(a) Capital employed - Goodwill + current liabilities

A = { 1, 2, 3, 7}, B = { 3, 0, -1, 7}
A x B = {(1, 3), (1, 0), (1, - 1), (1, 7),(2,3), (2, 0), (2, -1), (2, 7), (3,3), (3, 0), (3, - 1), (3, 7), (7, 3)., (7, 0), (7, -1), (7 ,7)}
R₄ = {(7, - 1), (0, 3), (3, 3), (0, 7)}
In this (0, 3) and (0, 7) ∈ R₄
But (0, 3) and (0, 7) are not the elements of A x B.
Hence R₄ is not a relation from A to B.

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

A table

x	0	1	2	3
f(x)	1	3	5	7

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

Sum of the roots \(=\left( \frac { -3 }{ 2 } \right) \)
product of the roots (∝β) = (-1)
Required equation = x² - (∝ + β)x + ∝β = 0
\({ x }^{ 2 }-\left( \frac { -3 }{ 2 } \right) x-1=0\)
2x² + 3x - 2 = 0

\(\sqrt { ({ 4x }^{ 2 }-9x+2)(7{ x }^{ 2 }-13x-2)({ 28x }^{ 2 }-3x-1) } \)
\(=\sqrt { (x-2)(4x-1)(x-2)(7x+1)(4x-1) } \)
\(\\ =\sqrt { { (x-2) }^{ 2 }{ (4x-1) }^{ 2 }{ (7x+1) }^{ 2 } } \)
|(x - 2)(7x + 1)(4x - 1)|

AB = 4 cm,
AC = 6 cm,
BD  = 1.6 cm,
CD =  2.4 cm.

\( \frac{A B}{A C}=\frac{4}{6}=\frac{2}{3} \)
\(\frac{B D}{D C}=\frac{1.6}{2.4}=\frac{2}{3} \)
\(\frac{A B}{A C}=\frac{B D}{D C} \)
By the converse of the Angle Bisector theorem AD is the bisector of \(\angle\)A

Given straight line is 2x - 3y + 8 = 0
Slope m = \(\frac { -2 }{ -3 } =\frac { 2 }{ 3 } \)
Since product of slope is −1 for perpendicular lines, slope of any line perpendicular to 2x - 3y + 8 = 0 is \(\frac { -1 }{ \frac { 2 }{ 3 } } =\frac { -3 }{ 2 } \)

Here the largest value = 650
The smallest value = 400
 \(\therefore\) Range = L- S
= 650 - 400
= 250
Range R = 250

x² - 3x - 2 = 0
x² - 3x = 0 (Shifting the Constant to RHS)
x² - 3x + \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) = 2 + \({ \left( \frac { 3 }{ 2 } \right) }^{ 2 }\) (Add [\(\frac {1}{2}\)(co-efficient of x)]² to both sides)
\({ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }=\frac { 17 }{ 4 } \) (writing the LHS as complete square)
\(x-\frac { 3 }{ 2 } =\pm \frac { \sqrt { 17 } }{ 2 } \) (Taking the square root on both sides)
x = \(\frac { 3 }{ 2 } +\frac { \sqrt { 17 } }{ 2 } \) or x = \(\frac { 3 }{ 2 } -\frac { \sqrt { 17 } }{ 2 } \)
Therefore, x = \(\frac { 3+\sqrt { 17 } }{ 2 } , \frac { 3-\sqrt { 17 } }{ 2 } \)

It is a geometric progression with
\(a=5, r=\frac{-3}{5} \neq 1\)
Sum upto n terms S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-3}{5}-1}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-3-5}{5}}\)
\(=\frac{5\left[\left(\frac{-3}{5}\right)^{n}-1\right]}{\frac{-8}{5}}\)
\(=\frac{-25}{8}\left[\left(\frac{-3}{5}\right)^{n}-1\right] \)
\(=\frac{25}{8}\left[1-\left(\frac{-3}{5}\right)^{n}\right] \)

in the right triangle ABC [see figure. (a)]
tan \(\theta \) =\(\frac { opposite\ side\ }{ adjacent\ side\ } =\frac { 4 }{ 5 } \)
= tan^-1(0.8)
\(\theta \) = \(38.7°\)(since tan \(38.7°\) = 0.8011)
\(\angle \)BAC = \(38.7°\)

Here Largest value L = 28
Smallest value S = 18
Range R = L - S
R = 28 - 18 = 10 Years

