Suppose the bisector of ALP intersect CD in R and
the bisector of PLP intersect CD in S. Prove that LRS + RSL = 90˚.
M
M
M
X
Y
A
X˚
Y˚
B
L
C
S
R
Proof:
Let ALN = NLP
= x˚ and
BLM = MLP = y˚
ALB = 180˚
ALN + NLP + PLM + MLB = 180˚
X +
x + y +
2x + 2y
y = 180
= 180 ˚
X + y = 180 = 90˚
2
⇒ NLM = 90˚
But NLM = RLS = 90˚ * V. O . A+
In the ∆ LRS
P
LRS + RSL + LRS = 180˚
[Sum of all the angles of ∆le]
LRS + RSL + 90˚ = 180 ˚
LRS + RSL = 180˚ - 90˚
LRS + RSL
= 90˚
8 In the adjoining figure. AB and CD are parallel lines. The transversals
PQ and RS interest at U on the line AB. Given that DWU = 110˚ and
CVP = 70 , find the measure of QUS.
S
Q
X
U
A
B
110˚
C
(
70˚
(
)
V
P
Ans: Given - DWU
D
W
R
= 110˚ , CVP = 70˚
ProofUVP = KVP = 70˚ ( V.O. A)
VWU + UWD = 180˚ ( Adjacent Angles)
V + W + U = 180˚ ( Sum of angles of triangle is 180˚)
U = 180˚ - 70˚ - 70˚ = 40˚
X = U = 40˚ ( V. o . A)
∴ QUS = 40˚
9. What is the angle between the hour’s hand and minute’s hand
Of a clock at i) 1.40 hours , ii) 2.15 hours? ( Use 1˚ = 60 minute)
10. How much would hour’s hand moved from its position to 12 noon when the
time is 4.24 p.m?
11.Let AB be a line segment and let C be the midpoint of AB . Extend AB to D
such that B lines between A and D. Prove that AD + BD = 2CD.
12.Let AB and CD be two lines intersecting BOD . Prove that the extension of 0X
to the left of 0 bisects AOC.
13.Let 0X be a ray and let 0A and 0B be two rays on the same side of 0X and 0B.
Let 0C be the bisector of AOB. Prove that AOX + XOB = 2 XOC.
14. Let 0A and 0B be two rays and let OX be a ray between 0A and 0B such that
AOX > XOB . Let 0C be the bisector of AOB. Prove that
AOX - XOB = 2 Cox.
15. Let 0A , 0B , 0C be the three rays such that 0C lies between 0A and 0B .
Suppose the bisectors of AOC and COB are perpendicular to each other. Prove
that B, O , A are collinear.
16. In the adjoining figure, AB
A
DB . Prove that ABC + DCB + CDE = 180˚.
B
D
C
E
17. Consider two parallel lines and a transversal . Among the measurement of
transversal. Among the measurement of triangles formed, how many distinct
numbers are there?
(Answer will be given in next issue)
CHAPTER - 1
UNIT1 – 2
EXERCISE 1. 2. 2
1. Express the following statement mathematically.
i)
Square of 4 is 16
Ans: 4² = 16
ii)
Square of 8 is 64
Ans: 8² = 64
iii)
Square of 15 is 225
Ans : 15² = 225
2. Identify the perfect squares among the following numbers:
2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900,100, 1000, 100000.
Ans: Perfect squares : 6² = 36 , 7² = 49,
9² = 81 ,
13² = 169, 25² = 625,
30² = 900 , 10² = 100
3. Make a list of all perfect squares from 1 to 500.
Ans: Perfect squares: 1,4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,
256, 289, 324, 361, 400, 441, 484.
4. Write 3 – digit number ending with 0, 1, 4, 5, 6,9 , one for each digit, but
none of them is a perfect square.
Ans: 110, 111, 114, 115, 116, 119 ( ending with 0, 1, 3, 5, 6, 9, but not a perfect
square) .
5. Find numbers from 100 to 400 that and with 0, 1, 4, 5, 6, or 9 which are
perfect squares.
Ans: Perfect squares from:
100 to 400 – 100 ( 10²)
144(12²),
169 ( 13²),
196 (14²),
225 (15²),
289 (17²),
324 (19²),
400(20²).
EXERCISE 1. 2. 3
1. Find the sum 1 + 3 + 5 + …… + 51 ( the sum of all odd numbers from 1
to 51) without actually adding them.
Ans: 1 + 3 + 5 + …… +51 = 51 + 1 = 26
26²
= 676
2
Sum of ‘n’ odd numbers = n²
Sum of 26 odd numbers = 26² = 676
2. Express 144 as a sum of 12 odd numbers.
Ans: 144 = 12²
Sum of ‘n’ odd numbers = n²
Sum of 12 odd numbers = 12²
1+ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
12² = 144
3. Find the 15 th and 16 th triangular numbers and find their sum. Verify the
statement 8 for this sum.
