(a) Capital employed - Goodwill + current liabilities

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47 b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.

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

A table

It is not a function as the vertical line PQ cuts the graph at two points

\( \frac{\cos \theta}{1+\sin \theta} =\sec \theta-\tan \theta \)
\(\text { LHS } =\frac{\cos \theta}{1+\sin \theta} \)
\(=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta} \)
[Multiplying the Numerator and Denominator by 1 - sin\(\theta\)]
\( =\frac{\cos \theta(1-\sin \theta)}{1^{2}-\sin ^{2} \theta} \)
\(\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta} \)
\(\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{(1-\sin \theta)}{\cos \theta} \)
\(=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta-\tan \theta \)
= RHS

9y² - 6√2y + 2 = 0
a = 9, b = -6√2, c = 2
∆ = b² - 4ac
= (-6√2)² - 4 x 9 x 2
= 36 x 2 - 72
= 72 - 72 = 0
∴ The roots are real and equal.

Given points (1, - 5) and (4, 2)
Mid-point of the line joining the points (1, - 5), (4, 2)
\(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{1+4}{2}, \frac{-5+2}{2}\right) \)
\(=\left(\frac{5}{2},-\frac{3}{2}\right) \)
Equation of a straight line passing through
\(\left(\frac{5}{2},-\frac{3}{2}\right) \) and parallel to Y-axis is x = a
\(\text { i.e., } x=\frac{5}{2} \Rightarrow 2 x-5=0\)

t₂ - t₁ = 5\(\sqrt { 2 } \) - 3\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \)
t₃ - t₁ = 7\(\sqrt { 2 } \) - 5\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \)
t₄ - t₃ =9\(\sqrt { 2 } \) - 7\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \) 
Thus, the differences between consecutive terms are equal. Hence the terms of the sequence 3\(\sqrt { 2 } \), 5\(\sqrt { 2 } \), 7\(\sqrt { 2 } \), 9\(\sqrt { 2 } \) ..... are in A.P

 f(x) =\(\frac { x+6 }{ 3 } \), g(x) = 3 - x
fog(x) = f(g(x)) = f(3 - x)  = \(\frac { 3-x+6 }{ 3 } \) = \(\frac { 9-x }{ 3 } \) ..(1)
gof(x) = g(f(x)) = g\(\left( \frac { x+6 }{ 3 } \right) =3-\frac { x+6 }{ 3 } \)
=\(\frac { 9-x+6 }{ 3 } =\frac { 3-x }{ 3 } \)..(2)
f o g ≠ g o f

P(A) = 0.5, P(A ∩ B) = 0.3
We have P(A U B) ≤ 1
P(A) + P(B) - P(A ∩ B) ≤ 1
0.5 + P(B) - 0.3 ≤ 1
P(B) ≤ 1 - 0.2
P(B) ≤ 0.8
Therefore, probability of B getting selected is allmost 0.8.

Standard deviation = 12
Coefficient of variation = 25.6
Coefficient of variation\(\text { C.V }=\frac{\sigma}{x} \times 100 \%\)
\(25.6 =\frac{1.2}{\bar{x}} \times 100 \% \)
\(\bar{x} =\frac{1.2}{25.6} \times 100=4.687 \)
Mean \(\bar{x} =4.69 \)

If the range = 36.8 and
the smallest value =13.4
Range R = L - S
36.8 = L-13.4
= 36.8 + 13.4 = 50.2
The largest value L = 50.2

Compare x² + 2x - 2 with the standard form ax² + bx + c = 0
a = 1, b = 2, c = -2
x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\)
substituting the values of a, b and c in the formula we get,
x = \(\frac { -2\pm \sqrt { { \left( 2 \right) }^{ 2 }-4\left( 1 \right) \left( -2 \right) } }{ 2\left( 1 \right) } =\frac { -2\pm \sqrt { 12 } }{ 2 } =-1\pm \sqrt { 3 } \)
Therefore, x = \(-1+\sqrt { 3 } , -1-\sqrt { 3 } \)

Given straight line is 3x - 7y = 11
gives 3x - 7y - 11 = 0
Slope m = \(\frac { -3 }{ -7 } =\frac { 3 }{ 7 } \)
Since parallel line have same slopes, slope of any line parallel to
3x - 7y = 11 is \(\frac { 3 }{ 7 } \)

\(a=4, r=\frac{8}{4}=2\)
n^th term of a G.P. t_n = ar^n-1
8192 = 4r^n-1
\(r^{n-1} =\frac{8192}{4} \)
\(r^{n-1} =2048=2^{11} \)
n - 1 = 11
n = 11 + 1 = 12
Number of terms in the given G.P. is 12

Required Ratio = C.S.A. of the sphere: C.S.A. of the cylinder : C.S.A. of the cone
\(4\pi { r }^{ 2 }:2\pi rh:\pi rl,\ (l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 2r }^{ 2 } } =\sqrt { 2r } units)\)
\(=4:2:\sqrt { 2 } =2\sqrt { 2 } :\sqrt { 2 } :1\).