\( \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} =\sec \theta+\tan \theta \)
\(\mathbf{L H S} =\sqrt{\frac{1+\sin \theta}{1-\sin \theta}} \)
\(=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}\)
[Multiplying the Numerator and denominator by \(\sqrt{1-\sin \theta}\)]
\( =\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\sqrt{\frac{\cos ^{2} \theta}{(1-\sin \theta)^{2}}} \quad\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{\cos \theta}{1-\sin \theta} \)
\(=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \)
[Multiplying Numerator and denominator by \(1+\sin \theta\)]
\( =\frac{\cos \theta(1+\sin \theta)}{1^{2}-\sin ^{2} \theta}=\frac{\cos \theta(1+\sin \theta)}{\cos ^{2} \theta} \)
\({\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]\left[1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \)
\(=\frac{1+\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta+\tan \theta=\text { RHS }\)

Given points (- 3, - 4), (7, 2) and (12, 5)
Let the points be A (- 3, - 4),8 (2, 2) and C (12, 5)
Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4-2}{-3-7}=\frac{-6}{-10}=\frac{3}{5} \)
Slope of BC = \(\frac{2-5}{7-12}=\frac{-3}{-5}=\frac{3}{5} \)
Slope of AB = Slope of BC
The points A, B and C are collinear

Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m²
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m²

First term = a = 20; common difference = d = 8
Arithmetic Progression is a, a + d , a + 2d , a + 3d,....
In this case , we get 20, 20 + 8, 20 + 2 (8), 20 + 3(8),....
So, the required A.P is 20, 28, 36, 44,.... 

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

In \(\Delta \)LOP & \(\Delta \)QMO,
\(\angle \)OLP = \(\angle \)MQO 90°
and \(\angle \)LOP =\(\angle \)OMQ (corresponding angles)
(by AA criterion of similarity)
\(\Delta \)LOP ~ \(\Delta \)QMO

f(x) = x² - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)² - 2 = 2; f( -1) = ( -1)² - 2 = -1
f(0) = (0)² - 2 = - 2 ; f(3) = (3)² - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.

        

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

σ  = 21.2, \(\bar { x } \)  = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

Let a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

Let the ten's digit of the number be x. It is given that the product of the digits is  12.
Unit's digit \(\frac{12}{x}\)
Number =10x+\(\frac{12}{x}\)
It 36 is added to the number the digits interchange their places.
\(\therefore 10x+\frac { 12 }{ x } +36=10\times \frac { 12 }{ x } +x\)
\(\Rightarrow 10x+\frac { 12 }{ x } +36=\frac { 120 }{ x } +x\)
\(\Rightarrow 9x-\frac { 108 }{ x } +36=0\)
⇒9x² - 108 + 36x = 0
⇒X²+ 4x - 12 = 0
⇒ (x + 6)(x - 2) = 0 (∵ (x + 6) ≠ 0 as x >0)
x=-6,2
But a number can never be (-ve). So, x = 2. The
number is 10x2+\(\frac{12}{2}\)=26

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n² - 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac{n}{2}\) 2(2a + (n -1)d)
1050 = \(\frac{14}{2}\)(20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 - 1) x 10
= 20
∴ 20th term = 200.

A = {1,2,3,4,5}
B = {1,2,3,4, ... }
f: A \(\rightarrow\)B, f(x) = x²
\(\therefore\)  f(1) = 1² = 1
f(2)  = 2²= 4
f(3) = 3² = 9
f(4)  = 4²= 16
f(5) = 5²= 25
\(\therefore\) Range of f = {1, 4, 9, 16, 25}
Elements in A have been connected with different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Given that intercepts are equal.
a = b
Equation of the line in intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{a}+\frac{y}{a}=1 \)
x + y = a
This passes through (- 8, 4)
-8 + 4 = a
a = -4
b = -4
Equation of the straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-4}+\frac{y}{-4}=1 \)
x + y + 4 = 0.