Ans: 1+2
3+3
6+ 4
10 + 5
15+6
I
II
III
IV
V
21 + 7
28 +8
36+9
45+10
55+11
VI
VII
VIII
IX
X
66+12
78+13
91+14
105+15
XI
XII
XIII
XIV
120+16
136+17
XV
XVI
n + ( n + 1) = ( n + 1 ) 2
15th ∆ no + 16th ∆ no = (16)2
120 + 136 = 256
4. What are the remainders of a perfect square when divided by 5.
Ex. Problem
Write 5th and 6th triangular numbers and verify the formula of statement 8.
Ans: 1+2
I
3+3
6+4
II
15+6
21+ 7
V
VI
III
n+ ( n+1) = ( n + 1)2
5th ∆ no + 6th ∆ no = ( 6)
2
15 + 21 = 36 ∴ Statement 8 is true
Methods for squaring a number:
1842 = (180 + 4)2 *( Using identity)
=180 2 + 4 2 + 2 x 180 x 4
= 32,400 + 16 + 1,440
= 33,856
*No. ending with 5 –
10+5
IV
1.9 x 10 5 2
9025
2. 7x8 5 2
3. 5x6 5 2
5625
3025
EERSICE 1.2.4
1. Find the squares of :
i)
31 = (30 + 1)2
(a + b)2 = a2 + b2 + 2ab
(30+1)2 = ( 30) 2 + (1)2 + 2 x 30 x 1
= 900 + 1 + 60
= 961
ii)
(72)2 = (70 + 2) 2
(70 + 2 ) 2 = (70)2 + (2)2 + 2 x 70 x 2
= 4,900 + 4 + 280
= 5184
iii)
37 = (30 + 7)2
(30 + 7)2 = (30)2 + (7)2 + 2 x 30 x 7
= 900 + 49 + 420
= 1369
iv)
166 = ( 160 + 6)2
(160 + 6)2 = ( 160) 2 + (6)2 + 2 x 60 x 6
= 25,600 + 36 + 1920
= 27,556
2. Find the squares of :
i)
8 x 9 5 = 7225
ii)
11 x12 5 = 13225
iii)
16 x 17 5 = 27225
3. Find the square of 1468 by writing this as 1465 + 3
Ans: 1465 + 3
( a + b)2 = a2 + b2 + 2ab
(1465 + 3 )2 = ( 1465)2 + (3)2 + 2 x 1465 x 3
= 21,46,225 + 9 + 8790
= 2146234 + 8790
= 21,55,024
EXERCISE 1.2.5
1. Find the squares root of the following numbers by
factorization:
i)
196
2 196
2 98
7 49
√ 196
7 7
=√22x72 = 2x7
1
√ 196
ii)
= 14
256
2 256
2 128
2
64
2
32
2
16
2
8
2
4
2
2
1
√256 = √ 2 2 x 2 2 x 2 2 x 2 2 = 2x 2x2x2
√256 = 16
iii)
10404
2 10404
2 5202
3 2607
3 867
17 287
17
17
1
√10404 = √ 2 2 x 3 2 x 17 2 =2 x 3 17
= 102
iv)
1156
2 1156
2
578
17 289
17
17
1
√1156 = √2 2 x 17 2 = 2 x 17
√1156 = 34
v)
13225
5 13225
5 2645
23 529
23
23
1
√13225 = √5 2 x 23 2 = 5 x 23
√13225 = 115
2. Find the squares of :
i)
100 + 36 = 10 + 6 = 16
2 100
2 50
5 25
5
5
1
√100 = √2 2 x 5 2 =
2x5
√100 = 10
3 36
3 12
2 4
2 2
1
ii)
√1360 + 9
√36 = √3 2 x 2 2 = 3 x 2
=6
√1369 = 37
37 1369
37
√1369
= √ 372 = 37
37
1
iii)
√2704 + √144 +√ 289
52 + 12 + 17
64 + 17
= 81
2 2704
2 1352
2 676
2 338
2 169
13 13
1
√2704 =√ 22 x 3 2 x 13 2 = 2 x 2 x 13 =52
2 144
2 72
2 36
2 18
3 9
3 3
1
√144 = √2 2 x 2 2 x 3 2 = 2 x 2 x 3 = 12
17 289
17 17
1
iv)
√ 289 = √17 2 = 17
225 - 25 = 15 – 5 = 10
5 225
4 45
3 9
√225 = √5 2 x 3 2 = 2 x 2 = 15
3 3
1
5 25
√25 = √ 5 2 = 5
5 5
1
v)
1764 - 1444 = 42 - 38 = 4
2 1764
2 882
3 441
3 147
7 49
7
7
1
√1764 = √ 2 2 x 3 2 x 72 = 2 x 3 x 7
√1764 = 42
2 1444
2 722
19 361
19 19
√1444= √ 2 2 x 19 2
1
√1444 = 38
vi)
169 x 361
2
13 x 19
117
13x
247
13 169
13 13
1
√ 169 = √13 2 = 13
=2x9
19 361
19 361
1
√ 361
= √ 19 2 19
3. Square yard has area 1764m2 . From a corner of this
yard, a square part of area 784 m2 is taken out for
public utility. The remaining portion is divided in to
5 equal square parts. what is the parameter of these
equal parts?
