(i) sin \(\theta\) = 0.9858  = 0.9857 + 0.0001
From the sine table 0.9857 = 80^o 18'
Mean difference 1 = 2′
0.9858 = sin 80⁰20'
sin\(\theta\) = 0.9858 = sin 80⁰20'
\(\theta\) = 80⁰20'
(ii) cos \(\theta\) = 0.7656 =  0.7660 - 0.0004
From the natural cosine table
0.7660 = 40°0′
Mean difference 4 = 2′       
0.7656 = 40°0′         
cos \(\theta\) = 0.7656 = cos 40°2'
\(\theta\) = 40°2'                                                                                          

In the quadrilateral ABCD, join one of the diagonals, say AC.
Area of \(\triangle\)ABC = \(\frac{1}{2}\)\(\times\) base \(\times\) height
=\(\frac{1}{2}\)\(\times\)8\(\times\)15\(\times\) 60 cm²
By Pythagoras theorem, in right angled triangle ABC,
AC² = AB² + BC²
= 8² +15² = 64 + 225 = 289 cm
Therefore, AC =\(\sqrt{289}\) =17cm
Now, for\(\triangle\)ACD, let us consider a = 17 cm, b =12 cm, c =25 cm
then, s = \(\frac{a+b+c}{2}=\frac{17+12+25}{2}=\frac{54}{2}\) = 27cm

Area of \(\triangle\)ACD =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt{27(27-17)(27-12)(27-25)}\)
=\(\sqrt{27\times10\times15\times2}\)
=\(\sqrt{3\times3\times3\times2\times5\times5\times3\times2}\)
= 3 × 3 × 2 × 5 = 90cm²
Therefore, Area of quadrilateral ABCD
=Area of \(\triangle\)ABC + Area of \(\triangle\)ACD
= 60 + 90 = 150 cm²

The given system of equations are
3x − 4y = 10 \(\Rightarrow\) 3x − 4y −10 = 0 .....(1)
  4x + 3y = 5 \(\Rightarrow\) 4x + 3y − 5 = 0 .....(2)
For the cross multiplication method, we write the co-efficients as

\(\frac { x }{ (-4)(-5)-(3)(-10) } =\frac { y }{ (-10)(4)-(-5)(3) } =\frac { 1 }{ (3)(3)-(4)(-4) } \)
\(\frac { x }{ (20)-(-30) } =\frac { y }{ (-40)-(-15) } =\frac { 1 }{ (9)-(-16) } \)
\(\frac { x }{ 20+30 } =\frac { y }{ -40+15 } =\frac { 1 }{ 9+16 } \)
\(\frac { x }{ 50 } =\frac { y }{ -25 } =\frac { 1 }{ 25 } \)
Therefore, we get \(x=\frac { 50 }{ 25 } ;y=\frac { -25 }{ 25 } \)
x = 2; y = -1
Thus the solution is x = 2, y = –1.
Verification :
3x–4y = 10 ...(1)
3(2)–4(–1) = 10
6 + 4 = 10
10 = 10 True
4x + 3y = 5 ...(2)
4(2) + 3(–1) = 5
8–3 = 5
5 = 5 True

Let A(−2, −1) and B(4, 8) are the given points.

Let P(a,b) and Q(c,d) be the points of trisection of AB, so that AP = PQ = QB .
By the formula proved above,
P is the point.
\(\left( \frac { { x }_{ 2 }+{ 2x }_{ 1 } }{ 3 } ,\frac { { y }_{ 2 }+{ 2y }_{ 1 } }{ 3 } \right) =\left( \frac { 4+2(-2) }{ 3 } ,\frac { 8+2(-1) }{ 3 } \right) \) = (0, 2)
Q is the point
\(\left( \frac { { 2x }_{ 2 }+{ x }_{ 1 } }{ 3 } ,\frac { {2 y }_{ 2 }+{ y }_{ 1 } }{ 3 } \right) =\left( \frac { 2(4)-2 }{ 3 } ,\frac { 2(8)-1 }{ 3 } \right) \) = (2, 5)

(i) Let us prepare a table to find the ordered pairs of points for the line y = 3x −1.
We shall assume any value for x, for our convenience let us take −1, 0 and 1.
When x = −1, y = 3(–1)–1 = –4
When x = 0 , y = 3(0)–1 = –1
When x = 1, y = 3(1)–1 = 2

The points (x,y) to be plotted :
(−1, −4), (0, −1) and (1, 2).

