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

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

\(\bar { { x }_{ 1 } } \) =\(\frac { 140 }{ 5 } \) =28
σ₁ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 880 }{ 5 } } \)
= \(\sqrt { 176 } \)
= 13.27
CV₁ = \(\frac { \sigma }{ \bar { x_{ 1 } } } \) x 100
CV₁ = \(\frac { 13.27 }{ 28 } \) x 100
= 47.39%
CV₁ < CV₂

\(\bar { { x }_{ 2 } } =\frac { 150 }{ 5 } \) =30
σ₂ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 1600 }{ 5 } } \)
= \(\sqrt { 320 } \)
 =17.89
CV₁ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100
CV₂  = \(\frac { 17.89 }{ 30 } \) x 100
= 59.63%
∴ Team A is more consistent.

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 numbers be ∝, β
Sum of the roots = ∝ + β = 15          ...(1)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { 3 }{ 10 } \quad \quad \quad ...(2)\)
\(\\ \frac { +\alpha }{ \alpha \beta } =\frac { 3 }{ 10 } \)
10(∝+ β)= 3∝β    ....(3)
30∝β=10x15=150
Products of the roots =∝β=50 ....(4)
∴ From (1) & (4), we have
x²-15x+50=0
(x-10)(x-5)=0⇒x=10,5

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.

If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the midpoint of AC are same as the co-ordinates of the mid-point of BD. i.e.
\(\left( \frac { -2+4 }{ 2 } ,\frac { -1+b }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,\frac { 0+2 }{ 2 } \right) \)
⇒ \(\left( 1,\frac { b-1 }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,1 \right) \)
⇒ \(\frac { a+1 }{ 2 } \) = 1 ⇒ a + 1 = 2 ⇒ a = 1
⇒ \(\frac { b-1 }{ 2 } \) = 1 ⇒ b - 1 = 2 ⇒ b = 3

f(x) = 1 + x
g(x) = (2x - 1)
fog(x) = f(g(x) =  f(2x - 1)
= 1 + 2x - 1 = 2x ...(1)
gof(x) = g(f(x) = g(1 + x) = 2(1 + x) - 1
= 2 + 2x - 1
= 2x + 1      ...(2)
(1) \(\neq \)(2)
\(\therefore\) fog(x) \(\neq \) gof(x)
It is verified

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.

\(\frac{\tan ^{2} \theta-1}{\tan ^{2} \theta+1}=1-2 \cos ^{2} \theta\)
\( \text { LHS } =\frac{\tan ^{2} \theta-1}{\tan ^{2} \theta+1} \)
\(=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-1}{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1}\)
\( =\frac{\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta}} \)
\(= \frac{\sin ^{2} \theta-\cos ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)} \)
\(= \frac{\sin ^{2} \theta-\cos ^{2} \theta}{1}\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right] \)
\(=\left(1-\cos ^{2} \theta\right)-\cos ^{2} \theta \)
\(=1-\cos ^{2} \theta-\cos ^{2} \theta \)
\(= 1-2 \cos ^{2} \theta=\text { RHS } \)

Given intercepts are  -5, \(\frac 34\)
\(a=-5, b=\frac{3}{4}\)
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-5}+\frac{y}{\left(\frac{3}{4}\right)}=1 \)
\(\frac{x}{-5}+\frac{4 y}{3}=1 \)
3x - 20y = -15
3x - 20y + 15 = 0

LHS (AB)C
AB = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }=\left[ \begin{matrix} 1-2+2 & -1-1+6 \end{matrix} \right] =\left[ \begin{matrix} 1 & 4 \end{matrix} \right] \)
(AB)C = \({ \left[ \begin{matrix} 1 & 4 \end{matrix} \right] }_{ 1\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1+8 & 2-4 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(1)
RHS = A(BC)
BC = \({ \left[ \begin{matrix} 1 & -1 \\ 2 & 1 \\ 1 & 3 \end{matrix} \right] }_{ 3\times 2 }\times { \left[ \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right] }_{ 2\times 2 }=\left[ \begin{matrix} 1-2 & 2+1 \\ 2+2 & 4-1 \\ 1+6 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] \)
A(BC) = \({ \left[ \begin{matrix} 1 & -1 & 2 \end{matrix} \right] }_{ 1\times 3 }{ \left[ \begin{matrix} -1 & 3 \\ 4 & 3 \\ 7 & -1 \end{matrix} \right] }_{ 3\times 2 }\)
A(BC) = \(\left[ \begin{matrix} -1-4+14 & 3-3-2 \end{matrix} \right] =\left[ \begin{matrix} 9 & -2 \end{matrix} \right] \) ....(2)
From (1) and (2), (AB)C = A(BC).