Ans. Area of square = a2
Area of given square yard = 1764 m2
Area used for public utility = 784 m2
∴ Area of remaining portion = 1764 –
784
980 m2
If the area is divided into 5 parts, then the area of each
Square
= 980 = 196 m2
5
Length of each side = √ 196
a-
= 14m
Perimeter of square = 4a
= 4 x 14
= 56m
4. Find the smallest positive interger with which one has to
multiply each of the following numbers to get a perfect
square.
Ans
i) 847
11 847
11 77
7 7
1
847 = 11 2 x 7
∴ 7 is the smallest positive number to be added to 847 to get a
perfect square.
ii)450
2 450
5 225
5 45
3 9
450 = 5 2 x 3 2 x 2
3 3
1
∴ 2 is the smallest positive interger to be multiplied with 450 to get a perfect
square.
iii) 1445
5
1445
17 289
17
17
1
1445 = 17 2 x 5
∴ 5 is the smallest positive interger to be multiplied with 1445 to get a perfect
square.
iv)
1352
2
1352
2
676
2
338
13 169
13
1352 = 2 2 x 13 2 x 2
13
1
∴ 2 is the smallest positive interger to be multiplied with 1352 to get a perfect
square.
5. Find the largest perfect square factor of each of the following numbers:
i)
48
2 48
2 24
2 12
2 6
3 3
1
2x2x2x2x3
4 x 4 x 3
16 x 3
∴ 16 is the largest perfect square factor.
ii)
11280
2 11280
2 5640
2 2820
2 1410
5
705
3
141
47 47
2 x 2 x 2 x 2 x 5 x 3 x 47
1
4 x 4 x 5 x 3x 47
16 x 5 x 3 x 47
∴ 16 is the largest perfect square factor.
iii)
729
3 729
3 243
3
81
3
27
3
9
3
3
1
3x3x3x3x3
9 x 9 x9
81 x 9
729
∴ 729 is the largest perfect square factor.
iv)
1352
2 1352
2
676
2
338
13 169
13 13
1
2 x 2 x 2 x 13 x 13
4 x 2 x 169
676 x 2
∴ 676 is the largest perfect square factor.
6. Find the proper positive factor of 48 and a proper positive multiple of
48 which add up to a perfect square. Can you prove that there are infinitely
many such pairs?
Ans: The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….
1) 48 + 1
49 = 7 2
2 ) 192 + 4
192 = 14 2
3)288 + 1
289 = 11 2
4 ) 96 + 4
100 = 10 2
5)96 + 48
144 = 12 2
6) 48 + 16
64 = 8 2
7)240 + 16
256 = 16 2
Yes, we can prove that there are infinitely many such fairs.
EXERCISE 1.2.6
1. Find the nearest integer to the square of the following numbers.
Ans: i) 232
225 < 232 < 256
15 2
162
∴ Square root of 232 is nearest to 15
ii)600
576 < 600 < 625
242
252
∴ Square
root of 600 s nearest to 24.
iii)728
678 < 728 < 729
262
272
∴ Square root of 728 is nearest to 27.
iv)824
784 < 824 < 841
282
292
∴Square root of 824 is nearest to 27
v)1729
1684 < 1729 < 1764
412
422
∴Square root of 1729 is nearest to 42.
2. A piece of land is in the shape of a square and its area is 1000m2. This
has to be fenced using barbed wire. The barbed wire is available only in
integral lengths. What is the minimum length of the barbed wire that
has to be bought for this purpose.
Ans: Area of the land = 1000m2
Area = a2 = 1000
Perimeter = 4a
Squaring , (Perimeter)2 = ( 4a)2
= 16a2
= 16 x 1000
P2 = 16,000
P
= 16000
√15876 < √16000 < √16129
126
127
∴ Nearest number is 126
But this is not enough to cover the hard.
∴Length of barbed wire required = 127 m.