\(\frac { { b }^{ 2 }+3b-28 }{ { b }^{ 2 }+4b+4 } \div \frac { { b }^{ 2 }-49 }{ { b }^{ 2 }-5b-14 } \)
\(\frac { { b }^{ 2 }+3b-28 }{ { b }^{ 2 }+4b+4 } \times \frac { { b }^{ 2 }-5b-14 }{ { b }^{ 2 }-49 } \)

\(=\frac { b-4 }{ b+2 } \)

City A			City B
x1	d1 = x - \({ \overline { x } }_{ 1 }\)	d12	x2	d2 = x - \({ \overline { x } }_{ 2 }\)	d22
20	0	0	10	-5	25
22	2	4	20	5	25
19	-1	1	18	3	9
23	-3	9	12	-3	9
16	-4	16	15	0	0
100	6	30	75	0	68

\(\bar { { x }_{ 1 } } =\frac { 100 }{ 5 } =20\)
\(\sigma =\sqrt { \frac { \sum { { d }^{ 2 } } }{ n } } \)
\(=\sqrt { \frac { 30 }{ 5 } } \)
\(=\sqrt { 6 } \)
= 2.44
\(CV=\frac { \sigma }{ \bar { x } } \times 100\)
\(=\frac { 2.44 }{ 20 } \times 100\)
= 12.2
\(\bar { { x }_{ 2 } } =\frac { \sum { x } }{ n } =\frac { 75 }{ 5 } =15\)
\(\sigma =\sqrt { \frac { \sum { { d }^{ 2 } } }{ n } } \)
\(=\sqrt { \frac { 68 }{ 5 } } \)
\(=\sqrt { 13.6 } \)
= 3.68
\(CV=\frac { \sigma }{ \bar { x } } \times 100\)
\(=\frac { 3.86 }{ 15 } \times 100\)
\(=\frac { 368 }{ 15 } \)
C.V. of city A < C.V of city B.
∴ City A is more consistents.

F^o = (c^o x 1.8) + 32
σ_c = 5°C
σ_F = (1.8 x 5°C) . 9°F
Adding value to data doesn't affect standard deviation.
New variance = σ²_F = 81°F.

Mean x = 60
Standard deviation  = 15
Wrong scores = 40 and 27
Correct scores = 45 and 72.
Old Mean \(=\frac{\sum x_{i}}{n}=60\)
\(\frac{\sum x_{i}}{100}=60\)
\(Old\ \Sigma x_{i} \) = 60 x 100 = 6000
\(- Correct \ \Sigma x_{i}\) = 6000 - wrong scores + correct scores
= 6000 - (40 + 27) + (45 + 72)
= 6000 - 67 + 117
\(Correct \ \Sigma x_{i}\) = 6050
Correct Mean = \(\text { Correct } \frac{\Sigma x_{i}}{n}=\frac{6050}{100}\)
Correct Mean = 60.5
Old Standard deviation \(\sigma=15\)
\(\sigma =\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}} \)
\(15 =\sqrt{\frac{\sum x_{i}^{2}}{n}-(60)^{2}} \)
Squaring on both sides
\(225 =\frac{\Sigma x_{i}^{2}}{n}-3600 \)
\(225+3600 =\frac{\sum x_{i}^{2}}{100} \)
\(\frac{\text { old } \Sigma x_{i}^{2}}{100} =3825 \)
\(Old \ \Sigma x_{i}^{2} \) = 3825 x 100
\(Old \ \sum x_{i}^{2} \) = 382500
\(Correct \ \sum x_{i}^{2}\) = 382500 - (wrong scores)² + (correct scores)²
= 382500 - 40² - 27² + 45² + 72²
= 382500 - 1600 -729 + 2025 + 5184
= 382500 - 2329 + 2025 + 5184
= 389709 - 2329
\(Correct \ \sum x_{i}^{2}\) = 387380
Correct \(\sigma =\sqrt{\frac{387380}{100}-(60.5)^{2}} \)
\(=\sqrt{3873.80-3660.25} \)
\(=\sqrt{213.55} \)
\(\sigma =14.61 \)
Correct = 14.61; Correct Mean = 60.5

Let a man is standing on the deck of a ship at a point E
Such that AE = 40 m.
AE = BD = 40m
Also AB = ED
Let BC be the height of the hill.
(i) In right triangle \(\triangle\)ABE
\( \tan 30^{\circ} =\frac{A E}{A B} \)
\(\frac{1}{\sqrt{3}} =\frac{40}{A B} \)
\(A B =40 \sqrt{3} m \)
AB = 40 x 1.732 - 69.28 m
Distance of the hill from the ship = 69.28 m
(ii) In the right triangle \(\triangle\)CDE
\( \tan 60^{\circ} =\frac{C D}{E D} \)
\(\sqrt{3} =\frac{C D}{40 \sqrt{3}} \quad[\because A B=E D=40 \sqrt{3} \mathrm{~m}]\)
\(C D=40 \sqrt{3} \times \sqrt{3}=40 \times 3=120 \mathrm{~m}\)
Now height of the hill = BC = BD + DC
= 40 + 120 = 160m
Height of the hill = 160 m
Distance of the hill from ship = 69.28 m.

