A' = {a, c, e, g}
(A')' = U - A'
= {a, b, c, d, e, f, g, h} - {a, c, e, g}
= {b, d, f, h}

{1, 3, 6, 9, 11, 12}

{3, 4, 6, 7, 9, 11}

D is a null set

R = The set of natural numbers less than 5

not a set

Let M be the set of all students take part in music.
Let D be the set of all students take part in drama.
n( M) = 25, n(D) = 30 and n(M\(\cap \)D) = 8
By using venn diagram
From the venn diagram we get.
(i) Number of students take part in only music = 17
(ii) Number of students take part in only drama = 22
(iii) Total number of students in the class = 17 + 8 + 22 = 47

Alter :
(i) The number of students who take part in only music
= n(M) - n(M\(\cap \)D)
= 25 - 8 = 17
(ii) The number of students who take part in only drama
= n(D) - n(M\(\cap \)D)
= 30 - 8 = 22
(iii) The total number of students
n(M∪D) = n(M) + n(D) - n(M\(\cap \)D)
= 25 + 30 - 8 = 47

Number of zeros = 0 (The curve is not intersecting the x-axis)

When y = 0
f(y) = 6y-3y² + 3
f(0) = 6(0) - 3(0)²+ 3
= 0 - 0+ 3 = 3

(A – B)′

Given A = {1, 2, 6}, B = {2, 3, 4}
A∪B = {1,2,3, 4,6}.

A= {a,b}
Subsets of A are \(\phi \),{a}, {b} and {a, b}.

Given PQ = PR
∴ ㄥR = 50° (opposite angles are equal)
Steps for construction:
Step 1: Draw the ΔABC with the given measures.
Step 2: Construct the perpendicular bisector of any two sides (QR arid PR) and let them meet at S. S is the circumcenter of Δ PQR.
Step 3: With S as centre SP = SQ = SR as radius. Draw the circumcircle.


Circum radius = 3.5 cm.
Construction of Orthocentre of a Triangle:
Orthocentre: The orthocentre is the point of concurrency of the altitudes of a triangle. Usually it is denoted by H.

(i) A = The set of shapes which have four sides. ,
B = {square, rectangle, rhombus}
B ⊆ A ( B is the subset of A)
(ii) A - Set of natural numbers; B = {set of all even numbers}
B ⊆ A
(iii) M = A set offamily members; N = {Daughter, Son}.
N ⊆​ ​​​​​​M

Let the point A  (-1, -1), (1, 1) and (\(-\sqrt{13},\sqrt{13}\))
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(1+1)^2+(1+1)^2}\)
\(\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}\)
BC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}\)
\(\sqrt{3+1+3+1}=\sqrt{8}\)
AC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}+1)^2}\)
\(=\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}}\)
\(=\sqrt{4+4}=\sqrt{8}\)

AB= BC=AC= \(\sqrt{8}\)
\(\therefore\)ABC is an equilateral triangle.

p(x₁) = 2x³ + ax² + 4x - 12
When it is divided by x - 3,
p(3) = 2(3)³ + a(3)² + 4(3) - 12
= 54 + 9a + 12 - 12
= 54 + 9a .....(R₁)
p(x2) = x³ + x² - 2x + a
When it is divided by x - 3,
p(3) = 3³ + 3² - 2(3) + a
= 27 + 9 -6 +a
= 30 + a  ......(R₂)
The given remainders are same (R₁ = R₂)
\(\therefore\) 54 + 9a = 30 + a
9a - a = 30 - 54
8a = -24
\(\therefore\)  a = - 24/8 = - 3
Consider R₂,
Remainder = 30 - 3
= 27

Let the height of the parallelogram be " h "
Base of the parallelogram = 5x + 4
Area of a parallelogram = 25x² - 16
Base \(\times\) Height = 25x² - 16
(5x + 4) \(\times\) h = 25x²-16
h = \(\frac { { 25 }x^{ 2 }-16 }{ 5x+4 } \) (or)
h = \(\frac { \left( { 5x }^{ 2 } \right) -{ 4 }^{ 2 } }{ 5x+4 } =\frac { \left( 5x+4 \right) \left( 5x-4 \right) }{ 5x+4 } \)
Height of the parallelogram = 5x - 4

Let \(\triangle APB\) and parallelogram ABCD lie on base AB and between parallels AB and DC.
Draw perpendicular line from A to CD, It meets at E.
 \(\therefore\) AE \(\perp\) DE. 
Area of parallellogram ABCD = Base \(\times\) height
AB \(\times\) AE
Let ABC be the triangle. Base AB vertical height A.E
Area of \(\Delta\)ABC = \(\frac{1}{2}\times\)base\(\times\)height
= \(\frac{1}{2}\times\) AB \(\times\) AE
Hence proved

In the given rhombus, the diagonal AC and BD bisect and intersect at right angles at O.
In right \(\triangle AOB\)
OB² = AB²- AO² (By pythagoras theorem)
= 13²-12²
= 169 - 144
OB² = 25
\(OB=\sqrt {25}=5\)
\(\therefore\) Diagonal BD = 5 \(\times\) 2 = 10 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.

Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x - 2 units
Area of the rectangle = (3x + 2 ) (3x - 2)
= 9x² - 6x + 6x - 4
= 9x²- 4
When x = 20
Area of the rectangle = 9(20)2 - 4
= 9(400) - 4
= 3600 - 4 = 3596 sq.units.

Here A = (7, 10), B = (-2, 5), C = (3,-4)
\(AB=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( -2-7 \right) ^{ 2 }+\left( 5-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -9 \right) ^{ 2 }+\left( -5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\(A{ B }^{ 2 }=106\)............1
\(BC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-\left( -2 \right) \right) ^{ 2 }+\left( -4-5 \right) ^{ 2 } } \)
\(=\sqrt { \left( 5 \right) ^{ 2 }+\left( -9 \right) ^{ 2 } } \)
\(=\sqrt { \left( 81+25 \right) } \)
\(=\sqrt { 106 } \)
\({ BC }^{ 2 }=106\)...............2
\(AC=\sqrt { \left( { x }_{ 2 }-{ x }_{ 1 } \right) ^{ 2 }+\left( y_{ 2 }-y_{ 1 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 3-7 \right) ^{ 2 }+\left( -4-10 \right) ^{ 2 } } \)
\(=\sqrt { \left( -4 \right) ^{ 2 }+\left( -14 \right) ^{ 2 } } \)
\(=\sqrt { \left( 16+196 \right) } \)
\(=\sqrt { 212 } \)
\({ AC }^{ 2 }=212\)..........3
From (1), (2) & (3) we get
AB²+BC² = 106 +106 = 212 = AC²
AB² + BC² = AC² 
ΔABC is a right angled triangle, right angled at B.