3. A student was asked to find√ 961 . He read it wrongly and found √691
to the nearest integer. How much small was his number form the
correct answer?
Ans.
√ 961 = 31
√691
√676 < √691 < √729
26
27
26 is nearest number to √ 691
∴ difference = 31 – 26 = 5
EXERCISE 1.2.7
1. Looking at the pattern, fill in the gaps in the following:
2
3
4
-5
-
23=8
33= -
- = 64
-=-
63 = -
Ans: 33 = 27 , 43 = 64
(-5)3 = -125 , 6, 63 = 216
83 = 512 -9, (-9)3 = -729.
8
- =-
- = - 729
2. Find the cubed of the first five odd natural numbers and the cubes of
the first five even natural numbers. What can you say about parity of
the odd cubes and even cubes?
Ans:
Odd Cubes
Even Cubes
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
63 = 216
73 = 343
83 = 512
93 = 729
103 = 1000
Parity
Cubes of odd numbers is always odd.
Cubes of even numbers is always even.
3. How many perfect cubes you can find from 1 to 100 ? How many from –
100to 100?
Ans: 1 1to 100
-
4 perfect cubes
-100 to 100
-
4 perfect cubes
4. How many perfect cubes are there from 1 to 500? How many are perfect
square among cubes?
Ans: There are 7 perfect cubes from 1 to 500. Only 13 and 43 are the squares
among these cubes.
5. Find the cubes of 10, 30 , 100, 1000. What can you say about the zeros
at the end?
Ans: 103 = 1000
303 = 27000
1003 = 1000000
10003 = 10000000000
The number of zero of a cube are 3 times, the no. of zero of numbers.
6. What are the digits in the unit’s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8 , 9,
10? Is it possible to say that a number is not a perfect cube by looking at
the digit in unit’s place of the given number, just like you did for
squares?
Ans:
Numbers
The digit in the unit’s place
1
1
2
8
3
7
4
4
5
5
6
6
7
3
8
2
9
9
10
0
EXCERSISE 1.2.8
1. Find the cube root by prime factorization.
i)
10648
2 10648
2
5324
2
2662
11 1331
11
11
1
3
ii)
√ 10648 = 3√ 2 2 x 112 = 2x11
46656
2 46656
2 23328
= 22
2 11664
2
5832
2
2916
2
1458
9
729
9
81
9
9
1
3
√15625 = 3√ 53 x 53 = 5x5 = 25
2.Find the cube root of the following by looking at the last digit
and using estimation.
i)
9112 5
43 < 91 < 53
403 < 91125 < 503
45
ii)
16637 5
53 < 166 < 63
503 < 1166375 < 603
55
iii)
70496 9
83 < 704 < 93
803 < 704969 < 903
89
3.Find the nearest integer to the cube root of each of the following.
i)
331776
216000 < 331776 < 343000
603
√331776 lies between 60 and 70.
3
ii)
403
46656
27000 < 46656 < 64000
303
√466656 lies between 30 and 40.
3
iii)
403
373248
216000 < 373248 < 343000
√ 373248 lies between 60 and 70.
3
UNIT 2
CHAPTER – 2
EXERCISE 2.2.2
1. Resolve into factors:
i)
x2 + xy
ii)3x2 – 6x
x(x+y)
3x(x-2)
iii)(1.6)a2 – (0.8)a
iv) 5-10m – 20m
0.8a (2a – 1)
5(1-2m – 4n)
II. Factories.
i)
a2+ ax + ab + bx
a(a + x) (a + b)
(a + x) ( a + b)
ii)
3ac + 7bc – 3ad - 7bd
3ac + 7bc – 3ad – 7bd
c(3a + 7b) – d(3a + 7b)
(3a + 7b) (c – d)
iii)
3xy – 6yz – 3xt + 6zt
3y ( x-2z) – 3t(x-2z)
(x-2z) (3y- 3t)
iv)
y3 -3y3 + 2y – 6 – xy + 3x
y2 (y-3) + 2(y -3) – x(y-3)
(y-3) ( y2 + 2 – x)
3.Factorise
i)
4a2 – 25
(2a)2 – ( 5)2
(2a + 5) (2a – 5)
x2 – 9
16
ii)
(x)2 - ( 3 ) 2
4
( x + 3 ) ( x – 3)
4
4
iii)
x4 – y4
(x2)2- (y2)2
(x2 + y2) ( x2 – y2)
iv)
73 2 - 21
10
10
73+21
10 10
2
73– 21
10 10
= 94 52
10 10
= 1222
25
v)
(0.7)2 – (0.3)2
(0.7 + 0.3) (0.7 – 0.3)
(1) (0.4) = 0.4
vi)
(5a – 2b)2 – (2a – b)2
[5a – 2b + ( 2a – b )] [ 5a – 2b – ( 2a – b)]
(5a – 2b + 2a – b ) ( 5a – 2b – 2a + b)
(7a – 3b) ( 3a – b)
EXERCISE 2.2.3
1. In the following you are given the product pq and the
sum p+ q. Determine p and q.