Given vertices of a quadrilateral are A (-5, 7),
B (- 4, k), C (- 1, - 6) and D (4, 5)
Area of quadrilateral is 72 sq. units
\(\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right]=72\)
( -5 + 1)(k - 5) - (-4 - 4)(7 + 6) = 144
-4k + 20 + 104 = 144
-4k = 144 - 124 = 20
\(k=\frac{-20}{4}=-5\)

In \(\Delta AOB\) OD is the bisector of \(\angle AOB\)
\(\therefore \frac { OA }{ OB } =\frac { AD }{ DB } \)          ...(1)
In \(\Delta BOC\) OE is the bisector of \(\angle BOC\)
\(\therefore \frac { OB }{ OC } =\frac { BE }{ EC } \)         ...(2)
In \(\Delta COA\) OF is the bisector of \(\angle COA\)
\(\therefore \frac { OC }{ OA } =\frac { CF }{ FA } \)        ...(3)
Multiplying the corresponding sides of (1), (2) and (3), we get
\(\frac{O A}{O B} \times \frac{O B}{O C} \times \frac{O C}{O A} \mid=\frac{A D}{D B} \times \frac{B E}{E C} \times \frac{C F}{F A}\)
\(1=\frac { AD }{ DB } \times \frac { BE }{ EC } \times \frac { CF }{ FA } \)
\(\Rightarrow \) DB x EC x FA = AD x BE x CF
 AD x BE x CF = DB x EC x FA
Hence proved.

Hollow metallic cylinder
External radius R = 4.3 cm
Internal radius r = 1.1 cm
Length = height = h = 4 cm
Volume \(=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \mathrm{h} \text { cu. units } \)
\(=\pi\left((4.3)^{2}-(1.1)^{2}\right)(4) \)
\(=\pi(18.49-1.21) 4 \)
\(=69.12 \pi \mathrm{cm} 3 \)
Solid cylinder
height h = 12 cm
radius r = ?
volume \(=\pi r^{2} h\ sq. units \)
\(=\pi r^{2}(12) \)
Given, Hollow cylinder is melted to form solid cylinder
Volume of cylinder = Volume of hollow cylinder
\(\pi r^{2}(12) =69.12 \pi \)
\(r^{2} =\frac{69.12}{12}=5.76 \)
r = 2.4 cm
Diameter of cylinder  = 2r = 4.8 cm

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.

Diameter of a solid cylinder = 1.4 cm
Radius of a solid cylinder = \(\frac{1.4}{2}=0.7 \mathrm{~cm}\)
Height of a solid cylinder = 2.4 cm
Volume of the cylinder = \(\pi r^{2} h \text { cu. units }\)
\(=\frac{22}{7} \times 0.7 \times 0.7 \times 2.4\)
Radius of cone = 0.7 cm
Height of cone = 2.4 cm
volume of cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 0.7 \times 0.7 \times 2.4 \)
Volume of the remaining solid = Volume of cylinder - Volume of cone
\(=\frac{22}{7} \times 0.7 \times 0.7 \times 2.4-\frac{1}{3} \times \frac{22}{7} \times 0.7 \times 0.7 \times 2.4 \)
\(=\frac{22}{7} \times 0.7 \times 0.7 \times 2.4\left(1-\frac{1}{3}\right) \)
\(=\frac{22}{7} \times 0.7 \times 0.7 \times 2.4 \times \frac{2}{3} \)
= 2.464 cm³

b=-42
a=49

f(x) = \(\sqrt { 1+\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } } \)
\( f(x)=\sqrt{1-t} \)
\(where\ t=\sqrt{1-\sqrt{1-x^{2}}}\)
\(1-t \geq 0 \)
\(t \leq 1 \)
\(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
Squaring \(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
\(-\sqrt{1-x^{2}} \leq 0 \)
\(\sqrt{1-x^{2}} \geq 0 \)
\(1-x^{2} \geq 0 \)
\(x^{2} \leq 1 \)
= x [-1, 1] i.e., {- 1, 0, 1}

