Let AD = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\( \frac{A D}{D B}=\frac{A E}{E C} \)
\(\frac{x}{7.2}=\frac{1.8}{5.4} \)
\(x=\frac{1.8 \times 7.2}{5.4}\)
x = 2.4
∴ AD = 2.4 cm

Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{1.5}{3}=\frac{1}{x}\)
\(x=\frac{3 \times 1}{1.5}\)
x =  2
∴ EC = 2 cm

Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So,   AP = BP
\(\Rightarrow\)   AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
       \(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\)   x² + 49 - 14x + y² + 1 - 2y
                = x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)              8x - 8y = 16
\(\Rightarrow\)                 x - y = 2
                                [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

Given points are A(0,0) and B(36,15) 
Now, using distance formula, we have
AB = \(\sqrt {(36-0)^{2}+(15-0)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

Let us take position of the town A as che origin and the position of the town B as the point B IS) in the coordinate axis as Shown in figure. 
Now, |AB|= \(\sqrt{(36^{2}+(15)^{2}}\)
=\(\sqrt{{36}^{2}+{15}^{2}}\) = \(\sqrt{1296+225}\)
=\(\sqrt{1521}\) =39 units

We have :
a₂ – a₁ = 11 – 5 = 6,
a₃ – a₂ = 17 – 11 = 6,
a₄ – a₃ = 23 – 17 = 6
As a_{k + 1} – a_k is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, a = 5 and d = 6.
Let 301 be a term, say, the nth term of this AP.
We know that
a_n = a + (n – 1) d
So,   301 = 5 + (n – 1) x 6
i.e.,   301 = 6n – 1
So,  \(n=\frac{302}{6}=\frac{151}{3}\)
But n should be a positive integer . So, 301 is not a term of the given list of numbers.

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have a_n = a + (n – 1) d
So,  a₁₀ = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.

We know that 105 is the first and 994 is the last three-digit numbers divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119,... , 994.
Clearly, the successive difference of the terms is same. So, above list of numbers forms an AP, with first term (a) = 105
and common difference (d) =112- 105 =7
Let there be n terms in the AP.
Then, nth term = 994
\(\Rightarrow\) 105 + (n - 1) 7 = 994 [\(\because\) a_n = a + (n -1)d]
 \(\Rightarrow\) 7 (n -1) = 994 - 105
\(\Rightarrow\) 7 (n - 1) = 889
\(\Rightarrow\) n = 127 + 1 = 128
Hence, there are 128 numbers of three-digit which are divisible by 7.

It is in AP with common difference d=−6+10=4, and a=−10
Next three terms are 
a+(5−1)d=6,
a+(6−1)d=10,
a+(7−1)d=14
Yes, d = 4 and next three terms are 6, 10, 14.

Here, a = 24, d = 21 – 24 = –3, S_n = 78. We need to find n.
We know that    \(\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]\)
So,  \(78=\frac{n}{2}[48+(n-1)(-3)]=\frac{n}{2}[51-3 n]\)
or   3n² – 51n + 156 = 0
or    n² – 17n + 52 = 0
or    (n – 4) (n – 13) = 0
or    n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.

In \(\Delta\)ACB,    LM || CB         [given]
\(\Rightarrow \quad \frac{A M}{M B}=\frac{A L}{L C}\)    ....(i)
[ by basic proportionality theorem]
In \(\Delta\)ACD, LN ||CD    [given]
\(\Rightarrow \quad \frac{A N}{N D}=\frac{A L}{L C}\)    ....(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac{A M}{M B}=\frac{A N}{N D} \Rightarrow \frac{M B}{A M}=\frac{N D}{A N}\)
[on taking reciprocal of the terms]
\(\Rightarrow \quad \frac{M B}{A M}+1=\frac{N D}{A N}\) + 1  [adding 1 on both sides]
\(\begin{aligned} & \Rightarrow \quad \frac{M B+A M}{A M}=\frac{N D+A N}{A N} \\ \end{aligned}\)

\(\begin{aligned} \Rightarrow \quad \frac{A M}{A M+M B}=\frac{A N}{A N+N D} \end{aligned}\)
[on taking reciprocal of the terms]
\(\therefore \quad \frac{A M}{A B}=\frac{A N}{A D} \quad\left[\begin{array}{c} \because A D=A N+N D \\ \text { and } A B=A M+M B \end{array}\right]\)
Hence proved.