i)
pq= 18 and p + q = 11
+9
Product = 18
+2
Sum = 11
∴ p = 9, q = 2
ii) pq= 32 and p + q = - 12
-8
Product = 32
-4
Sum= -12
∴ p = - 8 , q = -4
iii)
pq = - 24 and p + q = 2
+6
Product = 24
-4
Sum =
iv)pq = -12 and p + q = 11
+12
Product = - 12
-1
Sum =
11
∴p = + 12 , q = -1
v)pq = -6 and p + q = -5
-6
2
Product = - 6
+1
Sum= - 7
∴ p= - 11 , q = + 4
II.
Factorise
i)
x2 + 6x + 8
+4x
x2 + 4x + 2x + 8
x(x + 4) +2 (x + 4)
Product = 8x2
x(x+4) +2 (x + 4)
(x + 4) ( x + 2)
+2x
Sum = 6x
ii)
x2 +4x + 3
x2 + 3x + 1x + 3
x( x + 3) + 1 ( x + 3)
+3x
Product = 3x2
(x + 3) ( x + 1)
+ 1x
Sum = + 4x
iii)
a2 + 5a + 6
+3x
a2 + 3a + 2a + 6
a(a + 3) + 1 ( a + 3)
Product =6a2
(a + 3) (a + 2)
+2a
Sum = +5a
iv)
a2 – 5a + 6
-2a
a2 – 2a – 3a + 6
a(a – 2) – 3( a- 2)
Product = 6a2
(a – 2) ( a – 3)
-3a
Sum = - 5a
v)
a2 – 3a – 40
-8a
a2 – 8a + 5a – 40
a(a – 8) + 5(a -8)
Product = -40a2
(a – 8) ( a+ 5)
+5a
Sum = - 3a
vi)
x2 – x – 72
-9x
x2 – 8x + 5x – 40
x(x – 9) + 8 ( x – 9)
Product = - 72x2
( x – 9) ( x + 8 )
+8x
Sum=
-x
ADDITIONAL PROBLEM ON “ FACTORISATION”
1. Choose the correct answer :
a. 4a + 12 b is equal to
a. 4a
b. 12b
c. 4(a + 3b)
d. 3a
(c )
b. The perfect of two numbers is positive and their sum negative only
when
a. Both are positive
b. both are negative
c. one positive the other negative
d. one of them equal to zero
( b)
c. F a c t 2 + 6x + 8 , we get
a. ( x + 1) ( x + 8)
b. ( x + 6) ( x + 2)
c. ( x + 10 ) ( x + 2)
d. ( x + 4) ( x + 2)
(d)
d. The denominator of an algebraic fraction should not be
a. 1
b. 0
c. 4
d.7
(b)
e. If the sum of two integers is – 2 d their product is – 24 the number
are
a. 6 and 4
b. 6 and 4
c. – 6 and – 4
d. 6 and – 4
( b)
f . The difference ( 0.7) 2 – ( 0.3)2 simplifies to
a. 0.4
b. 0.04
c. 0.49
2. Factorise the following:
i)
x2 + 6x + 9
x2 + 3x + 3x + 9
x( x+ 3) + 3 ( x + 3)
(x + 3) ( x + 3)
= ( x + 3)2
ii)
1- 8x + 16 x2
16x2 – 8x + 1
16x2 – 4x – 4x +1
4x (4x – 1) – 1(4x – 1))
=(4x – 1) 4x – 1)
iii)
4x2 – 81y2
(2x)2 – ( 9y)2
(2x + 9y) ( 2x – 9y)
iv)
4a2 + 4ab + b 2
4a2 + 2ab + 2ab + b 2
2a(2a + b) + b(2a + b)
(2a + b) ( 2a + b)
v)
A2 b2 + c 2 d2 – a2 c2 – b2 d2
A2 b2 – a2 c2 + c2 d2 – b2 d2
d. 0.56
(a)
A2 (b2 –c 2) + a2 (c 2 – b2)
A2 ( b2 – c2 ) – d2 (b2 – c 2)
=( b2 – c2) ( a2 – d2)
3. Factories the following:
i) x2 + 7x + 12
+4x
x2 + 4x + 3x + 12
x( x + 4 ) + 3( x + 4)
P=12 x 2
(x + 4) ( x + 3)
+3x
Sum =
7x
