When a coin is tossed thrice, the outcome will be

The sample space s = {(HHH), (THH), (HTH), (HHT), (HTT), (THT), (TTH), (TTT)}
n(S) = 3
Let A be the event of getting two consecutive tails
4 = {HTT, TTH, TTT}
n(A) = 3
\(\Rightarrow P=\frac { n\left\{ F \right\} }{ n\{ O\} } =\frac { 3 }{ 8 } \)
Probability of getting two consecutive tails = \(\frac{3}{8}\)

Largest value L = 67; Smallest value S =18
Range R = L = S = 67 - 18 = 49
Coefficient of range = \(\frac { L-S }{ L+S } \)
Coefficient of range = \(\frac { 67-18 }{ 67+18 } =\frac { 49 }{ 85 } \) = 0.576

(- 6, 1) and (-3, 2)
The slope \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2-1 }{ -3+6 } =\frac { 1 }{ 3 } \)

Here θ = 30⁰
Slope m = tan θ
Therefore, slope m = tan 30⁰ = \(\frac{1}{\sqrt{3}}\)

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

Given Set =  {(x, y) / y = x + 3, x \(\in \) {0, 1, 2, 3, 4, 5}}
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
When x = 2,y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
Relation R = {(0, 3), (1,4), (2,5), (3,6), (4,7), (5,8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Given that A = {1,3,5} and B = {2,3}
(i) A x B = {1,3,5} x {2,3} = {(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)}     ...(1)
B x A = {2,3} x {1,3,5} = {(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)}         ...(2)
(ii) From (1) and (2) we conclude that A x B ≠ B x A as (1,2) ≠ (2,1) and (1,3) ≠) (3,1). etc
(iii) n(A) = 3; n (B) = 2.
From (1) and (2) we observe that, n (A x B) = n (B x A) = 6;
we see that, n(A) x n(B) = 3 x 2 = 6 and n (B) x n (A) = 2 x 3 = 6
Hence, n (A x B) = n (B x A) = n(A) x n(B) = 6.
Thus, n(A x B) = n (B x A) = n(A) x n(B).

Statement
A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.
Proof

In \(\Delta ABC\) ,D is a point on AB and E is a point on AC
To prove : \(\cfrac { AD }{ DB } =\cfrac { AE }{ EC } \) 
Construction: Draw a line DE || BC

No.	Statement	Reason
1.	\(\angle ABC=\angle ADE=\angle 1\)	Corresponding angles are equal because DE || BC
2.	\(\angle ACB=\angle AED=\angle 2\)	Corresponding angles are equal because DE || BC
3.	\(\\ \angle DAE=\angle BAC=\angle 3\)	Both triangles have a common angle
4.	\(\Delta ABC\sim \Delta ADE\)	By AAA similarity
	\(\frac { AB }{ AD } =\frac { AC }{ CE } \)	Corresponding sides are proportional
	\(\frac { AD+DB }{ AD } =\frac { AE+EC }{ AE } \)	Split AB and AC using the points D and E.
	\(1+\frac { DB }{ AD } =1+\frac { EC }{ AE } \)	On simplification
	\(\frac { DB }{ AD } =\frac { EC }{ AE } \)	Cancelling 1 on both sides
	\(\frac { AD }{ DB } =\frac { AE }{ EC } \)	Taking reciprocals
		Hence proved

When we roll two dice, the sample space is given by
S = \(\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \)
n(S) = 36
(i) Let A be the event of getting the sum of outcome values equal to 4.
Then A = {(1, 3),(2, 2),(3, 1)}; n(A) = 3.
Probability of getting the sum of outcomes equal to 4 is  P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(ii) Let B be the event of getting the sum of outcome values greater than 10.
Then B = {(5,6),(6,5),(6,6)}; n(B) = 3
Probability of getting the sum of outcomes greater than 10 is P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(iii) Let C be the event of getting the sum of outcomes less than 13. Here all the outcomes have the sum value less than 13. Hence C = S
Therefore, n(C) = n(S) = 36
Probability of getting the sum value less than 13 is P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 36 } \) = 1

