S = {HH, HT, TH, TT}
(i) probability of two heads = \(\frac { 1 }{ 4 } \)
(ii) probability of one head = \(\frac { 1 }{ 2 } \)
(iii) probability of at least one head = \(\frac { 3 }{ 4 } \)
(iv) probability of at most one head = \(\frac { 3 }{ 4 } \).

\(x=\sqrt { { 13 }^{ 2 }-{ 12 }^{ 2 } } \)
\(\sqrt { 169-144 } =\sqrt { 25 } =5\)
\(sin\theta =\cfrac { 5 }{ 13 } ;cos\theta =\cfrac { 12 }{ 13 } ;tan\cfrac { 5 }{ 12 } ;cosec\theta =\cfrac { 13 }{ 5 } ;\)
\(sec\theta =\cfrac { 13 }{ 12 } ;cot\theta =\cfrac { 12 }{ 5 } \)

3a = 150
a = \(\frac { 150 }{ 3 } \) = 50 m

Centre of the circle \(=\left( \frac { 10+2 }{ 2 } ,\frac { 11+3 }{ 2 } \right) =(6,7)\)

\(\cfrac { 1 }{ 3 } x-\cfrac { 5 }{ 2 } =6\)
\(\cfrac { 1 }{ 3 } x=6+\cfrac { 5 }{ 2 } =\cfrac { 12+5 }{ 2 } =\cfrac { 17 }{ 2 } \)
\(x=\cfrac { 17 }{ 2 } \times 3=\cfrac { 51 }{ 2 } \)

When two dice are rolled
Sample space
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) Event of the sum is equal to 1 = 0
∴ Probability = \(\frac { 0 }{ n(S) } \) = 0
(ii) Event of the sum is equal to 4
B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
P(B)= \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 12 } \)
(iii) Event of the sum is equal to less than 13
C= {(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(C) = 36
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 6 } \).

In the figure AP 丄 CD and BQ 丄 CD
⇒ PQ = AB = 10, AP = BQ = h
Assume DP = x  ⇒ QC = 15 - (10 + x)
= 15 - 10 - x = 5 - x
In Δ ADP,  AP² = 7²- x²= 49 - x²
In Δ BQC, BQ² = 8²- (5 - x)² = 64 - (25 - 10x + x²) .
= 64 - 25 + 10x - x²
=39 - x²+  10x
Equating AP² = BQ²
49 - x² = 39 - x² + 10x

⇒ AP=\(\sqrt { { 7 }^{ 2 }-{ x }^{ 2 } } =\sqrt { 49-1^{ 2 } } =\sqrt { 48 } \)
∴ Area of the trapezium = \(\frac{1}{2}\) h(a+b) sq.units
=\(\frac { 1 }{ 2 } \times \sqrt { 48 } \times (10+15)=\frac { 1 }{ 2 } \times \sqrt { 3\times 16 } \times 25\)

= 2 \(\times\) 25 \(\times\) 1.732 = 86.6 m²

\(sin\ \theta =\cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(cos\ \theta =\cfrac { b }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(b\ sin\ \theta =b\times \cfrac { a }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } -\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ... (1)
\(a\ cos\ \theta =a\times \cfrac { b }{ \sqrt { a^{ 2 }+{ b }^{ 2 } } } =\cfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)    ...(2)
(1) = (2)\(\Rightarrow\) LHS = RHS
Hence proved

By By the Pythagoras theorem
BC² = AC² + AB²
= \(\left( \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \right) ^{ 2 }-{ a }^{ 2 }\)
= a²+b²-a²
= b²
BC = b

\(\frac { 11 }{ 3 } \) = 3.66
Arrow D best shows the position of \(\frac { 11 }{ 3 } \) on the number line.

27 = 3 \(\times\) 3 \(\times\) 3; 
therefore 27 = 3³

x^o=\(\cfrac { 1 }{ 2 } \angle BOC\)
=\(\cfrac { 1 }{ 2 } \) (50^o+30^o)
=\(\cfrac { 1 }{ 2 } \) \(\times\) 80 =40^o

The distance between the two chord
FE = OF + OE
OE = \(\sqrt { { 5 }^{ 2 }-3^{ 2 } } =\sqrt { 25-9 } \)
= \(\sqrt { 16 } \) = 4 cm
OF = \(\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } =\sqrt { 25-16 } =\sqrt { 9 } \) = 3 cm
\(\therefore\) DE = 4cm + 3cm = 7cm
Distance between the chord is 7 cm.