ii)x2 + x – 12
x2 – 3x + 4x – 12
+4x
Product =12x
x(x-3) + 4(x – 3)
(x–3)(x+4)
-3
Sum
= +x
iii)x2 – 3x – 18
x2 – 6x + 3x - 18
-6x
Product = 18x2
x( x – 6 )+ 3 (x – 6)
( x – 3 ) ( x – 7)
+3x
Sum
=
iv)x2 + 4x – 192
x2 – 3x + 7x – 21
-3x
+7x
P= +21x2
x( x – 3) + 7 ( x – 3)
( x – 3 ) ( x – 7)
-3x
Sum =
v)x2 – 4x – 192
x2 – 16x + 12x – 192
+12x
P = + 192 x
x( x – 16 ) + 12 ( x – 16)
( x – 16) ( x + 12 )
+4x
-16x
Sum =
-4x
vi)x4 - 5x2 + 4
x4 – 4x2 – 1x2 + 4
x2 ( x2 – 4) – 1 ( x2 – 4)
-4x2
P = + 4x4
(x2 – 4) ( x2 – 1)
= [x2 – (2)2 ] [ x2 – ( 1)2]
-1 x2
Sum =
- 5x2
(x + 2) ( x – 2) ( x + 1) ( x -1)
vii)x4 – 13x2y2 + 36y4
x4 – 9x2y2 – 4x2 y2 + 36y2
x2(x2 – 9y) – 4y2 (x2 – 9y2)
(x2 – 9y2) (x2 – 4y2)
[x2 – (3y)2] [x2 – (2y)2]
( x + 3y) ( x – 3y) ( x + 2y) ( x – 2y)
4. Factorise the following:
-9x2y2
P=+36x4y4
-4x2y2
i)
2x2 + 7x +6
2x2 + 4x + 3x + 6
24x
Product = 12x2
2x( x+ 2) + 3 ( x + 2)
( x + 2) ( 2x + 3)
ii)
+3x
Sum =
3x2 – 17x + 20
3x2 – 12 x + 5x + 20
-22x
P = + 60x2
3x ( x – 4) – 5(x – 4)
( x – 4) ( 3x – 5)
iii)
iv)
v)
6x2 – 5x – 14
6x2 – 12x + 7x -14
6x (x-2) + 7 ( x-2)
(x – 2) (6x – 7)
4x2 + 12xy + 5y2
4x2 + 2xy + 10xy + 5y2
2x ( 2x+y) 5y (2x + y)
(2x + y) (2x + 5y)
4x4 – 5x2 +1
4x4 – 4x2 – 1x2 + 1
X2(x2 – 1) -1 ( +1x2 -1 )
(x2 – 1) (4x4 – 1)
[(x2) – (1)2] [(2x)2 – (1)2]
(x+1) (x -1)(2x+1) (2x-1)
7x
+ 5x
Sum = - 17x
-12xy
2 2
P= x y
Sum =
+ 7x
-5x
+2xy
P=20x2y2
+10xy
Sum= 12xy
-4x2
P=4x4
-1x2
Sum= -5x2
CHAPTER – 3
UNIT – 2
EXERCISE 3.2.1
1. Match the following
i)
(a) Equilateral triangle
ii)
(b) Acute angled triangle
iii)
( c) Right angled triangle
iv)
(c) Obtuse angled triangle
Ans: i. c
ii. d.
iii. a
iv. b.
2. Based on the sides , classify the following triangles ( figures not
drawn to the scales)
i)
ii)
3cm
5cm
7cm
4cm
Scalene Triangle
5cm
4cm
Scalene Triangle
iii).
iv)
6.5 cm
6cm
Scalene Triangle
3.5 cm
6.5cm
6.5cm
6cm
Isosceles Triangle
v).
vi)
3cm
5.6cm
2.5cm
4.3 cm
Scalene Triangle
vii).
3.2 cm
Scalene Triangle
viii).
9cm
4.1cm
5cm
3cm
3cm
3cm
6cm
Scalene Triangle
Equilatereal Triangle
ix).
x).
6cm
5cm
6cm
5cm
8cm
3.5cm
Isosceles Triangle
Isosceles Triangle
EXERCISE 3.2.2
1. In a triangle ABC , if A = 55˚ and B = 40˚, find C
Ans: In ∆ ABC
A + B +
C = 180˚ ( Theorem 3)
55˚ + 40˚ + C = 180˚
C = 180˚ - ( 55˚ + 40˚ )
C = 85˚
2. In a right angled triangle, if one of the other two angle is 35˚ , find the
remaining angle.
