Number of cards n(S) = 52
No. of King cards n(A) = 4
No. of Queen cards n(B) = 4
No. of Jack cards n(C) = 4
Probability of drawing a King card
\(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Queen card
=\(\frac { n(B) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Jack card
=\(\frac { n(C) }{ n(S) } =\frac { 4 }{ 52 } \)
∴ The Probability of drawing a King or a Queen or a Jack from a deck of cards
= p(A) + P(B)+ P(C) =\(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 4+4+4 }{ 52 } =\frac { 12 }{ 52 } =\frac { 3 }{ 13 } \).

(S) Days in a week = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
n(S) = 7
∴ No. of days in week = 7
(A) Event of selecting Sunday = {Sunday}
n(A) = 1
∴ Probability of selecting Sunday = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 7 } \).

Area of the quadrilateral ABCD
= Area of the Δ ABD + Area of the Δ BCD
Sides of the triangle ABD are 13 cm, 14 cm, 15 cm.
s = \(\frac { 13+14+15 }{ 2 } \)cm
= \(\frac { 42 }{ 2 } \) = 21 cm
Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 21(21-13)(21-14)(21-15) } \)
=\(\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 7056 } \)=84 cm²
sides of the triangle
BCD are 12 cm, 9 cm, 15 cm
∴ s = \(\frac { 12+9+15 }{ 2 } =\frac { 36 }{ 2 } \) = 18 cm
Area =\(\sqrt { 18(18-2)(18-9)(18-15) } \)
=\(\sqrt { 18\times 6\times 9\times 3 } =\sqrt { 2916 } \) = 54 cm²
∴ Area of the quadrilateral
= 84 cm² + 54 cm² =138 cm²

x = 3, x = 5 
2x-y-4 = 0
y = 2x-4

The points to be plotted
(-1, - 6), (0, - 4), (1, - 2), (2, 0), (3,2), (5, 6)
Area of the quadrilateral = \(\cfrac { 1 }{ 2 } \times h(a+b)=\cfrac { 1 }{ 2 } \times 2(2\div 6)\)
Area of the quadrilateral = 8 sq.units

(i) sides: 10 cm, 24 cm, 26 m
Using Heron's formula
Area of the triangle = \(\sqrt { s(s-a)(s-b)(s-c) } \) sq. units
s = \(\frac { a+b+c }{ 2 } =\left( \frac { 10+24+26 }{ 2 } \right) cm=\frac { 60 }{ 2 } \) = 30 cm
∴ Area =\(\sqrt { 30(30-10)(0-24)(0-26) } \)
= \(\sqrt { 30\times 20\times 6\times 4 } =\sqrt { 600\times 24 } =\sqrt { 14400 } \) = 120 cm²
(ii) Sides: 1.8 m, 8m, 8.2 m
s = \(\\ \frac { 1.8+8+8.2 }{ 2 } =\frac { 18 }{ 2 } \) = 9
∴ Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 9(9-1.8)(9-8)(9-8.2) } =\sqrt { 9\times 7.2\times 0.8 } \)
= \(\sqrt { 51.84 } \) = 7.2 m²

L.H.S = cos 3A= cos 3 (30°)
= cos 90°
=0 ---- (1)
R.H.S = 4 cos³ A- 3 cosA
= 4 cos³ 30° - 3 cos 30°
= \(4\left( \cfrac { \sqrt { 3 } }{ 2 } \right) ^{ 3 }-3\times \cfrac { \sqrt { 3 } }{ 2 } \)

(1) = (2). Hence it is verified

In the figure,
\(\triangle\)OPM, \(\triangle\)OQN are similar triangles. In similar triangles the sides are in the same proportional.
\(\cfrac { QN }{ PM } =\cfrac { QO }{ PO } \)
\(\cfrac { h }{ 5 } =\cfrac { 35 }{ 25 } \)
\(h=\cfrac { 5\times 35 }{ 25 } \)

h = 7m

x₁ y₁    x₂ y₂   m : n 
A (4, -3), B (9, 7), 3 : 2
By section formula \(P\left( \frac { { mx }_{ 2 }+{ nx }_{ 1 } }{ m+n } ,\frac { { my }_{ 2 }+{ ny }_{ 1 } }{ m+n } \right) =P(x,y) \)

