In Δ POQ, DE || OQ    [given]
\(\therefore \frac{P E}{E Q}=\frac{P D}{D O}\) ...(i)
[by basic proportionality theorem] 
In ΔPOR,    DF || OR     [given]
\(\therefore \frac{P F}{F R}=\frac{P D}{D O}\) ...(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get 
\(\frac{P E}{E Q}=\frac{P F}{F R}\)
In ΔPQR, we have \(\frac{P E}{E Q}=\frac{P F}{F R}\)
∴  EF || QR
[by Converse of basic proportionally theorem]
Hence proved.

According to the question, there are three sections of each class, so the number of trees that are planted by students of class I, class lI, class III,... and class XII, are respectively, 1 \(\times\)3, 2 \(\times\)3,3 \(\times\)3, ... and 12 \(\times\)3
Thus, we get the following list of numbers, 3, 6, 9, .., 36
Clearly, it forms an AP.
Here, the first term, a = 3
common differencc, d = 6 - 3 = 3
and the last term, l = 36
Let the last term of this AP be its nth term.
Then, a_n = a + (n - 1) d = l
\(\Rightarrow\)  3 + (n - 1) (3) = 36 \(\Rightarrow\) (n - 1) 3 = 33
\(\Rightarrow\) n - 1 = 11 \(\Rightarrow\)  n = 12
Hence, the number of the trees planted by the students
= Sum of 12 terms of above AP
\(\begin{array}{ll} =\frac{12}{2}(3+36) & {\left[\because S_n=\frac{n}{2}(a+l)\right]} \end{array}\)
= 6 \(\times\) 39 = 234

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.

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

Let a be the first term and d be the common difference of given AP.
Given, a₄ + a₈ = 24
\(\therefore\) (a + 3d) + (a + 7d) = 24   [\(\because\) a_n = a + (n - 1)d]
\(\Rightarrow\) a + 5d = 12   ....(i)
[dividing both sides by 2]
and a₆ + a₁₀ = 44
\(\Rightarrow\) (a + 5d) + (a + 9d) = 44
\(\Rightarrow\) 2a + 14d = 44 \(\Rightarrow\) a + 7d = 22 .... (ii)
[dividing both sides by 2]
On subtracting Eq. (i) from Eq. (ii), we get
2d = 10 \(\Rightarrow\) d = 5
On putting d = 5 in Eq. (i), we get
a + 25 = 12 \(\Rightarrow\) a = -13
Hence, the first three terms are
a,(a + d) and (a + 2d)
i.e. -13, (-13 + 5) and (-13 + 2 \(\times\)5)
i.e. -13, -8 and -3.

In ∆ ABC and Δ PQR,
\(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} \text { and } \frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}\)
That is, \(\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}\)
So, Δ ABC ~ Δ RQP (SSS similarity)
Therefore, ∠ C = ∠ P (Corresponding angles of similar triangles)
But ∠ C = 180° - ∠ A - ∠ B (Angle sum property)
= 180° - 80° - 60° = 40°
So, ∠ P = 40°

Proof : Let the prime factorisation of a be as follows :
a = p₁ p₂ . . . p_n , where p₁ ,p₂ , . . ., p_n are primes, not necessarily distinct. Therefore, a² = ( p₁ p₂ . . . p_n )( p₁ p₂ . . . p_n ) = p²₁ p²₂ . . . p ²_n .
Now, we are given that p divides a ² . Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a ² . However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁ ,p₂ , . . ., p_n . So p is one of p₁ , p₂ , . . ., p_n .
Now, since a = p₁ p₂ . . . p_n ,p divides a.
We are now ready to give a proof that \(\sqrt 2\) is irrational.
The proof is based on a technique called ‘proof by contradiction’. 

5 cm

87

Let AB is lighthouse
AB=150 m
Initially boat is at C and after 2 minutes it reaches at  D.
In right ∆ABC
\(\frac { AB }{ BC } \)=tan45^o
⇒ \(\frac { 150 }{ BC } \)=1 ⇒ BC=150 m
In right ∆ABD, \(\frac { AB }{ BD } \)=tan 30^o
⇒ \(\frac { 150 }{ BD } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow BD=150\sqrt { 3 } \)
Distance covered in 2 minutes=BD-BC= \( 150\sqrt { 3 } -150=150(\sqrt { 3 } -1)\)m
Speed=Distance covered/time taken=\(\frac { 150(\sqrt { 3 } -1) }{ 2 } \)
=75 x (1.732-1)=54.9 m/minutes

Construction: Join AD and BC
Proof: The tangent drawn from an internal point to a circle are equal in length.
If A is external point for circle hving centre O.
AB = AD    .....(i)
If C is external point then
BC = CD  .....(ii)
Now, B is external point for circle having centre O
AB = BC  .....(iii)
So, from (i), (ii) and (iii), we get
AB = BC = CD
So, AB = CD Hence proved.