Ans:
A = 35˚ , B = 90˚ , ( In a right angled triangle, one angle should be
90˚), in a right angled triangle
A + B + C = 180˚ ( Theorem )
35˚ + 90˚ + C = 180˚
C = 180˚ - ( 35˚ + 90˚ )
C = 180˚ – 125˚
C = 55˚
3. If the vertex angle of an isosceles triangle is 50, find the other angles.
Ans: A = 50˚ , Let B and C be x ( Because in isosceles ∆ two angles are
equal)
A + B + C = 180˚ ( Theorem)
50˚ + x + x = 180˚
2x = 180˚ – 50˚
X = 130˚
2
X = 65˚
∴ A = 50˚ , B = 65˚ and C = 65˚
4. The angles of triangle are in the ratio 1:2:3 Determine the three angles.
Ans: 1:2:3
Let the angles be 1x, 2x and 3x
1x + 2 x + 3x = 180˚ ( Theorem 3)
6x = 180˚
x = 180˚
6
x = 30˚, 2x = 2 x 30 = 60˚
3x = 3 x 30 = 90˚
5. In the adjacent triangle ABC, find the value of x and calculate the
measure of all the angles of the triangle.
Ans: In∆ ABC
A + B + C = 180˚
( Theorem 3)
x+ 15 + x + 15 + x + 30˚ = 180˚
30x + 30 = 180˚
3x = 180˚ - 30
x = 150
3
x = 50˚
∴ A = x + 15 + 50 + 15 = 65˚
∴ B = x – 15 = 50 – 15 = 35˚
∴ C = x + 30 = 50 + 30 = 80˚
6. The angles of triangle are arranged in ascending order of their
magnitude. If the difference between two consecutive angles is 10˚ , find
the three angles.
Ans: Let the angles be x , x + 10 and x + 20
x + x +10 + 20 = 180˚ ( Theorem 3)
3x + 30 = 180˚
3x
x
x
= 180˚ – 30
= 150
3
= 50
∴ x + 10 = 50 + 10 = 60˚
∴ x + 20 = 50 + 20 = 70˚
∴ The angles are 50˚ , 60˚ and 70˚
EXERCISE 3.2.3
1. The exterior angles obtained on producing the base of a triangle both
ways are 104˚ and 136˚. Find the angels of the triangle.
Ans:
A
136
B
C
104
T
Given – ACT is triangle with the lines
BC and TM extending both ways.
BCA = 136˚ , ATM = 104˚
Proof : BCA + ACT = 180˚
ACT = 180˚ – 136˚ = 44˚
CTA + ATM = 180˚ ( Pro.1)
CTA = 180˚ – 104˚
M
CTA = 76˚
In∆ ACT, A + C + T = 180˚
( Interior angle theorem)
A + 76˚ = 180˚
A + 180˚ – ( 44˚ + 76˚ )
A + 180˚ – 120˚
A = 60˚
∴The Angles in the∆ ACT are,
A = 60˚ , C = 44˚ , T = 76˚
2. Sides BC, CA and AB of a triangle ABC are produced in an order,
forming exterior angles ACD, DAE and CBF. Show that ACD +
DAE + CBF = 360˚.
Ans:
E
E
A
B
D
C
F
ACB + ACD = 180˚ ( Proposition 1)
BAC + CAE = 180˚ ( Proposition 1)
FBC + ABC = 180˚ ( Proposition 1)
Adding above equations.
ACB + ACD + BAC + CAE + FBC + ABC = 360˚ + 160˚
ACD + CAE + CBF + { ACB + BAC + ABC } = 360˚ + 180˚
ACD + BAE + CBF + 180˚ ( Interior angle theorem ) = 360˚ + 180˚
ACD + BAE + CBF = 360˚ ( Axiom 3)
3. Compute the value of x in each of the following figures:
Ans:
i)
Given – ABC is an isosceles single and CD is extended
ABC = 50˚
A
To prove - ACD
Proof : In∆ ABC
50˚
C = B = 50˚ ( AB = AC )
B
x
C
D
BCA + ACD = 180˚ ( Proposition 1)
ACD = 180˚- 50˚
P
T
180˚
ACD = 130˚
∴ x = 130˚
S
160˚
x
Q
R
ii)
Given – PTR = 130˚ , SQT = 106˚
To prove – QRT
Proof : SQT + TQR = 180˚ ( Prop . 1)
TQR = = 180˚ - 106˚
TQR = 74˚
PTR + PTQ = 180˚ ( Prop .1 )
PTR = 180˚ - 130˚
PTR = 50˚
In ∆PTQ , R + T + Q = 180˚
( Interior angle theorem)
R = 180˚ - ( 50˚ + 74˚ )
R = 56 ˚
QRT = 56˚ , x = 56˚
iii)
Given – XRY = 65˚ , RST = 100˚
To Prove – RZS
Proof : ZRS = XRY = 65˚
(Vertically opposite angle)
ZSR + RST = 180˚ ( Prop . 1)
ZSR = 180˚ – 100˚
ZSR = 80˚
In ∆RZS , R + Z + S = 180˚ ( Interior angle theorem)
Z =180˚ - ( 65˚ + 80˚ )
x
y
65˚
Z = 180˚ – 145˚
R
Z = 35˚
x
RZS = 35˚
∴ X = 35˚
iv)
100˚
z
s
T
Given – ABE = 120˚ , BED = 112˚
To prove – BEC
A
Proof : ABE + EBC = 180˚ ( Proo.1) 120˚ B
EBC = 180˚ - 120˚
EBC = 60˚
EGB + BCD = 180˚ ( Prop.1)
ECB = 180˚ – 120˚
x
E
112˚
C
ECB = 68˚
In ∆ EBC , E + B + C = 180˚ ( Interior angle theorem)
E = 180˚ - ( 60˚ + 68˚ )
E = 52˚
BEC = 52˚
∴ x = 52˚
v)
Given – ACT is an isosceler triangle.