In \(\triangle PQR, \angle 1=\angle 2\)  [given]
\(\Rightarrow \) PR = PQ  [since, sides opposite to equal angles of a triangle are not equal]
Given that, \(\frac { QR }{ QS } =\frac { QT }{ PR } \Rightarrow \quad \frac { QR }{ QS } =\frac { QT }{ PQ } \)
[\(\because \)  PQ = PR, proved above]
\(\Rightarrow \frac { QS }{ QR } =\frac { PQ }{ QT } \)  
[on taking of the terms reciprocals]  ... (i)
In \(\triangle PQS\) and \(\triangle TQR\), we have
\(\angle PQS=\angle TQR\)     [common]
and \(\Rightarrow \frac { QS }{ QR } =\frac { QP }{ QT } \)  [from Eq.(i)]
\(\therefore \triangle PQS\sim \triangle TQR\)    [by SAS similarity criterion]
Hence proved.

Let us join AC to intersect EF at G

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in \(\Delta\) ADC,
EG || DC (As EF || DC)
So, \(\frac{AE}{ED}=\frac{AG}{GC}\)
Similarly, from  \(\Delta\)CAB,
\(\begin{aligned} & \frac{C G}{A G}=\frac{C F}{B F} \\ \end{aligned}\)
\(\begin{aligned} & \frac{A G}{G C}=\frac{B F}{F C} \end{aligned}\)
Therefore, from (1) and (2),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)

No

AB²=(-2-7)²+(5-10)² = (-9)²+(-5)² =81+25 =106
BC²=(3-(-2))²+(-4-5)²=(5)²+(-9)²=25+81=106
AC²=(3-7)²+(-4-10)²=(4)^2₊(14)²=16+196=212
Since AB²+BC²=AC²
∴ ABC is a right triangle.
AB = \(\sqrt { 106 } \) and BC = \(\sqrt { 106 } \)
∵ AB = BC
∴ ABC is an isosceles right triangle. 

A(1,2), B(5,4), C(3,8) and D(-1, 6)
\(AB=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20};\ BC=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
\(CD=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}=\sqrt{20};\ DA=\sqrt{(-2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}\)
Here AB=BC=CA=DA
\(AC=\sqrt{2^2+6^2}=\sqrt{40}\ and\ BD=\sqrt{(-6)^2+(2)^2}=\sqrt{36+4}=\sqrt{40}\)
All sides of quadrilateral are equal and diagonals are equal.
ABCD is square.

Given, a = 9, d = 17 - 9 = 8,
Sn = 636
Sn = \(\frac{n}{2}[2a+(n-1)d]\)
\(\Rightarrow\)    636 = \(\frac{n}{2}[2\times9+(n-1)8]\Rightarrow636\times 2=n[18+8n-8]\)
\(\Rightarrow\)    636 X 2 = n ( 10 + 8n ) \(\Rightarrow\) 636 x 2 = 2n ( 5 + 4n )
\(\Rightarrow\)    \(\frac{636\times2}{2}\) = 5n + 4n²    \(\Rightarrow\) 4n² + 5n - 636 = 0
\(\Rightarrow\)   4n² + 53n + 48n - 636 = 0 \(\Rightarrow\) 4n( n + 53 ) - 48 ( n + 53 ) = 0
\(\Rightarrow\)     ( 4n - 48 ) ( n + 53 ) = 0 \(\Rightarrow\) Either 4n = 48 or n + 53 = 0
\(\Rightarrow\)     n = \(\frac{48}{4}\) or n = -53 (rejected)
\(\Rightarrow\)   n = 12
Hence   n = 12

Basic Proportionality theorem It states that if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides of the triangle in proportion.
Given Let ABC be a triangle.