\(P(x,y)=\left( \frac { 3(9)+2(4) }{ 3+2 } ,\frac { 3(7)+2(-3) }{ 3+2 } \right) \)
\(=\left( \frac { 27+8 }{ 5 } ,\frac { 21-6 }{ 5 } \right) =\left( \frac { 35 }{ 5 } ,\frac { 15 }{ 5 } \right) =(7,3)\)

In a parallelogram diagonals bisect each other and diagonals are not equal.
Mid point of DE \(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
(-3, 6)\(=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \)
\(\frac { x_{ 1 }+{ x }_{ 2 } }{ 2 } =-3\ \ \ \ \ \frac { y_{ 1 }+y_{ 2 } }{ 2 } =6\)
x₁ + x₂ = -6........(1)   
y₁ + y₂ = 12.......(2)
Mid point of EF\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
(0, 7)=\(\left( \frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \right) \)
\(\frac { x_{ 2 }+{ x }_{ 3 } }{ 2 } =0\ \ \ \frac { y_{ 2 }+{ y }_{ 3 } }{ 2 } =7\)
x₂+x₃= 0..............(3)
Mid point of FD=\(\left( \frac { { x }_{ 3 }+{ x }_{ 1 } }{ 2 } ,\frac { { y }_{ 3 }+{ y }_{ 1 } }{ 2 } \right) =(-3,-2)\)
x₃ + x₁ = -6.......(5)     
y₃ + y₁ = -4..........(6)

∴ D (x₁, y₁) = (-6, -3)
Mid points of the diagonals are equal in parallelogram
∴ We have to prove this
Mid point of AC \(=\left( \frac { (-3)+(-3) }{ 2 } ,\frac { 6+(-2) }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
Mid Point of BD \(=\left( \frac { -6+0 }{ 2 } ,\frac { -3+7 }{ 2 } \right) =\left( \frac { -6 }{ 2 } ,\frac { 4 }{ 2 } \right) =(-3,2)\)
∴ Mid point of AC = Mid point of BD
∴ ABCD is a parallelogram

(i) 

(ii) 

(iii) 

\(\therefore\) The points to be plotted:
(0, -4), (1, -2), (2, 0), (3,2)

(iii)
2y = - 6
\(y=\cfrac { -6 }{ 2 } \Rightarrow y=3\)

(v) - 9 + 9 - 3x = 0 -9

\(x=\cfrac { 9 }{ 3 } \Rightarrow x=3\\ \)

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

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

Here, length (l) = 25 m, breadth (b) =15 m, height (h) = 5 m.
Area of four walls = LSA of cuboid
= 2(l + b) × h
= 2(25 +15) × 5
= 80 × 5 = 400 m²
Area of the floor = l × b

= 25 ×15
= 375 m²
Total renovating area of the hall = (Area of four walls + Area of the floor) = (400 + 375) m² = 775 m²
Therefore, cost of renovating at the rate of Rs.80 per m² = 80 × 775
= Rs. 62,000

Given the dimensions of the cuboid;
that is length (l) = 7.5 m, breadth (b) = 3 m and height (h) = 5 m.
TSA = 2(lb + bh + lh)

= 2[(7.5 × 3) + (3 × 5) + (7.5 × 5)]
= 2(22.5 +15 + 37.5)
= 2 × 75
= 150 m²
LSA = 2(l + b) × h
= 2(7.5 + 3) × 5
= 2 ×10.5 × 5
= 105 m²

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

Given linear equations are
kx + 2y = 3 ......(1)
2x - 3y = 1  .......(2)
\(\left[ \begin{matrix} { a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0 \\ { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0 \end{matrix} \right] \)
Here a₁ = k, b₁ = 2, a₂ = 2, b₂ = −3 ;
For unique solution we take \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \); therefore \(\frac { k }{ 2 } \neq \frac { 2 }{ -3 } \);\(k\neq \frac { 4 }{ -3 } \), that is \(k\neq -\frac { 4 }{ 3 } \)