Total number of cards = 52
n(S) = 52
Let 'A' be the event of getting a red king card
n(A) = 2
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let 'B' be the event of getting black queen card
n(B) = 2
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{2}{52}\)
Since A and B are mutually exclusive events
\( A \cap B =0 \)
\(\mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cup B) =\frac{2}{52}+\frac{2}{52}-0 \)
\(\mathrm{P}(A \cup B) =\frac{4}{52}=\frac{1}{13} \)
Probability of getting either a red kirrg or a black queen = \(\frac{1}{13} .\)

When two dice are rolled together, there will be 6 x 6 = 36 outcomes. Let S be the sample space. Then n(S) = 36.
Let A be the event of getting a doublet and B be the event of getting face sum 4.
Then A = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
B = {(1,3),(2,2),(3,1)
Therefore, A∩B = {(2,2)}
Then, n(A) = 6, n(B) = 3, n(A∩B) = 1
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 6 }{ 36 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 1 }{ 36 } \)
Therefore, P (getting a doublet or a total of 4) = P(AUB)
P(AUB) = P(A) + P(B) - P(A∩B)
=\(\frac { 6 }{ 36 } +\frac { 3 }{ 36 } -\frac { 1 }{ 36 } =\frac { 8 }{ 36 } =\frac { 2 }{ 9 } \)
Hence, the required probability is \(\frac { 2 }{ 9 } \).

\(\bar { { x }_{ 1 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 110 }{ 5 } \)
= 22
\({ \sigma }_{ 1 }=\sqrt { \frac { { \Sigma d }_{ 1 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 40 }{ 5 } } \)
= \(\sqrt { 8 } \)
\(\bar { { x }_{ 2 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 75 }{ 5 } \)
= 15
\({ \sigma }_{ 2 }=\sqrt { \frac { { \Sigma d }_{ 2 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 20 }{ 5 } } \)
= \(\sqrt { 4 } \)
= \(2\sqrt { 2 } \)
\(CV_{ 1 }=\frac { { { \sigma }_{ 1 } } }{ { x }_{ 1 } } \times 100\)
= \(\frac { 2.83 }{ 22 } \times 100\)
= 12.86
\({ CV }_{ 2 }=\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \times 100\)
= \(\frac { 2 }{ 15 } \times 100\)
= 13.33
\(\therefore\)Co-efficient of variation of City A is less than C.V of City B.
\(\therefore\) City A is more consistent.

Let AB be the statue CD be the building Given
BD = 35 m
DB = CE = 35 m
In right triangle ECB
\(\tan 34^{\circ}=\frac{E B}{C E}\)
EB = CE x tan 34^o
= 35 x 0.6745 m
EB = 23.61 m
In right AEC
\(\tan 24^{\circ}=\frac{A E}{E C}\)
AE = tan 24^o x EC
= 0.4452 x 35
AE = 15.58 m
(1) + (2) + AE + EB = 15.58 + 23.61
AB = 39.19 m
Height of the statue = 39.19 m.

Distance d = t²- 0.75t,
Given that d = 11.25 = t²- 0.75t.
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(=\frac { (+0.75)\pm \sqrt { { (-0.75) }^{ 2 }-4\times 1\times -11.25 } }{ 2\times 1 } \)
\(=\frac { +0.75\pm \sqrt { 0.5625+45 } }{ 2 } \)
\(=\frac { +0.75\pm \sqrt { 45.5625 } }{ 2 } \)
\(=\frac { +0.75\pm 6.75 }{ 2 } \)
\(=\frac { 7.50 }{ 2 } or\frac { -6 }{ 2 } \)
= 3.75 or-3 It is not possible
∴ t = 3.75 s.