(ii) Let us prepare a table to find the ordered pairs of points for the line y =\(\left( \frac { 2 }{ 3 } \right) x+3\)
Let us assume −3, 0, 3 as x values.
(why?)
When x = -3, \(y=\frac { 2 }{ 3 } (-3)+3=1\)
When x = 0, \(y=\frac { 2 }{ 3 } (0)+3=3\)
When x =3, \(y=\frac { 2 }{ 3 } (3)+3=5\)

The points (x, y) to be plotted: (-3, 1), (0, 3) and (3, 5).

(i) \(\sqrt { 10 } \) is an irrational number ( since 10 is not a perfect square number).
(ii) \(\sqrt { 49 } =7=\frac { 7 }{ 1 } \) a rational number(since 49 is a perfect square number).
(iii) 0.025 is a rational number (since it is a terminating decimal).
(iv) \(0.7\overline { 6 } \) = 0.7666…. is a rational number ( since it is a non – terminating and recurring decimal expansion).
(v) 2.505500555…. is an irrational number (since it is a non – terminating and non–recurring decimal).
(vi) \(\frac { \sqrt { 2 } }{ 2 } =\frac { \sqrt { 2 } }{ \sqrt { 2 } \times \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } \) is an irrational number ( since 2 is not a perfect square number).

Let T be the set of families buy Tamil news Papers
Let E be the set of families buy English news Papers
n(TUE) = 100
n(T) = 65 and n(E) = 55
Let "X" be the number families buy both newspapers.
By using venn-diagram
From the venn-diagrarn we get
65 - x + x + 55 - x = 100
120 - x = 100
120 -100 = x
20 = x
(i) Families buy both newspapers = 20
(ii) Families buy Tamil newspapers only = 65 - x
= 65 - 20
= 45
(iii) Families buy English newspapers only = 55 - x
= 55 - 20
= 35.

(i) (x³+x²-7x-3) ÷ (x-3)
Let p (x) = x³+x²-7x-3
q (x) = x - 3
To find the zero of x - 3 :
p(x) in standard form ((i.e.) descending order)
Co-efficients are 1 1 -7 -3

Quotient is:  x² +4x+5
Remainder is :12
(ii) (x³+2x²-x-4) ÷ (x+2)
p(x) = x³+2x²-x-4
Co-efficients are 1 2 -1 -4
To find zero of x+2, put x+2 = 0; x = -2

Quotient is: (x²-1)
Remainder is: -2
(iii) (3x³-2x²+7x-5) ÷ (x+3)
To find zero of the divisor (x + 3), put x + 3 = 0;  x = - 3
Dividend in Standard form 3x³-2x²+7x-5
Co-efficients are 3 -2 7 -5
Synthetic Division

Quotient is: 3x²-11x+40
Remainder: is -125
(iv) (8x⁴-2x²+6x+5) ÷ (4x+1)
To find zero of the divisor 4x + 1, put 4x + 1 = 0; 4x = -1; x = \(-\frac{1}{4}\)
Dividend in Standard form 8x⁴-2x²+6x+5
Co-efficients are 8 0 -2 6 5
Synthetic Division

8x⁴-2x²+6x+5 = \((x+\frac{1}{4})(8x^3-2x^2-\frac{3x}{2}+\frac{51}{8})+\frac{109}{32}\)
=\(\frac{4x+1}{ ̶4̶}\times ̶4̶(2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
= (4x+1)\((2x^3-\frac{x^2}{2}-\frac{3x}{8}+\frac{51}{32})+\frac{109}{32}\)
Quotient is: \(2x^2-{x^2\over2}-{3x\over8}+{51\over32}\)
Remainder is: \({109\over32}\)

Compare with ax² + bx + c
Here, a = 2, b = 15, c = −27
product ac = 2\(\times\)–27 = –54, sum b = 15

The required factors are –3 and 18
∴ we split the middle term as 18x and –3x
2x² + 15x - 27 = 2x² +  18x − 3x - 27
= 2x (x + 9) − 3(x + 9)
= (x + 9) (2x − 3)
Therefore, (x + 9) (2x − 3) and are the factors of 2x² + 15x - 27