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

Th e slope of line r is m₁  \(=\frac { 8-2 }{ 5+2 } =\frac { 6 }{ 7 } \)
The slope of line θ is m₂  \(=\frac { 0-7 }{ -2+8 } =\frac { -7 }{ 6 } \)
The product of slopes \(=\frac { 6 }{ 7 } \times \frac { -7 }{ 6 } =-1\)
That is, m₁m₂ = -1

A = {1, 2, 3}, B = {4, 5, 6, 7}; f = {(1, 4),(2, 5),(3, 6)}
Then f is a function from A to B and for different elements in A, there are different images in B. Hence f is one–one function. Note that the element 7 in the co-domain does not have any pre-image in the domain. Hence f is not onto.
Therefore f is one–one but not an onto function.

The ratio of the corresponding sides of similar triangles is same as the ratio of their perimeters.
Since, \(\Delta ABC\sim \Delta PQR\)
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } =\frac { 36 }{ 24 } \)
\(\frac { AB }{ PQ } =\frac { 36 }{ 24 } \ \frac { AB }{ 10 } =\frac { 36 }{ 24 } \)
\(AB=\frac { 36\times 10 }{ 24 } =15cm\)

sec\(\theta \) - cos\(\theta \)  = \(\frac { 1 }{ cos\theta } -cos\theta =\frac { 1-co{ s }^{ 2 }\theta }{ cos\theta } \)
= \(\frac { si{ n }^{ 2 }\theta }{ cos\theta } \) [since 1 - cos²\(\theta \) = sin²\(\theta \)]
= \(\frac { sin\theta }{ cos\theta } \times sin\theta =tan\theta sin\theta \) 

cot 85° + cos 75°
= cot(90° - 5°) + cos(90° - 15°)
= tan 5° + sin 15°

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Total number of cards = 52
Event of selecting a spade card = A
Event of selecting a heart card = B
n(A) = 13, n(B) = 13
P(A) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \), P(B) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \).
A ⋂ B refers to a card is both spade and heart. It is not possible

A, B are exclusive events.
P(A,B) = P(A) + P(B) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

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 length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

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.

1,-1,-3, -5, ...
t₂ - t₁ = -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

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.

\(\sqrt { { 37x }^{ 2 }-28{ x }^{ 3 }+4{ x }^{ 4 }+42x+9 } =?\)

\(\therefore \sqrt { { 37x }^{ 2 }-28{ x }^{ 3 }+4{ x }^{ 4 }+42x+9 } \)
=|2x² - 7x - 3|

Let C and D are two ships.
Let AB be the height of the light house.
\(\mathrm{CD}=200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right) m\)
In the right ABD
\(\tan 60^{\circ}=\frac{A B}{B D} \)
\(\sqrt{3}=\frac{A B}{B D} \)
\(\mathrm{BD}=\frac{A B}{\sqrt{3}} \)
In the right triangle ABC
\(\tan 45^{\circ} =\frac{A B}{B C} \)
\(1 =\frac{A B}{B C} \)
\(B C =A B \)
\((1)+(2) \Rightarrow B D +B C=\frac{A B}{\sqrt{3}}+A B \)
\(\mathrm{CD}=A B\left(\frac{1}{\sqrt{3}}+1\right) \quad[\because \mathrm{CB}+\mathrm{BD}=\mathrm{CD}] \)
\(\frac{C D}{\left(\frac{1}{\sqrt{3}}+1\right)}=\mathrm{AB} \)
\(A B=\frac{200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}{\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)}\)
AB = 200 m
Height of the light house is 200 m.