In the given triangle, line I is parallel to BC which intersect AB at D and AC at E.
To prove \(\frac{A D}{D B}=\frac{A E}{E B}\)
Construction Join BE and CD. Draw perpendicular lines DM and EN to the sides AC and AB of \(\Delta\)ABC respectively, as shown in the figure.
Proof Consider the given \(\Delta\)ABC.
\(\Delta\)ADE and \(\Delta\)DEB have equal heights i.e. EN.
So, area of \(\Delta\)ADE = \(\frac{1}{2} \times A D \times E N\)
Area of \(\Delta\)DEB = \(\frac{1}{2} \times D B \times E N\)
Thus, \(\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle D E B)}=\frac{A D}{D B}\)              ...(i)
Similarly, consider the \(\Delta\)ADE with base AE and height DM.
\(\begin{aligned} & \text { Area of } \triangle A D E=\frac{1}{2} \times A E \times D M \\ \end{aligned}\)
\(\begin{aligned} & \text { Area of } \triangle C D E=\frac{1}{2} \times E C \times D M \\ \end{aligned}\)
\(\begin{aligned} & \frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle C D E)}=\frac{A E}{E C} \end{aligned}\)             ...(ii)
It is known that triangles with the same base and between same parallel lines have an equal area
Here, \(\Delta\)BDE and \(\Delta\)CDE have the same base DE and lie between the parallel lines DE and BC
\(\Rightarrow\)    \(A(\triangle D E B)=A(\triangle D E C)\)    ...(iii)
From Eqs. (i), (ii) and (iii), we get
\(\begin{array}{rlrl} \frac{A(\triangle A D E)}{A(\triangle D E B)} =\frac{A(\triangle A D E)}{A(\triangle D E C)} \\ \end{array}\)
\(\Rightarrow \quad \frac{A D}{D B}=\frac{A E}{E C}\)             Hence proved

S_n = 3n² + 4n.
a₁=S₁= 3(1)² + 4(1) = 7
a₁ + a₂ = S₂ = 3(2)² + 4(2)
= 12 + 8 = 20
a₂ = S₂ - S₁ = 20 - 7 = 13
\(\Rightarrow\)  a + d = 13
\(\Rightarrow\) 7 + d = 13
\(\therefore\) d = 13 - 7 = 6
\(\therefore\) A.P. becomes 7, 13, 19, ......
Now, a_n = a + (n - 1)d
\(\Rightarrow\) = 7 + (n - 1)(6)
 = 7+ 6n - 6
 = 6n + 1
\(\Rightarrow\) a_n = 6n+ 1

Here, |AB|= \(\sqrt { (-2-2)^{ 2 }+(2-3)^{ 2 } } \)
= \(\sqrt { (-4)^{ 2 }+(-1)^{ 2 } } \)
=\(\sqrt { 16+1 } =\sqrt { 17 } \)  Units
|BC|=  \(\sqrt { (-1+2)^{ 2 }+(-2-2)^{ 2 } } \)
= \(\sqrt { (1)^{ 2 }+(-4)^{ 2 } } \)
=\(\sqrt { 1+16 } =\sqrt { 17 } \)  units
|CD|= \(\sqrt { (3+1)^{ 2 }+(-1+2)^{ 2 } } \)
=\(\sqrt { 4^{ 2 }+1^{ 2 } } \)
=   \(\sqrt { 17 } \)=units
|DA|=\(\sqrt { (2-3)^{ 2 }+(3+1)^{ 2 } } \)
=  \(\sqrt { (-1)^{ 2 }+4^{ 2 } } \)=\(\sqrt { 1+16 } \)
= \(\sqrt { 17 } \) units
\(\Rightarrow \) AB=BC=CD=DA= units
Now,Diagonal |AC| =\(\sqrt { (-1-2)^{ 2 }+(-2-3)^{ 2 } } \)
= \(\sqrt { (-3)^{ 2 }+(-5)^{ 2 } } \)
=\(\sqrt { 9+25 } =\sqrt { 34 } \)  units
Diagonal |BD|= \(\sqrt { (3+2)^{ 2 }+(-1-2)^{ 2 } } \)
= \(\sqrt { { 5 }^{ 2 }+(-3)^{ 2 } } \)
= \(\sqrt { 25+9 } =\sqrt { 34 } \) units
\(\Rightarrow \) Diagonal AC=Diagonal BD= units 
Hence ABCD is a square