(i) am+bm+cm
am+bm+cm
m(a+b+c) factored form
(ii) a³-a²b
a². a - a². b group in pairs
a² \(\times\) (a-b) factored form
(iii) 5a −10b −4bc +2ac
5a-10b+2ac-4bc
5(a-2b)+2c(a-2b)
(a-2b)(5+2c)
(iv) x + y −1−xy
x-1+y-xy
(x-1)+y(1-x)
(x-1)-y(x-1)   [(a-b)= -(b-a)
(x-1)(1-y)

We know that (x +a)(x +b)(x +c) = x³ + (a +b +c)x² + (ab +bc + ca) x +abc     .....(1)
(i) (x+5)(x+6)(x+4)
= x³ + (5 + 6 + 4) x²+ (30 + 24 + 20) x + (5)(6)(4)
= x² + 15x²+74x + 120
[Replace a by 5 b by 6 c by 4 in (1)]
(ii) (3x-1)(3x+2)(3x-4)
(3x)³+(-1+2-4)(3x)²+(-2-8+4)(3x)+(-1)(2)(-4)
= 27x³+(-3)9x²+(-6)(3x)+8
= 27x³- 27x²-18x + 8
[Repalce x by 3x, a by -1, b by 2, c by -4 in (1)]

(i) \(\sqrt{3}\times\sqrt{5}\times\sqrt{2}=\sqrt{3\times5\times2}=\sqrt{30}\)
(ii) \(\sqrt{35}\div\sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}\)
(iii) \(\sqrt[3]{27}\times\sqrt[3]{8}\times\sqrt[3]{125}=\sqrt[3]{27\times8\times125}=\sqrt[3]{3^3\times2^3\times5^3}=3\times2\times5=30\)
(iv) \((7\sqrt{a}-5\sqrt{b})(7\sqrt{a}+5\sqrt{b})=(7\sqrt{a})^2-(5\sqrt{b})^2=49a-25b\) = 21\(\sqrt 3\)
(v) \([\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}]\div\sqrt{\frac{16}{81}}\)

=\([\sqrt{\frac{15^2}{27^2}}-\sqrt{\frac{5^2}{12^2}}]\times\sqrt{\frac{9^2}{4^2}}\)
=\((\frac{15}{27}-\frac{5}{12})\times\frac{9}{4}=(\frac{5}{9}-\frac{5}{12})\times\frac{9}{4}\)
=\((\frac{20-15}{36})\times\frac{9}{4}=\frac{5}{ ̶3̶6̶}\times\frac{ ̶9̶}{4}=\frac{5}{16}\)

A - Chess
B - Carrom
C - Table Tennis

n(A) = 22
n(B) = 21
n(C) = 15
n(A⋂C) = 10
n(B∩C) = 8
n(A∩B∩C) = 6
y = 22 - (x + 6 + 4) = 22 - (x + 10)
= 22 - x - 10 = 12 - x
z = 21 - (x + 6 + 2) = 21 - (8 + x)
= 21 - 8 - x = 13 - x
y + z + 3 + x + 2 + 4 + 6 = 35

x = 40 - 35 = 5
(i) Number of students who pay only chess and Carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 - x = 12 - 5 = 7
(iii) Number of students who play only carrom = 13 - x = 13 - 5 = 8

Let f(x) = ax³ + 4x² + 3x –4 and g(x) = x³ – 4x + a,When f(x) is divided by (x–3), the remainder is f(3).
Now f(3) =a(3)³ + 4(3)² + 3(3) - 4
= 27a + 36 + 9 – 4
f(3) = 27a + 41 (1)
When g(x) is divided by (x–3), the remainder is g(3).
Now g(3) = 3³ – 4(3) + a
= 27 – 12+ a
= 15 + a (2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 + a
27a – a = 15 – 41
26a = –26
\(a={-26\over 26}=-1\)
Substituting a = –1,in f(3), we get
f(3) =
= – 27 + 41
f(3) = 14
∴ The remainder is 14.