Let AB the lighthouse. Let C and D be the positions of the two ships.\(\times \)
Then, AB = 200m.
\(\angle ACB=30°,\angle ADB=45°\)
In right triangles BAC, tan30°= \(\frac { AB }{ Ac } \) 
\(\frac { 1 }{ \sqrt { 3 } } =\frac { 200 }{ AC } \ gives\ AC=200\sqrt { 3 } \)      ...(1)
In the right triangle BAD,tan45°= \(\frac { AB }{ AD } \)
\(1=\frac { 200 }{ AD } \) gives AD = 200     ...(2)
Now, CD = AC + AD = \(200\sqrt { 3 } +200\) [by(1) and (2)]
CD = 200\((\sqrt { 3 } +1)\) = 200 x 2.732 = 546.4
Distance between two ships is 546.4m

Let \(\triangle\)ABC be B triangle points D, E, F are angular bisectors of \(\angle A, \angle B \text { and } \angle C\) respectively. By angular bisector theorem we have
\( \frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \mathrm{AB}=\frac{B D \times A C}{D C} \)
\(\frac{A C}{B C}=\frac{A F}{F B} \Rightarrow \mathrm{AC}=\frac{A F \times B C}{F B} \)
\(\frac{A E}{E C}=\frac{A B}{B C} \Rightarrow \mathrm{AB}=\frac{A E \times B C}{E C}\)
From (1) and (3), we have
\(\frac{B D \times A C}{D C}=\frac{A E \times B C}{E C}\)
Now substituting (2) in (4) we have
\( \frac{B D \times\left(\frac{A F \times B C}{F B}\right)}{D C} =\frac{A E \times B C}{E C} \)
\(\frac{B D \times A F \times B C}{D C \times F B} =\frac{A E \times B C}{E C} \)
\(B D \times A F \times E C =\frac{A E \times B C \times D C \times F B}{B C} \)
\(B D \times A F \times C E =E A \times F B \times D C \)
\(\therefore \frac{B D \times A F \times C E}{E A \times F B \times D C}=1\)
Hence by Ceva's theorem we conclude that the angle bisectors of a triangle are concurrent.

External surface area of the cock = Surface area of frustum + CSA of hemisphere


CSA of frustum = π(R + r)l sq. units.
Here R = \(\frac{5}{2}cm\)
\(r=\frac { 2 }{ 2 } =1cm\)
\(l=\sqrt { ({ R-r) }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { (2.5-1) }^{ 2 }+{ 6 }^{ 2 } } \)
\(=\sqrt { { 1.5 }^{ 2 }+36 } \)
\(=\sqrt { 2.25+36 } \)
\(=\sqrt { 38.25 } \)
\(\cong 6.18\)
∴ CSA of the frustum \(=\frac { 22 }{ 7 } \times 3.5\times 6.1=\frac { 469.7 }{ 7 } \)
= 67.1 cm²
CSA of hemisphere = 2π²
\(=2\times \frac { 22 }{ 7 } \times 1\times 1\)
= 6.28cm²
∴ Total external surface area
= 67.1 + 6.28
= 73.38cm²
= 73.39 cm²(approx.)

Radius of hemisphere = 6 cm
Volume of hemisphere \(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
\(=\frac{2}{3} \pi(6)^{3} \)
\(=\frac{2}{3} \pi(216) \)
\(=144 \pi \mathrm{cm}^{3} \)
base of cone = 6 cm
Height of the cone = 12 cm
Volume of the cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \pi(6)^{2}(12) \)
= 144 cm³
volume of the solid = Volume of cone + Volume of hemisphere
\(=144 \pi+144 \pi=288 \pi\)
Volume of water displaced
= Volume of the solid placed in the cylinder
\(=288 \pi=288 \times \frac{22}{7}\)
= 905.14 cm³

\(\therefore \sqrt { \frac { { x }^{ 2 } }{ { y }^{ 2 } } -10\frac { x }{ y } +27-10\frac { y }{ x } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } } =\)\(\left| \frac { x }{ y } -5+\frac { y }{ x } \right| \)

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

Since, \(\triangle\)QAQC is a right triangle at C, AQ² = AC² + QC²  ...(1)
Also, \(\triangle\)BPC is a right triangle at C, BP² = BC²+ CP²        ...(2)
\(\triangle\) ABCC is a right triangle at C, AB² = AC² + BC²              ....(3)
From (1) and (2), AQ²+  BP² = AC²+ QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)²+ ABC² + (2CP)²
=  4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²) (From equation (3))
4(AQ² + BP²) = 5AB²