Let AD be the electronic pole and BC be the tower.
From the right triangle \(\triangle\) ABC
\(\tan 60^{\circ}=\frac{B C}{A B}\)
\(\sqrt{3}=\frac{15}{A B}\)

\(A B=\frac{15}{\sqrt{3}}\)         ....(1)
\(\tan 30^{\circ}=\frac{C E}{D E}\)
\( \frac{1}{\sqrt{3}}=\frac{C B-E B}{A B} \quad[\because \mathrm{DE}=\mathrm{AB} \text { and } \mathrm{CE}=\mathrm{CB}-\mathrm{EB}] \)
\(\frac{1}{\sqrt{3}}= \frac{15-D A}{A B} \quad[\because \mathrm{EB}=\mathrm{DA}] \)
\(\mathrm{AB} =(15-\mathrm{AD}) \sqrt{3}\)         ...(2)
\( \frac{15}{\sqrt{3}} =(15-A D) \sqrt{3} \)
\(15 =(15-A D) \sqrt{3} \cdot \sqrt{3}\)
= (15 - AD)3
15 = 45 - 3AD
3AD = 45-15
3AD = 30
\(\mathrm{AD}=\frac{30}{3}=10 \mathrm{~m}\)
Height of the electric pole = 10 m

Let CF be the height of the man; AD be the height of the window; BC is the distance between the observer and the house.
From the right triangle \(\triangle\)CBD
\( \tan 45^{\circ} =\frac{D B}{B C} \)
\(1 =\frac{D B}{5} \)
DB = 5m     ...(1)
From the right triangle CBA
\( \tan 60^{\circ} =\frac{A B}{C B} \)
\(\sqrt{3} =\frac{A D+D B}{5} \)
\(5 \sqrt{3} =\mathrm{AD}+5 \quad[\because \text { from }(1) D B=5 \mathrm{~m}] \)
\(\mathrm{AD} =5 \sqrt{3}-5=5(\sqrt{3}-1) \)
\(\mathrm{AD} =5(1.732-1) \)
\( {[\text {Given } \sqrt{3}=1.732] }\)
= 5 x 0.732 = 3.660
Height of the window = 3.66 m

Given area of triangle is 5
Vertices of triangle are (2, 1), (3, - 2) and (x, y)
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=5\)
2(-2 - y) + 3(y - 1) + x(1 + 2) = 10
-4 - 2y + 3y - 3 + 3x = 10
3x + y = 17
Given that Y = x + 3, Substituting in (1)
3x + x + 3 = 17
4x = 4
x = 7/2
and y = 7/2 + 3
= 13/2
Third vertex is (7/2, 13/2)

Mean
\(\bar { x } =\frac { \Sigma { x }_{ i }{ f }_{ i } }{ N } =\frac { 432 }{ 48 } \) = 9 (Since N = Σf_i)
Standard deviation
σ =\(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } } =\sqrt { \frac { 124 }{ 48 } } =\sqrt { 2.58 } \)
σ ≃ 1.6
Assumed Mean method:
Let x₁, x₂, x₃, ......x₄ be the given data with frequencies f₁, f₂, f₃, ... fn respectively.
Let \(\bar { x } \) be their mean and A be the assumed mean
di = x_i - A
Standard deviation σ = \(\sqrt { \frac { \Sigma { f }_{ i }{ d }_{ i }^{ 2 } }{ N } -\left( \frac { \Sigma { f }_{ i }{ d }_{ i } }{ N } \right) ^{ 2 } } \).

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.

Given, n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
σ = \(\sqrt { \frac { 151 }{ 5 } -\left( \frac { 25 }{ 5 } \right) ^{ 2 } } =\sqrt { 30.2-25 } =\sqrt { 5.2 } \) ≃ 2.28
When we multiply each data by 4, we get the new values as 8, 12, 20, 28, 32.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 2416 }{ 5 } -\left( \frac { 100 }{ 5 } \right) ^{ 2 } } =\sqrt { 483.2-400 } =\sqrt { 83.2 } \)
σ = \(\sqrt { 16\times 5.2 } =4\sqrt { 5.2 } \) ≃ 9.12

Given that intercepts are in the ratio 2 : 5
\(\frac{a}{b} =\frac{2}{5} \)
\(a =\frac{2 b}{5} \)
Equation of the line in Intercepts form is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{x}{\left(\frac{2 b}{5}\right)}+\frac{y}{b}=1 \)
\(\frac{5 x}{2 b}+\frac{y}{b}=1 \)
5x + 2y = 2b
This passes through ( 1, - 4)
5(1) + 2(-4) = 2b
\(5-8=2 b \Rightarrow b=-\frac{3}{2} \)
\(a =\frac{2 b}{5}=\frac{2\left(-\frac{3}{2}\right)}{5}=-\frac{3}{5} \)
Equation of a straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \Rightarrow \frac{x}{\left(-\frac{3}{5}\right)}+\frac{y}{\left(-\frac{3}{2}\right)}=1\)
\(\frac{5 x}{-3}+\frac{2 y}{-3}=1\)
5x + 2y + 3 = 0.