\(f(x)= \begin{cases}x+2 & \text { if } x>1 \\ 2 & \text { if }-1 \leq x \leq 1 \\ x-1 & \text { if }-3<x<-1\end{cases}\)
i) f(3) = 3 + 2 = 5
ii) f(0) = 2
iii) f(-1.5) = -1.5 - 1 = -2.5
iv) f(2) + f( -2) = ( 2 + 2 ) + ( -2 -1)
= 4 - 3 = 1

A = {1, 2, 3, 4} ; B = {2, 5, 8,11,14}; f(x) = 3x − 1
f(1) = 3(1) –1 = 3 – 1 = 2; f(2) = 3(2) –1 = 6 –1 = 5
f(3) = 3(3) –1 = 9 –1 = 8; f(4) = 4(3) –1 = 12 –1 = 11
(i) Arrow diagram
Let us represent the function f :A ⟶ B by an arrow diagram  

(ii) Table form
The given function f can be represented in a tabular form as given below

x	1	2	3	4
f(x)	2	5	8	11

(iii) Set of ordered pairs
The function f can be represented as a set of ordered pairs as
f = {(1,2),(2,5),(3,8),(4,11)}
(iv) Graphical form
In the adjacent xy -plane the points
(1,2), (2,5), (3,8), (4,11) are plotted (Fig.1.20).

Given vertices are (- 4, - 2),(- 3, k), (3, - 2) and (2,3) and area of quadrilateral is 28 sq. units.
Area of quadrilateral \(=\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\right. \left.\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \)
\(\frac{1}{2}\) [(- 4 - 3) (k - 3) - (- 3 - 2) (- 2 + 2)] = 28
(-7) (k - 3) - (- 5) (0) = 56
-7k + 21 = 56
-7k = 56 - 21 = 35
\(k=\frac{35}{-7}=-5\)
k = -5

Before determining the area of quadrilateral, plot the vertices in a graph.
Let the vertices be A(8, 6), B(5, 11), C(-5, 12) and D(-4, 3).
Therefore, area of the quadrilateral ABCD
=\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
=\(\frac{1}{2}\) { (80 + 60  - 15 - 24) - (30 - 55 - 48 + 24)}
=\(\frac{1}{2}\) {109 + 49 }
=\(\frac{1}{2}\) { 158 } = 79 sq. units

From the table, we found that as x increases, y also increases. Thus, the variation is a direct variation.
Let y = kx, where k is a constant of proportionality.
From the given values, we have
\(k=\frac{3.1}{1}=\frac{6.2}{2}=\frac{9.3}{3}=\frac{12.4}{4}=\ldots=3.1\)
When you plot the points (1, 3.1) (2, 6.2) (3, 9.3), (4, 12.4), (5, 15.5), you find the relation y = (3.1)x forms a straight-line graph.
Clearly, from the graph, when diameter is 6 cm, its circumference is 18.6 cm.

1. Table

x	24	12	8	6	4	3	2	1
y	1	2	3	4	6	8	12	24

2. Variation:
Indirect Variation
3. Equation
xy = k
xy = 24 x 1 = 12 x 2 = 8 x 3 =... = 12
xy = 24
4. Points
(24,1),(12,2),(8,3),(6,4)
(4,6),(3,8),(2,12),(1,24)
5. Solution
From the graph
(i) If x = 3, then, y = 8
(ii) If y = 6 then, x = 4

Given the radius of the circle = 4 cm

Length of the tangents PA = PB = 10.2 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 4cm,
(2) Draw a line OP = 11 cm.
(3) Draw a \(\bot '\) bisector of OP, which cuts at M,
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 10.2 cm.

Given a triangle LMN, we are required to construct another triangle whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the \(\triangle\)LMN

Steps of construction:
1. Constructed a LMN with any measurement
2. Drawn a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Located 5 points M₁, M₂, M₃, M₄, M₅ on MX so that
MM₁ = M₁M₂ = M₂M₃ = M₃M₄ = M₄M₅
4. Joined, M₅N and drawn a line through M₄ parallel to M₅N to intersect MN at N.
5. Drawn' a line through N, parallel to the line NL to intersect ML at L
Then L'MN' is the required triangle each of whose sides is four-fifth of the corresponding sides of LMN