Let the length of the reactangle be '' l ''
The Breadth of the reactangle  = x + 3
Area of rectangle = Length \(\times\) breadth
x² + 7x + 12  = l (x + 3)  or 
l = \(\frac { { x }^{ 2 }+7x+12 }{ x+3 } \)
= \(\frac { \left( x+4 \right) \left( x+3 \right) }{ x+3 } =x+4\)
Length of the rectangle = x + 4

(i) Angle between band c = 30° (vertically opposite angles)
(ii) Angle between d and e = 180° - 75° = 105° (sum of the adjacent angles of a parallelogram is 180°)
(iii) Angle between d and f = 75° (opposite angles of a parallelogram)
(iv) Angle between c and f = 180° - 75° = 105° (Adjacent angles of a parallelogram)

Let DF be x.
FC = 20 - x (ABCD is a parallelogram AB = DC).
Given EF divides the parallelogram ABCD into two regions with equal area.
∴ Area of a trapezium ADFE = Area of a trapezium EBCF
(One pair of opposite sides are parallel)
\(\frac { 1 }{ 2 } \)h(AE + DF) = \(\frac { 1 }{ 2 } \)h(EB+FC)
\(\frac { 1 }{ 2 } \)h(3+x) = \(\frac { 1 }{ 2 } \)h(17+20-x)
3+x = 37-x
2x = 37 - 3 = 34
x = 34/2 = 17
Length of DF = 17 cm.

(i) Both pairs of opposite angles are equal in size are square and rectangle.
(ii) Both pairs of opposite sides are equal in length are square and rhombus.
(iii) Each diagonal is an angle bisector of square and rectangle.
(iv) The diagonals of the following quadrilateral bisect each other: Parallelogram, rectangle, square, and rhombus.
(v) Each pair of consecutive angles is supplementary are square, rectangle, rhombus, parallelogram.
(vi) The diagonals are equal in case of square and rectangle.
(vii) Square, rectangle, rhombus, and parallelogram can be divided into two congruent triangles.

\(3.4\bar { 5 } \) = 3.45454545...
 = 3.4545 (correct to 4 decimal places)
The number lies between 3 and 4

ABCD is a parallelogram AP and BP are bisectors of consecutive angles ∠A and ∠B.
Since the consecutive angles of a parallelogram are supplementary
∠A + ∠B = 180⁰
\(\frac{1}{2}\angle{A} +\frac{1}{2}\angle{B}=\frac{180^{0}}{2}\)
 ⇒ ∠PAB + ∠PBA = 90^O
In ΔAPB,
∠PAB + ∠APB + ∠PBA = 180^O (angle sum property of triangle)
∠APB = 180^O – [∠PAB + ∠PBA]
= 180^O – 90^O = 90^O
Hence Proved.

\(\sqrt{2}=1.414\)
\(\sqrt{3}=1.732\)
The two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\) are 1.514 and 1.632

(i) To plot (2, 4), draw a vertical line at x = 2 and draw a horizontal line at y = 4. The intersection of these two lines is the position of (2, 4) in the Cartesian plane. Thus, the Point A (2, 4) is located in the I quadrant of Cartesian plane.
(ii) To plot (–3, 5), draw a vertical line at x = –3 and draw a horizontal line at y = 5. The intersection of these two lines is the position of (–3, 5) in the Cartesian plane. Thus, the Point B (–3, 5) is located in the II quadrant of Cartesian plane.
(iii) To plot (–4, –5), draw a vertical line at x = –4 and draw a horizontal line at y =  -5. The intersection of these two lines is the position of (-4, 5) in the Cartesian plane. Thus, the Point C (–4, –5) is located in the III quadrant of Cartesian plane.
(iv) To plot (4, –2), draw a vertical line at x = 4 and draw a horizontal line at y = –2. The Intersection of these two lines is the position of (4,–2) in the Cartesian plane. Thus, the Point D (4,–2) is located in the IV quadrant of Cartesian plane

Step 1: 
Draw the ΔABC with the given measurements.

Step 2:
Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at S which is the circumcentre.
Step 3:
S as centre and SA = SB = SC as radius, draw the Circumcircle to passes through A, B and C. Circumradius = 3.9 cm.

Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of (AB and BC) any two sides and let them meet at S which is the circumcenter.
Step 3: With S as centre and SA = SB = SC as radius draw the circumcircle to passes through A, B and C.


Circum radius = 4.3 cm.