Given, AD and CE are altitudes which intersect each other at the point P.
In ΔPDC and ΔBEC,
∠PDC = ∠BEC           [each 90°]
and ∠PCD = ∠BCE   [Common angle]
\(\therefore\) ΔPDC ∼ ΔBEC     [by AA similarity criterion]

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.

In \(\triangle\)AOB and \(\triangle\)COD,
AB || CD

OAB = DCO and OBA = ODC (Alternate angles)
\(\triangle\)AOB ~\(\triangle\)COD    (AA similarity)
\(\therefore \frac { AO }{ BO } =\frac { CO }{ DO } .\\ \) (Corresponding sides of similar triangles)
or \(\frac { AO }{ CO } =\frac { BO }{ DO } \)

Mid-point of BC is D and coordinates of D are
\(D\left( \frac { 3-1 }{ 2 } ,\frac { -2+8 }{ 2 } \right) i.e.,D(1,3)\)
Now, \(|AD|=\sqrt { \left( 1-5 \right) ^{ 2 }+\left( 3+1 \right) ^{ 2 } } \)
\(=\sqrt { \left( -4 \right) ^{ 2 }+\left( 4 \right) ^{ 2 } } =\sqrt { 16+16 } \)
\(=4\sqrt { 2 } units\)

20 : 23

Here, a₃ = 15 and S₁₀ = 125
\(\because\) a₃ = 15
\(\therefore\) a + 2d = 15   ....(i)
                  [\(\because\) a_n = a + (n - 1)d]
Also, S₁₀ = 125
\(\begin{aligned} \therefore \quad \frac{10}{2}[2 a+(10-1) d] & =125 \\ \end{aligned}\)
\(\begin{aligned} {\left[\because S_n\right.} & \left.=\frac{n}{2}\{2 a+(n-1) d\}\right] \end{aligned}\)
\(\Rightarrow\) 2a + 9d = 25   [dividing by 5]   .....(ii)
On multiplying Eq. (i) by 2 and then subtracting Eq. (ii) from it, we get
2(a + 2d) - (2a + 9d) = 2 \(\times\)15 - 25
\(\Rightarrow\) 4d - 9d = 30 - 25
\(\Rightarrow\) -5d = 5
\(\Rightarrow \quad d=-\frac{5}{5}=-1\)
Now, a₁₀ = a + 9d
= (a + 2d) + 7d
= 15 + 7(-1)   [from Eq. (i)]
= 15 - 7 = 8
Hence, d = -1 and a₁₀ = 8.

Here, we have a₂ - a₁ = 0.22 - 0.2 = 0.02
and a₃ - a₂ = 0.222 - 0.22 = 0.002
Since, a₂ - a₁ \(\neq\) a₃ - a₂, therefore the given list of numbers does not form an AP.

Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that ∠ B = ∠ E, ∠ C = ∠ F and Ð A = ∠ D. That is, corresponding angles of these two
triangles are equal. What can you say about their corresponding sides. Note that \(\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{3}{5}=0.6\) What about \(\frac{\mathrm{AB}}{\mathrm{DE}} \text { and } \frac{\mathrm{CA}}{\mathrm{FD}} \) On measuring AB, DE, CA and FD, you will find that \(\frac{\mathrm{AB}}{\mathrm{DE}} \text { and } \frac{\mathrm{CA}}{\mathrm{FD}}\) are also equal to 0.6 (or nearly equal to 0.6, if there is some error in the measurement). Thus, \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}\) You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio (or proportion). This activity leads us to the following criterion for similarity of two triangles.

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.

\(AP=\frac { 3 }{ 7 } AB\Rightarrow AP:PB=3:4\)
\(x=\frac{6-8}{7}=-\frac{2}{7}\)
\(\therefore \ y=\frac{-12-8}{7}=-\frac{20}{7}\)
\(P=(-\frac{2}{7},-\frac{20}{7})\)

Let the coordinates of the fourth vertex D be  D(x, y). We know that diagonals of a parallelogram  bisect each other.
   
Therefore mid point of AC=midpoint of BD
\(\Rightarrow \) \(\left( \frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } ,\frac { { y }_{ 1 }+y_{ 3 } }{ 2 } \right) =\left( \frac { { x }+{ x }_{ 2 } }{ 2 } ,\frac { { y }+y_{ 2 } }{ 2 } \right) \)
\(\Rightarrow \)  \(\frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } =\frac { { x }+{ x }_{ 2 } }{ 2 } \)
and  \(\frac { y_{ 1 }+y_{ 3 } }{ 2 } =\frac { { y }+{ y }_{ 2 } }{ 2 } \)
\(\Rightarrow \) x₁+x₃=x+x₂ and y₁+y₃=y+y₂
\(\Rightarrow \)  x=x₁+x₃ +x₂ and y=y₁+y₃-y₂
Hence,the coordinates of the fourth vertex in terms of x₁,x₂,x₃,y₁,y₂ and y₃ are
(x₁+x₃-x₂,y₁+y₃-y₂)