(i) (a): Here, CD = \(\sqrt{(7-3)^{2}+(7-4)^{2}}\)
\(=\sqrt{4^{2}+3^{2}}\) = 5 units
Also, it is given that CE = 10 units
Thus, DE = CE - CD = 10 - 5 = 5 units (\(\because\) A, B, C, E are a line)
(ii) (b): Since, CD = DE = 5 units
\(\therefore\) Dis the midpoint of CE.
\(\therefore \quad \frac{x+3}{2}=7 \text { and } \frac{y+4}{2}=7 \)
\(\Rightarrow \quad x=11 \text { and } y=10 \Rightarrow x+y=21\)
(iii) (b)
(iv) (d): Let B divides AC in the ratio k:1, then

\(\frac{7}{4}=\frac{4 k+0}{k+1} \)
\(\Rightarrow 7 k+7=16 k \)
\(\Rightarrow 7=9 k \)
\(\Rightarrow k=\frac{7}{9}\)
Thus, the required ratio is 7 : 9
(v) (c): It can be easily verify that all the given points lie on the line represented by 3x - 4y + 7 = 0.

(i) (a): In \(\Delta\)PAB and \(\Delta\)PQR,
\(\angle\)P = \(\angle\)P (Common)
\(\angle\)A = \(\angle\)Q (Corresponding angles)
By AA similarity criterion, \(\Delta\)PAB \(\sim\) \(\Delta\)PQR
\(\therefore \frac{A B}{Q R}=\frac{P A}{P Q} \Rightarrow \frac{A B}{12}=\frac{6}{24} \Rightarrow A B-3 \mathrm{~m}\)
(ii) (d): Similarly, \(\Delta\)PCD and \(\Delta\)PQR are similar
\(\therefore \frac{P C}{P Q}=\frac{C D}{Q R} \Rightarrow \frac{14}{24}=\frac{C D}{12} \Rightarrow C D=7 \mathrm{~m}\)
(iii) (a): Area of whole empty land
= \(\frac{1}{2}\)x base x height = \(\frac{1}{2}\) x 1 x15= 90 m².
(iv) (b): Since,  \(\Delta\)PAB \(\sim\) \(\Delta\)PQR
\(\therefore \frac{\operatorname{ar}(\Delta P A B)}{a r(\Delta P Q R)}=\left(\frac{P A}{P Q}\right)^{2}=\left(\frac{6}{24}\right)^{2}=\frac{1}{16}\)
\(\Rightarrow \quad \operatorname{ar}(\Delta P A B)=\frac{1}{16} \times 90=\frac{45}{8} \mathrm{~m}^{2}\) \(\left[\because \operatorname{ar}(\Delta P Q R)=90 \mathrm{~m}^{2}\right]\)
(v) (d): Since,  \(\Delta\)PCD \(\sim\) \(\Delta\)PQR
\(\therefore \frac{\operatorname{ar}(\Delta P C D)}{\operatorname{ar}(\Delta P Q R)}=\left(\frac{P C}{P Q}\right)^{2}=\left(\frac{14}{24}\right)^{2}=\left(\frac{7}{12}\right)^{2}\)
\(\Rightarrow \operatorname{ar}(\Delta P C D)=\frac{90 \times 49}{144}=\frac{245}{8} \mathrm{~m}^{2}\)

Geeta's A.P. is -5, -2, 1,4, ...
Here, first term (a₁) = -5 and common difference (d₁) = -2 + 5 = 3
Similarly, Madhuri's A.P. is 187, 184, 181, ...
Here first term (a₂) = 187 and common difference (d₂) = 184 - 187 = -3
(i) (b): t₃₄ = a₂ + 33d₂ = 187 + 33(-3) = 88
(ii) (d): Required sum = 3 + (-3) = 0
(iii) (a): t₁₉ = a₁ + 18d₁ = (-5) + 18(3) = 49
(iv) (a) : \(S_{10}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]=\frac{10}{2}[2(-5)+9(3)]=85\)
(v) (b): Let n^th terms of the two A.P:s be equal.
\(\therefore\) -5 + (n - 1)3 = 187 + (n - 1)(-3)
\(\Rightarrow\) 6(n - 1) = 192 \(\Rightarrow\) n = 33