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 145+148+142+141+139+140 }{ 6 } =\frac { 855 }{ 6 } \)
x = 142.5 mm²

In the given data Rs. 800 occurs thrice. Hence the mode is Rs. 800

We know that (a+b)³ ≡ a³+3a²b+3ab²+b³
(2a+3b)³ = (2a)³+3(2a)²(3b)+3(2a)(3b)²+(3b)³
= 8a³+36a²b+54ab²+27b³

(0.1)^-4 = \(\left( \frac { 1 }{ 10 } \right) ^{ -4 }=\frac { 1 }{ \left( \frac { 1 }{ 10 } \right) ^{ 4 } } =\frac { 1 }{ \left( \frac { 1 }{ 10000 } \right) } \)

\((a)\frac { 2 }{ 3 } \)
\((b)\frac { 43 }{ 90 } \)
\((c)\frac { 1 }{ 999 } \)

(i) {5,9}
(ii) {1,2,3,4,5,7,8,9}

22

E = {x : x is an odd whole number less than z < 9}

p(x) = x³- 3x²+ 4x+ 50
g(x) = x - 3 
p(3) = 33 - 3(3)² + 4(3) + 50
=  27 - 27 + 12 + 50 = 62
\(\therefore\) The Remainder 62

not a set

It is not a trapezium

The shape of the geometrical figure Trapezium.

Let the dimensions of the cuboid be
l = 7x, b = 5x and h = 2x.
Given that volume of cuboid = 35840 cm³
l × b × h = 35840
(7x)(5x)(2x) = 35840
70x³ = 35840
x³ = \(\frac{35840}{70}\)
x² = 512
x =\(\sqrt [ 3 ]{ 8\times 8\times 8 } \)
x = 8 cm
Length of cuboid= 7x = 7 × 8 = 56cm
Breadth of cuboid= 5x = 5 × 8 = 40cm
Height of cuboid= 2x = 2 × 8 = 16 cm

In this experiment, each trial is a match where Team I faces Team II.
We are concerned about the winning status of Team I.
There are 10 trials in total; out of which Team I wins in the 1^st, 6^th and 9^th matches.
The relative frequency of Team I winning the matches = \(\frac { 3 }{ 10 } \)or 0.3

write 19⁰59' = 19⁰54' + 5'
From the table we have, cos19⁰54' = 0.9403 

Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 3 }{ 5 } \)
Slope of CD = \(\frac { change\ in\ y }{ change\ in\ x } =-\frac { 3 }{ 2 } \)
Slope of EF = \(\frac { 4 }{ 7 } \)
Slope of PQ = Undefined Slope of Rs. = 0.

\(\sqrt [ 5 ]{ 11 } +\sqrt [ 7 ]{ 11 } =\left( 5+7 \right) \sqrt { 11 } =12\sqrt { 11 } \)

Data in ascending order 27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63 
Number of terms = 13
If n is odd rhen Median = \(\left[\frac{\mathrm{N}+1}{2}\right]^{t h}\) observation
\(=\left(\frac{13+1}{2}\right)^{t h}\)observation
= 7^th observation
7^th observation = 32

A = {b,c,e,g,h}
B = {a,c,d,g,i}
C = {a,d,e,g,h}
B ∩ C = {a,d,g}
A - (B ∩ C) = {b,c,e,g,h} - {a,d,g} = {b,c,e,h}...(1)
A-B = {b,c,e,g,h} - {a,c,d,g,i} = {b,e,h}
A-C = {b,c,e,g,h} - {a,d,e,g,h} = {b,c}
(A - B)⋃(A - C) = {b,c,e,h}...(2)
From (1) and (2) it is verified that
A-(B ⋂ C) = (A - B) U (A - C)

Let the side AC be x.