Here,   S_n = 3n² + 5n                  ...(i)
Sn-1 = 3 ( n - 1 )² + 5 ( n - 1 )
= 3 ( n² + 1 - 2n ) + 5n - 5
= 3n² + 3 - 6n + 5n - 5 
= 3n² - n - 2                                ...(ii)
From (i), (ii), we obtain
S_n - S_n-1 = 3n² + 5n - ( 3n² - n - 2 )
a_n = 3n² + 5n - 3n² + n + 2
a_n = 6n + 2
Put   n = k, we have
a_k = 6k + 2 
\(\Rightarrow\)    164 = 6k + 2
\(\Rightarrow\)    6k = 162
\(\Rightarrow\)  k = 27

The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP . Let the number of rows in the flower bed be n.
Then a = 23, d = 21 – 23 = – 2, a_n = 5
As, a_n = a + (n – 1) d
We have,  5 = 23 + (n – 1) (– 2)
i.e.,   – 18 = (n – 1) (– 2)
i.e.,   n = 10
So, there are 10 rows in the flower bed.

Multiples of 5 lying between 101 and 999 are 105, 110, 115,......., 999 which are in AP.
Here  a = 105 and d = 5.
a_n = a + (n - 1)d
 995 = 105 + n(n -1)5
 890 = 5n - 5
 895 = 5n
 n = 179
S_n = \(\frac{n}{2}\)[a + l] = \(\frac{179}{2}\)[105 + 995] = \(\frac{179}{2}\) x 1100 = 98450.

From \(\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{CA}}{\mathrm{RP}}\)
But, \(\frac{\mathrm{CA}}{\mathrm{RP}}=\frac{\mathrm{AB}}{\mathrm{PQ}}\) [From (1)]
Therefore, \(\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{AB}}{\mathrm{PQ}}\) [From (6) and (7)]

It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a₂ − a₁ = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_n = 200
\(S_{n}=\frac{n}{2}[2 a+(n-1) d]\)
\(200=\frac{n}{2}[2(20)+(n-1)(-1)]\)
400 = n (40 − n + 1)
400 = n (41 − n)
400 = 41n − n²
n² − 41n + 400 = 0
n² − 16n − 25n + 400 = 0
n (n − 16) −25 (n − 16) = 0
(n − 16) (n − 25) = 0
Either (n − 16) = 0 or n − 25 = 0
n = 16 or n = 25
a_n = a + (n − 1)d
a₁₆ = 20 + (16 − 1) (−1)
a₁₆ = 20 − 15
a₁₆ = 5
Similarly,
a₂₅ = 20 + (25 − 1) (−1)
a₂₅ = 20 − 24
= −4
Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^th row is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^th row is 5.

Consider a trapezium ABCD such that \(AB\parallel DC\) and draw its diagonals AC and BD which intersect at point O.

In \(\triangle OCD\) and \(\triangle OAB\),
\(AB\parallel DC\)        [given]
\(\therefore \angle 1=\angle 3,\angle 2=\angle 4\)    [alternate interior angles]
Also, \(\triangle DOC=\triangle BOA\)     [vertically opposite angles]
\(\therefore \triangle OCD\sim \triangle OAB\)     [by AAA similarity criterion]
\(\Rightarrow \frac { OC }{ OA } =\frac { OD }{ OB } \)
[since, ratios of the corresponding sides of the similar triangles are equal]
\(\therefore \frac { OA }{ OC } =\frac { OB }{ OD } \)    [on taking reciprocal of the terms]
Hence proved.

(0, \(2\sqrt { 3 } \)) or (3, -\(\sqrt { 3 } \))

\(P({13\over 2},{9\over 2})\), Q(8,5), R\(({19\over 2}, {11\over2})\)

Let there are ( 2n + 1 ) stones. Middle stone is at B. There are n stones on one side of the middle stone and n stones on other side of the middle stone.
Let the man starts from A.
He picks the stone and travels to B.|
\(\therefore\)  Distance covered = 10n m
He comes back to pick the next stone and goes back to B.
Distance covered = 10 x ( n - 1) + 10 ( n - 1 )
= 2 x 10( n - 1)m
Simliarly, distance coverd to pick and place that stone at B.
= 2 X 10( n - 2) m and so on
Now total distance covered,
S₁ = 10 x n + 2 x 10( n - 1 ) + 2 x 10( n - 2 ) + .. up to n terms
S₁ = [ 2 x 10n + 2 x 10( n - 1 ) + 2 x 10 x ( n - 2 ) + .... up to n terms ] -10n
S₁ = 20 [ n + ( n - 1) + ( n - 2 ) + ... up to n terms] - 10n
S₁ = \(\left[ 20\times{n(n+1)\over2} -10n\right]m\)
Now to pick the stone at C, he will walk from B to C and then will come back to B, In order to pick all the stones, total distance covered will be S1 + distance covered from B to C.
= S₁ + 10n
Total distance covered = 3 km
\(\Rightarrow\)  S₁ + S₂ + 10n = 3000
\(\Rightarrow\)   \(\left[ 20\times{n(n+1)\over2} -10n\right]+\left[ 20\times{n(n+1)\over2} *10n\right]+10n=3000\)
\(\Rightarrow\)  10( n² + n ) - 10n + 10 ( n2 + n ) = 3000
\(\Rightarrow\)   20n² + 10n - 3000 = 0
\(\Rightarrow\)   2n² + n - 300 = 0
\(\Rightarrow\)  2n² + 25n -24n - 300 = 0
\(\Rightarrow\) n( 2n + 25 ) - 12 ( 2n + 25 ) = 0
\(\Rightarrow\)           ( 2n + 25 ) ( n - 12 ) = 0
\(\Rightarrow\) \(n={{-25}\over2}\)  or n = 12
rejecting \(n={-25\over2}\)
\(\therefore\) n = 12
\(\therefore\)  Total number of stones
= 2n + 1 = 2 x 12 + 1 = 25.