Internal diameter = 20 cm
External diameter = 28 cm
Internal radius = 10 cm
External radius = 14 cm
Total surface area \(=\pi\left(3 \mathrm{R}^{2}+\mathrm{r}^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(3(14)^{2}+(10)^{2}\right) \)
\(=\frac{22}{7}[588+100] \)
\(=\frac{22}{7} \times 688 \)
\(=\frac{15136}{7} \mathrm{~cm}^{2} \)
Cost of painting per sq.cm = Rs 0.14
Total cost \(=\frac{15136}{7} \times 0.14\)
= Rs. 302.72

Given AB = 12 cm, BC = 6 cm
In \(\triangle\)APQ and \(\triangle\)QRC
(PB || QR, corresponding angles)
( PQ || BR, corresponding angles)
By AA criterion of similarity, we have
\(\triangle A P Q \sim \triangle Q R C \)
\( \Rightarrow \frac{A P}{Q R}=\frac{P Q}{R C}=\frac{A Q}{Q C} \)
\(\frac{A P}{Q R}=\frac{P Q}{R C} \)
\(\frac{P Q}{A P}=\frac{R C}{Q R} \)
Now in \(\triangle\)APQ and \(\triangle\)ABC , we have
By AA criterion of similarity,
we have \(\triangle\)APQ - \(\triangle\)ABC
\(\frac{A P}{A B}=\frac{P Q}{B C}=\frac{A Q}{A C} \)
\(\frac{A P}{A B}=\frac{P Q}{B C} \)
\(\frac{P Q}{A P}=\frac{B C}{A B} \)
\(\frac{P Q}{A P}=\frac{6}{12} \)
Since PQRB is a rhombus, PQ = QR = RB = PB
\(\frac{P Q}{A B-P B}=\frac{6}{12} \)
\(\frac{P Q}{A B-P Q}=\frac{6}{12} \)
\(\frac{P Q}{12-P Q}=\frac{6}{12} \)
12 PQ = 6(12 - PQ)
12 PQ = 72- 6 PQ
12 PQ + 6 PQ = 72
18 PQ = 72
PQ = \(\frac{72}{18}=4\)
PQ = 4 cm
Since PQ = RB we have
PQ = RB = 4 cm

\(\frac { sinA }{ secA+tanA-1 } +\frac { cosA }{ cosecA+cotA-1 } =1\) 
\(=\frac { sinA(cosecA+cotA-1)+cosA(secA+tanA-1) }{ (secA+tanA-1)(cosecA+cotA-1) } \)
=\(\frac { sin\ A \ cosec \ A \ + \ sin \ A \ cot \ A \ - \ sin \ A\ +cos \ A \ sec \ A \ + cos \ A \ tan \ A \ - \ cos \ A }{ (sec \ A \ + \ tan \ A-1)(cosec \ A \ + \ cot \ A-1) } \)
=\(\frac { 1+cosA-sinA+1+sinA-cosA }{ \left( \frac { 1 }{ cosA } +\frac { sinA }{ cosA } -1 \right) \left( \frac { 1 }{ sinA } +\frac { cosA }{ sinA } -1 \right) } \)
=\(\frac { 2 }{ \left( \frac { 1+sinA-cosA }{ cosA } \right) \left( \frac { 1+cosA-sinA }{ sinA } \right) } \)
=\(\frac { 2sinAcosA }{ (1+sinA-cosA)(1+cosA-sinA) } \)
=\(\frac { 2 \ sin \ A \ cos \ A }{ [1+(sin \ A- \ cos \ A)][1-(sin \ A-cos \ A)] } =\frac { 2sinAcosA }{ 1-(sin \ A- \ cos \ A{ ) }^{ 2 } } \)
=\(\frac { 2sinAcosA }{ 1-(si{ n }^{ 2 }A+co{ s }^{ 2 }A-2sinAcosA) } =\frac { 2sinAcosA }{ 1-(1-2sinAcosA) } \)
=\(\frac { 2sinAcosA }{ 1-1+2sinAcosA } =\frac { 2sinAcosA }{ 2sinAcosA } =1.\)

sin² Acos² B + cos² Asin² B + cos² A + cos² B+  sin² Asin² B
= sin² Acos² B + sin² Asin² B + cos² A + cos² B + sin²Asin² B
= sin² A(cos² B + sin² B) + cos² A(sin² B + cos²B)
= sin² A(1) + cos² A(1) (since sin² B + cos² B = 1) 
= sin² A + cos² A = 1

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP 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 and PB = 8.7 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