Given points A (2, 2), B (- 2,- 3), C (1, - 3) and D (x, y)

Slope of a line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
Slope of AB = \(\frac{2+3}{2+2}=\frac{5}{4} \)
Slope of BC = \(\frac{-3+3}{-2-1}=0 \)
Slope of CD = \(-\frac{-3-y}{1-x} \)
Slope of AD = \(\frac{2-y}{2-x} \)
Since, the points form a parallelogram
AB is parallel to CD and BC is parallel to AD
Slope of AB = Slope of CD
\(\frac{5}{4}=\frac{-3-y}{1-x}\)
5(1 - x) = 4(-3 -y)
5 - 5x = -12 -4y
5x - 4y = 17
Slope of BC = Slope of AD
\(0=\frac{2-y}{2-x}\)
2 - y = 0
y = 2
Substituting in (1)
5x - 4(2) = 17
5x = 17 + 8 = 25
\(x=\frac{25}{5}=5\)
x = 5, y = 2

Construction
Step 1 : Draw a line segment QR = 5 cm.
Step 2 : At Q draw QE such that \(\angle\)RQE = 30^o.
Step 3 : At Q draw QF such that \(\angle EQF\) = 90^o
Step 4 : Draw the perpendicular bisector XY to QR which intersects QF at O and QR at G.
Step 5 : With O as centre and OQ as radius draw a circle.
Step 6: From G mark an arc in the line XY at M, such that GM = 42. cm.
Step 7 : Draw AB through M which is parallel to QR.
Step 8 : AB meets the circle at P and S.
Step 9 : Join QP and RP. Then\(\triangle\)PQR is the required triangle

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

A table

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}

x² + 7x + 10 here, a = -1, b = 7, c =10
if α and β are roots of the equation then,
α + β = \(\frac {-b}{a} = \frac {-7}{1}\) = -7; αβ = \(\frac {c}{a} = \frac {10}{1}\) = 10
\(\frac { \alpha }{ \beta } +\frac { \beta }{ \alpha } =\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 } }{ \alpha \beta } =\frac { { \left( \alpha +\beta \right) }^{ 2 }-2\alpha \beta }{ \alpha \beta } =\frac { 49-20 }{ 10 } =\frac { 29 }{ 10 } \)

Here \(\mathrm{a}=\frac{1}{3}, \mathrm{r}=\frac{t_{2}}{t_{1}}=\frac{1 / 9}{1 / 3}=\frac{1}{9} \times \frac{3}{1} \)
\(\mathrm{r}=\frac{1}{3} \)
n^th term of the G.P. \(t_{n}=a r^{n-1}\)
\(\frac{1}{2187} =\frac{1}{3} r^{n-1} \)
\(\frac{3}{2187} =r^{n-1} \)
\(\frac{1}{729} =r^{n-1} \)
\(\left(\frac{1}{3}\right)^{6} =r^{n-1} \)
\(\mathrm{n}-1=6 \quad\left[\because r=\frac{1}{3}\right]\)
n = 6 + 1 = 7
The number of terms in this G.P. is 7.

\(\frac { 4x }{ { x }^{ 2 }-1 } -\frac { x+1 }{ x-1 } =\frac { 4x }{ (x+1)(x-1) } -\frac { x+1 }{ (X-4) } \)
\(=\frac { 4x-(x+1)(x+1) }{ (x+1)(x-1) } \)
\(=\frac { -{ x }^{ 2 }+2x-1 }{ (x+1)(x-1) } =\frac { -({ x }^{ 2 }-2x+1) }{ (x+1)(x-1) } \)

\(=\frac { -(x-1) }{ (x+1) } =\frac { 1-x }{ 1+x } \)

In \(\Delta \)LOP & \(\Delta \)PRN,
we have
\(\angle \)PLO = \(\angle\)NRP = 90°
and \(\angle\)LPO = \(\angle\)PNR (corresponding angles)
By AA criterion of similarity
\(\Delta \)LOP~\(\Delta \)RPN

Here, the expression is defined for all real values of 'x'
i.e., x \(\epsilon\) R

5x ≡ 4 (mod 6)
5x - 4 = 6n for some integer n.
5x - 4 is a multiple of 6.
x = 2 is the least positive x.