ATC = 20˚
To prove - ACN
Proof : In ∆ ACT , A = T = 20˚ ( AC = CT )
D
In ∆ ACT , A + C + T = 180˚ ( Interior angle theorem)
ACN
= 180˚ - ( 20˚ + 20˚ )
A
ACN = 140˚
ACN = 40˚
∴ x = 40˚
X
N
20˚
C
T
4. In the figure , QT ⏊ PR , TQR = 40˚ and SPR = 30˚ , Find TRS
and PSQ
Ans: In ∆ TRQ,
QTR = 90˚ ( data)
TQR = 40˚ (data)
TQR = 180˚ - ( 90˚ + 40˚ ) (Remaining angle)
= 180˚ - 130˚
TRQ = 50˚
In ∆ PSR,
Ext PSQ = SPR + PRS ( Exterior angle theorem)
= 30˚ + 50˚
∴Ext PSQ = 80˚
5. An exterior angle of triangle is 120 and one of the interior opposite
angles is 30. Find the other angles of the triangles.
Ans:
ACB + ACD = 180˚ ( Proposition 1)
ACB = 180˚ - 120˚
ACB = 60˚ ( Third angle)
In ABC,
A
A + B + C = 180˚
( Interior angle theorem)
A = 180˚ ( 30˚ + 60˚ )
30˚
A = 180˚ - 90˚
B
120˚
C
D
A = 90˚ ( Other interior opposite angle)
6. Find the sum of all the angles at the five vertices of the adjoining
star.
Ans: 2 + 4 + 6 = 180˚
(Interior angle theorem)
1 + 8 + 10 = 180˚ ( Th .3)
3
2
8
2 + 9 + 5 = 180˚ ( Th.3)
7
1 + 8 + 4 = 180˚ ( Th.3)
3
+ 7 + 5 = 180˚ ( Th.3)
9
4
10
6
1
5
Adding above equations,
2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 6 + 7 + 8 + 9 + 10
= 5 x 180
2 ( 1 + 2 + 3 + 4 + 5 ) + 6 + 7 + 8 + 9 + 10 = 900
2 (1 +2
+ 3 + 4 + 5 ) + 540˚ ( Sum of angles of a
pertragon is 540˚) = 900
2 ( 1 + 2 + 3 + 4 + 5 ) = 900 - 540
1 + 2 + 3 + 4 + 5 = 360
2
1 + 2 + 3 + 4 + 5 = 180˚
ADDITIONAL PROBLEM ON THEOREMS
1. Fill in the blanks to make the following statements true.
a. The sum of the angles of a triangle is ( 180˚ )
b. An exterior angle of triangle is equal to the sum of
____ opposite
angles. ( Interior )
c. An exterior angle of a triangle is always ____ than either of the interior
opposite angle. ( Larger)
d. A triangle cannot have more than _____ right angle. (1)
2. Choose the correct answer from the given alternatives:
1. In a triangle , ABC , A = 80˚ and AB = AC , then B is
a. 50˚
b. 60˚
c. 40˚
d.70˚
(a)
2. In right angled triangle, A is right angle and B = 35˚ , then C is __
a. 65˚
b. 55˚
c. 75˚
d.45˚
(d)
3. In a triangle ABC , A = C = 45˚ , then the triangle is ___
a. right angle
b. Acute angled triangle
c. Obtuse angle triangle
d. Equilateral triangle
(a)
4. In an equilateral triangle, each exterior angle is _____
a. 60˚
b . 90˚
c.120˚
d. 150˚
(c )
5. Sum of three exterior angles of a triangle is _____
a. Two right angles
b. Three right angles
c. One right angle
d. Four right angles
(d)