(i) We know that cosec A =  sec(90⁰ A)
sec(90⁰ - A) sec(34⁰)
90⁰ - A =34⁰
We get A = 90° − 34°
A = 56⁰
(ii) We know that tan B = cot(90⁰ - B)
cot(90⁰ - B) = cot 47⁰
90⁰ - B = 47⁰
We get B = 90° − 47°
B = 43⁰

Let the vertices of a triangle be A(1, −6), B(−5, 2) and C(x₃, y₃)
Given the centroid of a triangle as (−2, 1) we get,
\(\frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } \) = -2
\(\frac { 1-5+{ x }_{ 3 } }{ 3 } \) = -2
−4 + x₃ = −6
x₃ = −2
\(\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \) = 1
\(\frac { 6+2+{ y }_{ 3 } }{ 3 } \) = 1
−4 + y₃ = 3
y₃ = 7
Therefore, third vertex is (−2, 7).

Given x – 2y = 7 …(1)
2x + 3y = 7 …(2)
When x = 5, y = −1 we get
From (1) x – 2y = 5 – 2(−1) = 5 + 2 = 7 which is RHS of (1)
From (2) 2x + 3y = 2(5) + 3(−1) = 10−3 = 7 which is RHS of (2)
Thus the values x = 5, y = −1 satisfy both (1) and (2) simultaneously. Therefore (5,−1) is a solution of the given equations.

Let BC = 13 and AB = 5
sec θ = \(\frac { hypotenuse }{ adjacentside } =\frac { BC }{ AB } =\frac { 13 }{ 5 } \)
By the Pythagoras theorem,
\(AC=\sqrt { { BC }^{ 2 }-{ AB }^{ 2 } } \)
= \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } \)
= \(\sqrt { 169-25 } \) = \(\sqrt { 144 } \) = 12
Therefore, \(sin\theta =\frac { AC }{ BC } =\frac { 12 }{ 13 } \) ; \(cos\theta =\frac { AB }{ BC } =\frac { 5 }{ 13 } \)
\(LHS=\frac { 2sin\theta -3cos\theta }{ 4sin\theta -9cos\theta } =\frac { 2\times \frac { 12 }{ 13 } =3\times \frac { 5 }{ 13 } }{ 4\times \frac { 12 }{ 13 } -9\times \frac { 5 }{ 13 } } =\frac { \frac { 24-15 }{ 13 } }{ \frac { 48-45 }{ 12 } } =\frac { 9 }{ 3 } =3\) = RHS

Given points are A(–3, 6) and B(1, –2), P(–2, 4) divide AB internally in the ratio m : n.
By section formula,
\(P(x, y)=P\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\)
= P(−2, 4) .......(1)
Here x₁ = −3, y₁ = 6, x₂ = 1, y₂ = −2
\((1)\Rightarrow \left( \frac { m(1)+n(-3) }{ m+n } ,\frac { m(-2)+n(6) }{ m+n } \right) \)
Equating x-coordinates, we get
\(\frac{m-3 n}{m+n}=-2\) or m− 3n = −2m− 2n
3m = n
\(\frac{m}{n}=\frac{1}{3}\)
m : n = 1: 3
Hence P divides AB internally in the ratio 1 : 3.

Let us form equations for the given statement.
Let us consider l and b as the length and breadth of the rectangle respectively.
Now let us frame the equation for the first statement
Perimeter of rectangle = 36
2(l+b) = 36
\(l+b=\frac { 36 }{ 2 } \)
l = 18 - b  .....(1)

Points: (2,16), (4,14), (5,13), (8,10)
The second statement states that the length is 2 metres more than three times the width which is a straight line written as l = 3b + 2 ... (2)
Now we shall form table for the above equation (2).

Points: (2, 8), (4, 14), (5,17), (8, 26)
The solution is the point that is common to both the lines. Here we find it to be (4,14).
We can give the solution to be b = 4, l = 14.

Verification :
2(l + b) = 36 ...(1)
2(14 + 4) = 36
2 \(\times\) 18 = 36
36 = 36 true
l = 3b + 2 ...(2)
14 = 3(4) +2
14 = 12 + 2
14 = 14 true