The height of the first building AB = 60 m. Now, AB = MD = 60 m
Let the height of the second building CD = h. Distance BD = 140 m
Now, AM = BD = 140 m
From the diagram,
\(\angle \)XCA = 30° =\(\angle \)CAM
In right triangle AMC, tan30° = \(\frac { CM }{ Am } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { CM }{ 140 } \)
CM=\(\frac { 140 }{ \sqrt { 3 } } =\frac { 140\sqrt { 3 } }{ 3 } \)
\(=\frac { 140\times 1.732 }{ 3 } \)
CM = 80.78
Now, h = CD = CM + MD = 80.78 + 60 = 140.78
Therefore the height of the second building is 140.78 m

When two coins are tossed together, the sample space is
S = {HH, HT, TH, TT} n(S) = 4
Let A be the event of getting different faces on the coins.
A = {HT, TH}; n(A) = 2
Probability of getting different faces on the coins is P(A) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

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

Let the x intercept be ‘a’ and y intercept be ‘– a’.
The equation of the line in intercept form is \(\frac { x }{ a } +\frac { y }{ b } =1\)
gives \(\frac { x }{ a } +\frac { y }{ -a } =1\) (Here b = – a)
Therefore, x − y = a ...(1)
Since (1) passes through (5, 7)
Therefore, 5 - 7 = a gives a = − 2
Thus the required equation of the straight line is x − y = − 2 ; or x − y + 2 = 0

Given circumference = 484 cm
\(2 \pi r =484 \)
\(2 \times \frac{22}{7} \times r =484 \)
\(r =\frac{484 \times 7}{44}=77 \mathrm{~cm} \)
height h = 105 cm
Volume of cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \)
= 652190 cm³

XY + CZ = 17cm
XZ + YW = 26cm
We know that diagonals if a rectangle bisect each other and the diagonals have equal length.
\(\therefore \text { Each diagonal }=\frac{26}{2}=13 \mathrm{~cm}\)
i.e., XZ = 13 cm and YW = 13 cm
Also given XY + YZ = 17 cm
Squaring on both sides (XY + YZ)² = 17²
\((\mathrm{XY})^{2}+(\mathrm{YZ})^{2}+2 \times(\mathrm{XY}) \times(\mathrm{YZ})=289\)
By Pythagoras theorem (XY)² + (YZ)² = XZ²
\(\therefore[\mathrm{XZ}]^{2}+2(\mathrm{XY}) \times(\mathrm{YZ})=289\)
132 + 2 x length x breadth = 289
2 x Area = 289 - 169
\(\text { Area }=\frac{289-169}{2}=\frac{120}{2}\)
The possible length and breadth are
(1,60) (2,30) (3,20) (4, 15), (5, 12) (6, 10).
In this pair the length and breadth should satisfy Pythagoras theorem for diagonal.
5,12 is the possible length and breadth.

Given, Slope = 5, y intercept, c =  −9
Therefore, equation of a straight line is y = mx + c
y = 5x − 9 gives 5x − y − 9 = 0

f(x) = x²- 1, g(x) = x - 2
f(a) = a² - 1
Given, (g o f) (a) = 1
g[f(a)] = 1
g[ a² - 1] = 1
a² - 1 - 2 = 1
a² - 3 = 1
a² = 1 + 3 = 4
a = ± 2

Each person requires 22 m² of floor area.
Required base area = 22 x 4 = 88 m²
\(\pi r^{2} =88 \)
\(r^{2} =\frac{88 \times 7}{22}=4 \times 7 \)
\(r =2 \sqrt{7} \mathrm{~m} \)
slant height = 19 m
height of the tent, h \(=\sqrt{l^{2}-r^{2}}\)
\(=\sqrt{(191)^{2}-(2 \sqrt{7})^{2}} \)
\(=\sqrt{361-28} =\sqrt{330}=18.25 \mathrm{~m} \)
Height of the tent = 18.25 m

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

Construction:
Steps (1) Draw a line segment PQ = 6.8 cm
Steps (2) At P, draw PE such that \(\angle QPE={ 50 }^{ 0 }\)
Steps (3) At P, draw PF such that \(\angle FPE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to PQ, which intersects PF at 'O' and PQ at G.
Steps (5) With O as center and OP as radius drawn a circle.
Steps (6) From P, marked an arc of 5.2 cm on PQ at D
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID
Steps (8) ID produced meets the circle at R now joining PR and QR, we get the required \(\triangle\)PQR.