(c) Assertion is correct but Reason is incorrect.

(a) Assertion Area of \(\triangle A B C\) 
\(=\frac{1}{2}\left|\frac{-3}{2}(-2-4)+6(4-3)+(-3)(3+2)\right|\) 
\(=\frac{1}{2}|9+6-15|=0\) 

(i) (b): The distance between A and C
\(=\sqrt{(8-4)^{2}+(5+3)^{2}}=\sqrt{4^{2}+8^{2}} \)
\(=\sqrt{16+64}=\sqrt{80}=4 \sqrt{5} \text { units }\)
(ii) (a): Let the coordinates of I be (x, y).

Then, by section formula
\(x =\frac{1 \times 8+2 \times 7}{1+2}=\frac{8+14}{3}=\frac{22}{3}\)
\(\text { and } y =\frac{1 \times 5+2 \times 3}{1+2}=\frac{5+6}{3}=\frac{11}{3}\)
Thus, the coordinates of I is \(\left(\frac{22}{3}, \frac{11}{3}\right)\)
(iii) (b): The mid -point of A and C
\(=\left(\frac{8+4}{2}, \frac{5-3}{2}\right)=(6,1)\)
(iv) (b): Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be
\(\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)\)
\(\text { Thus, we have }\left(\frac{8 k+4}{k+1}, \frac{5 k-3}{k+1}\right)=(7,3)\)
\(\Rightarrow \frac{8 k+4}{k+1}=7 \text { and } \frac{5 k-3}{k+1}=3 \)
\(\text { Consider, } \frac{8 k+4}{k+1}=7 \Rightarrow 8 k+4=7 k+7 \Rightarrow k=3\)
Hence, the required ratio is 3 : 1
(v) (a):\(\because\) B divides AC in the ratio 3 : 4.
\(\therefore\) A, B, C lie on a straight line.

(i) (a): Since \(\Delta\)ABC - \(\Delta\)PQR

\(\therefore \quad \frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{x}{20}=\frac{50}{10} \Rightarrow x=100\)
Thus, height of tower is 100 m.
(ii) (c): Since \(\Delta\)ABC - \(\Delta\)PQR

\(\therefore \quad \frac{100}{20}=\frac{x}{15} \Rightarrow x=\frac{1500}{20}=75 \mathrm{~m}\)
(iii) (d): Since, the shapes are similar

\(\therefore \quad \frac{x}{20}=\frac{20}{10} \)
\(\Rightarrow \quad x=\frac{20 \times 20}{10}=40 \mathrm{~m}\)
(iv) (b): Since, the shapes are similar, so, \(\frac{40}{100}=\frac{x}{40}\)
\(\Rightarrow\) x = 16m

(v) (d): Since, the shapes are similar, so \(\frac{20}{100}=\frac{x}{40}\)
\(\Rightarrow \quad x=\frac{20 \times 40}{100}=8 \mathrm{~m}\)

Number of pairs of shoes in 1^st, 2^nd, 3^rd row, ... are 3,5,7, ...
So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2
(i) (d): Let n be the number of rows required.
\(\therefore S_{n}=120 \)
\(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \)
\(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\)
\(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\)
So, 10 rows required to put 120 pairs.
(ii) (b): No. of pairs in 1ih row = t₁₇ = 3 + 16(2) = 35
No. of pairs in 10th row = t₁₀ = 3 + 9(2) = 21
\(\therefore\) Required difference = 35 - 21 = 14
(iii) (c) : Here n = 15
\(\therefore\) t₁₅ = 3 + 14(2) = 3 + 28 = 31
(iv) (a): No. of pairs in 30^th row = t₃₀ = 3 +29(2) = 61
(v) (c): No. of pairs in 5^th row = t₅ = 3 + 4(2) = 11
No. of pairs in 8^th row = t₈ = 3 + 7(2) = 17
\(\therefore\) Required sum = 11 + 17 = 28