Perimeter of a ∆ABC = 8m²-9m+7
AB+BC+AC = 8m²-9m+7
m² - m + 6 + 3m² - 2m - 4 + x = 8m²-9m+7
4m² - 3m + 2 + x = 8m²-9m+7
x = 8m²-9m+7-(4m²-3m+2)
= 8m²-9m+7-4m²+3m-2
= 4m²-6m+5
The third side AC = 4m²-6m+5

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac { { 360 }^{ 0 } }{ 30 } \)
=12⁰
3x =3 \(\times\) 12 = 36°
5x = 5 \(\times\) 12 = 60°
9x = 9 \(\times\) 12 = 108°
13x = 13 \(\times\) 12 = 156°
The required angles of quadrilateral are 36°, 60° 108° and 156°.

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M n S)
40 = 26 + 24 - n (M n S)
n (M n S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

(3, 5)

Let x = \(3.1\overline { 7 } \) = 3.1777 .......  (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 .........  (2)
(2)-(1)

\( =\frac{143}{45} \)

Given points are A(2, 3) and B(2, -4)
The point P lines on the x-axis.
∴ The point P is (x, 0)
AP  = \(\frac { 3 }{ 7 } \) AB
\(\frac { AP }{ AB } \) = \(\frac { 3 }{ 7 } \)
\(\frac { AP }{ PB } \) = \(\frac { 3 }{ 7 } \)  ........(1)
Distance =\(\sqrt { ({ x }_{ 2 }-{ x }_{ 1 })^{ 2 }+(y_{ 2 }-y_{ 1 })^{ 2 } } \)
AP = \(\sqrt { (x-2)^{ 2 }+(0-3)^{ 2 } } \)
= \(\sqrt { { x }^{ 2 }-4x+4+9 } \)
= \(\sqrt { { x }^{ 2 }-4x+13 } \)
BP =\(\sqrt { (x-2)^{ 2 }+(0+4)^{ 2 } } \)
= \(\sqrt { x^{ 2 }-4x+4+16 } \)
= \(\sqrt { x^{ 2 }-4x+20 } \)
From (1) we get
\(\frac { AP }{ PB } \) = \(\frac { 3 }{4 } \)
\(\frac { \sqrt { { x }^{ 2 }-4x+13 } }{ \sqrt { { x }^{ 2 }-4x+20 } } =\frac { 3 }{ 4 } \)    (Squaring on both sides)
\(\frac { { x }^{ 2 }-4x+13 }{ { x }^{ 2 }-4x+20 } =\frac { 9 }{ 16 } \)
16x²- 64x + 208 = 9x² - 36x + 180
7x² - 28x + 28 = 0
x²- 4x + 4 = 0
(x-2)²= 0
x-2 =0
x = 2
∴ The point P is (2, 0)

Let ABCD be the rhombus.
Its perimeter = 4 × side = 400 m
Therefore, each side of the rhombus = 100 m
Given the length of the diagonal AC = 120 m
In \(\triangle\)ABC, let a =100 m, b =100 m, c = 120 m

s = \(\frac{a+b+c}{2}=\frac{100+100+120}{2}\) = 160 m
Area of \(\triangle\)ABC =\(\sqrt{160(160-100)(160-100)(160-120)}\)
= \(\sqrt{160 \times 60\times 60 \times40}\)
= \(\sqrt{40 \times 2 \times \times2\times60\times60\times40}\)
= 40 × 2 × 60 = 4800 m²
Therefore, Area of the field ABCD = 2 × Area of \(\triangle\)ABC = 2 × 4800 = 9600 m²

Letp (x) x⁵ - 45x⁴ + 36x³ + 45x² - 36x- 1
Sum of co-efficients = 1 - 45 + 36 + 45 - 36 - 1 = 0
Thus x -1 is a factor of p (x)

\(\angle\)In the cyclic quadrilateral \(\angle\) A+\(\angle\)C = 180^o
y + 4^o + 3y^o + 8^o = 180^o
4y^o + 12^o = 80^o
4y^o = 180^o -12^o = 168
y =\(\cfrac { 168 }{ 4 } \) = 42
\(\angle\)B + \(\angle\)D = 180^o
8x + 12 = 180^o
8x = 180^o - 12^o = 168
x = \(\cfrac { 168 }{ 8 } \) = 21^o
\(\therefore\) \(\angle\)A = y + 4^o = 42^o + 4^o = 46^o
\(\angle\)C = 3y + 8 = 3 \(\times\) 42 + 8 = 126 + 8 = 134^o
\(\angle\)B = 3x + 6 = 3 \(\times\) 21 + 6 = 63 + 6 = 69^o
\(\angle\)D = 5x + 6 = 5 \(\times\) 21 + 6
= 105 + 6 = 111^o