Radius of sphere = 12 cm
Volume of sphere = \(\frac{4}{3} \pi r^{3} cu. units \)
= \(\frac{4}{3} \pi(12)^{3} \)
\(=2304 \pi \mathrm{cm}^{3}\)
Radius of cylinder = 8 cm
height = h cm
Volume of cylinder = \(\pi r^{2} h \) cu. units
= \(\pi(8)^{2} h \)
= \(64 \pi \mathrm{h} \mathrm{cm}^{3}\)
Given that sphere is melted and cast into a cylinder
Volume of cylinder = Volume of sphere
\(64 \pi h =2304 \pi \)
\(h =\frac{2304 \pi}{64 \pi}=36 \)
Height of the cylinder = 36 cm.

Diameter of the bowl = 14 cm
Radius r = 7 cm
Volume of hemisphere  \(=\frac{2}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \)
\(=\frac{2156}{3}=1718.67 \mathrm{~cm}^{3} \)
Radius of cylinder 'r' = 7 cm
Height 'h' = 6 cm
Volume of cylinder \(=\pi r^{2} h \text { cu. units } \)
\(=\frac{22}{7} \times 7 \times 7 \times 6 \)
= 924 cm³
capacity of the vessel = Volume of hemisphere + Volume of cylinder
= 718.67 + 924
= 1642.67 cm³

Because the given polynomial is a perfect square a - 16 = 0, b - 16 = 0
Therefore, a = 16, b = 16.

\(f(x)=\frac { x-1 }{ x+1 } ,x\neq 0\)
\(f(f(x))=f\left( \frac { x-1 }{ x+1 } \right) =\frac { \left( \frac { x-1 }{ x+1 } \right) -1 }{ \left( \frac { x-1 }{ x+1 } \right) +1 } \)
\(=\frac{\frac{\not x-1-x-1}{(\not x+1)}}{\frac{\not x-1+x+1}{(\not x+1)}}=\frac{-2}{2 x}=\frac{-1}{x}\)
Hence it is proved.

Radius of conical container = 10 m
Height of conical container = 15 m
Volume \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 10 \times 10 \times 15 \)
\(=\frac{11000}{7} m^{3} \)
water is released at the rate of 25 m³ / min
Time required to empty the container
\(= \frac{11000}{7} \)
\(= \frac{11000}{25}=62.85 \)
= 63 minutes (approx)

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

\(A=\left[ \begin{matrix} p & 0 \\ 0 & 2 \end{matrix} \right] ,B=\left[ \begin{matrix} 0 & -q \\ 1 & 0 \end{matrix} \right] ,C=\left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \)
\(BA={ C }^{ 2 }\Rightarrow \left[ \begin{matrix} 0 & -q \\ 1 & 0 \end{matrix} \right] \left[ \begin{matrix} p & 0 \\ 0 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \left[ \begin{matrix} 2 & -2 \\ 2 & 2 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 0 & -2q \\ p & 0 \end{matrix} \right] =\left[ \begin{matrix} (4-4) & (-4-4) \\ (4+4) & (-4+4) \end{matrix} \right] \)
\({ -2q=-8\\ q=4 }{ | }{ p=8\\ q=4 }\)

Construction:
(1) With O as the centre, draw a circle of radius 4.5 cm.
(2) Taken a point L on the circle through L drawn as chord LM
(3) Taken a point M distinct from L and N on the circle so that L, M, N are anti-clock wise direction. Joined LN and NM.
(4) Through 'L' drawn a tangent TT' such that \(\angle T L M=\angle M N L\)
(5) TT' is the required tangent.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 3 }{ 5 } \) of the corresponding sides of the triangle PQR

 
Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex P.
Located 3 points Q₁, Q₂ and Q₃ on QX so that Q Q₁ = Q₁ Q₂ = Q₂ Q₃
4. Joined Q₃R and drawn a line through Q₂ parallel to Q₃R to intersect QR at R'.
5. Drawn a line through R' parallel to the line RP to intersect QP at P'. Then PQR is the required triangle each of whose sides is two-thirds of the corresponding sides of PQR.