Let Assumed mean be 25

Mean \(\bar{x}=A+\frac{\Sigma fd}{\Sigma f}\)
= \(25+\frac{270}{63}\)
= 25 + 4.29 = 29.29

To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Relational between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 5 } \) = \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
= \(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \)
Rational numbers between\(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \)
=\(\frac { 1 }{ 2 } \left( \frac { 10+7 }{ 20 } \right) \)
=\(\frac { 17 }{ 40 } \)
Rational number between \(\frac { 1 }{ 2 } \) and \(\frac { 7 }{ 20 } \) =\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } \times \frac { 17 }{ 40 } \right) \)
=\(\frac { 1 }{ 2 } \left( \frac { 20+17 }{ 40 } \right) \)
=\(\frac { 37 }{ 80 } \)
Thus the rational numbers are \(\frac { 7 }{ 20 } ,\frac { 17 }{ 40 } and\frac { 37 }{ 80 } \)

We have a parallelogram ABCD in which diagonals AC bisect ㄥA.
ㄥDAC = ㄥBAC
(i) To prove that AC bisects LC
∵ ABCD is a parallelogram
∴ AB II DC and AC is a transversal
∴ ㄥ1 = ㄥ3 (Alternate interior angle)     .........(1)
.Also BC II AD and AC is a transversal.
∴  ㄥ2 = ㄥ4 (Alternate interior angle)  ...........(2)
But AC bisects ㄥA
∴ ㄥ1 = ㄥ2
From (1), (2) and (3) we get
ㄥ3 = ㄥ4
∴ AC bisects ㄥC.
(ii) To prove that ABCD is a rhombus.
In ΔABC, we have ㄥ1 = ㄥ4 [ ∵ ㄥ1 = ㄥ2 = ㄥ4]
∴ BC = AB (side opposite to equal angles are equal) (4)
Similarly AD = DC            ....... (5)
But ABCD is a parallelogram AB = DC (Opposite sides of a parallelogram)        .......(6)
From (4), (5) and (6) we have AB = BC = CD = DA.
Thus ABCD is a rhombus.

Distance = \(\sqrt{(x_2+x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(3-3)^2+(2+2)^2}\)
=\(\sqrt{0+4^2}=\sqrt{16}=4\)
BC =\(\sqrt{(-1-3)^2+(2-2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
CD =\(\sqrt{(-1-1)^2+(2-2)^2}\)
\(\sqrt{0+(-4)^2}=\sqrt{16}=4\)
AD =\(\sqrt{(-1-3)^2+(-2+2)^2}\)
\(\sqrt{(-4)^2+0}=\sqrt{16}=4\)
AB = BC = CD = DA = 4. All the four sides are equal.

\(\therefore \)ABCD is a Rhombus .................(1)
Diagonal AC =\(\sqrt{(3+1)^2+(-2-2)^2}\)
\(=\sqrt{4^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal BD =\(\sqrt{(3+1)^2+(2+2)^2}\)
\(=\sqrt{4^2+4^2}=\sqrt{16+16}=\sqrt{32}\)
Diagonal AC = Diagonal BD =\(\sqrt{32}\) ................ (2)
From (1) and (2) we getABCD is a square.

In \(\triangle \)ABC,
AB = 4 cm, AC = 3 cm, \(\angle\)A =  90°

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 4 cm, AC = 3 cm, \(\angle\)A = 90°
Step 2: Draw perpendicular bisectors of any two sides (AB and AC) to find the mid points of AB and AC.
Step 3: Draw the medians CD and BE. Let them meet at G.
Step 4: G is the centroid of the given triangle.

Construction :
Step 1: Draw\(\triangle \)ABC with AB = 9 cm. ㄥA = 115°,ㄥB = 40°.
Step 2: Construct angle bisectors of any two angles (B and C). Let them meet at I. I is the incentre of \(\triangle \)ABC.
Step 3: Draw perpendicular from I to any side (BC) to meet BC at D.
Step 4: Draw incircle, with I as centre and ID a radius. Measures the in radius.