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. So, for example, we regard 2 x 3 x 5 x 7 as the same as 3 x 5 x 7 x 2, or any other possible order in which these primes are written.
This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors.
In general, given a composite number x, we factorise it as x = p₁ p₂ ... p_n , where p₁ , p₂ ,..., p_n are primes and written in ascending order, i.e., p₁ ≤ p₂ ≤ . . . ≤ p_n . If we combine the same primes, we will get powers of primes. For example,
32760 = 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 = 2³ x 3² x 5 x 7 x 13
Once we have decided that the order will be ascending, then the way the number is factorised, is unique.
The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. 

Let a be the first term and d be the common difference of given AP.
Given,   \({a}_{m}=\frac {1}{n} \Rightarrow a + (m - 1)d = \frac {1} {n}\)   ....(i)
and       \({a}_{n}=\frac {1} {m} \Rightarrow a + (n-1)d = \frac {1} {m}\)    ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
  \((m-n)d=\frac {1}{n}-\frac{1}{m}\)
\(\Rightarrow \ (m-n)d=\frac {m-n}{mn}\)
 \(\Rightarrow d=\frac {1}{mn}\)
On putting \(d=\frac {1}{mn}\) in Eq. (i), we get
 \(a + (m-1)\frac {1}{m}=\frac {1}{n}`\)
\(\Rightarrow \quad a+\frac {1}{n}-\frac {1}{mn}=\frac{1}{n}\)
\(\Rightarrow a=\frac {1}{mn}\)
Now, sum of mn terms,
 \({S}_{mn}=\frac {mn}{2} \left[2a+(mn-1)d \right]\)
\(\left[\because {S}_{n}=\frac {n}{2}\{\ 2a + (n-1)d \}\ \right]\)
\(\Rightarrow \quad {S}{mn}=\frac {mn}{2}\left[\frac {2} {mn} + (mn - 1) \frac {1 } {mn }\right]=\frac {1} {2}(mn+1)\)

Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.
Let the circle touches the other sides AB and AC at points E and F, respectively.
We know that tangents drawn from an external point to the circle are equal in length.
\(\therefore\) CD = CF = 6 cm            [\(\because\) C is an external point]
BD = BE = 8 cm                [\(\because\) B is an external point]
and AF = AE = x cm (say  [\(\because\) A is an external point]
Area of \(\Delta\)OCB, \(A_1=\frac{1}{2} \times \text { Base } \times \text { Height }\)
\(\begin{aligned} & =\frac{1}{2} \times C B \times O D \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & \quad[\because C B=C D+B D=6+8=14] \end{aligned}\)
Area of \(\Delta\)OCA,
\(\begin{aligned} A_2 & =\frac{1}{2} \times A C \times O F \end{aligned}\)
\(\begin{aligned} =\frac{1}{2}(6+x) \times 4=(12+2 x) \mathrm{cm}^2 \end{aligned}\)
and area of \(\Delta\)OBA,
\(A_3=\frac{1}{2} \times A B \times O E=\frac{1}{2}(8+x) \times 4=(16+2 x) \mathrm{cm}^2\)
Thus, area of  \(\Delta\)ABC
= A₁ + A₂ + A₃ = [28 + (12 + 2x) + (16+ 2x)]
= (56 + 4x) cm²     ...(i)
Now, semi-perimeter of \(\Delta\)ABC=\(\frac{1}{2}\)(AB + BC + CA)
\(\Rightarrow \quad s=\frac{1}{2}(x+8+14+6+x)\)
\(\Rightarrow\) s = (14 + x) cm
Using Heron's formula,
area of \(\Delta\)ABC = \(\begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(14+x) \times x \times 8 \times 6} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x) 48 x} \end{aligned}\)     ...(ii)
From Eqs. (i) and (ii), we get
\(\sqrt{(14+x) 48 x}=56+4 x=4(14+x)\)
On squaring both sides, we get
(14 + x) 48 x = 4² (14 + x)²
\(\Rightarrow\) 3x = 14 + x
\(\Rightarrow\) 2x = 14
\(\Rightarrow\) x = 7
\(\therefore\)  Length of AC = 6 + x = 6 + 7 = 13 cm
and length of AB = 8 + x = 8 + 7 = 15 cm

(i) Number of logs stacked in each row form a sequence 20, 19, 18, 17,...., which is an AP with first term, a= 20 and common difference, d = 19 - 20 = -1.
Suppose number of rows is n, then S_n= 200
\(\begin{aligned} \Rightarrow \frac{n}{2}[2 \times 20+(n-1)(-1)] & =200 \\ \end{aligned}\)
\(\begin{aligned} {\left[\because S_n\right.} & \left.=\frac{n}{2}\{2 a+(n-1) d\}\right] \end{aligned}\)
\(\begin{array}{lr} \Rightarrow & 400=40 n-n^2+n \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-41 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-25 n-16 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n(n-25)-16(n-25)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & (n-25)(n-16)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n=16 \text { or } n=25 \end{array}\)
Hence, the number of rows is either 25 or 16.
When, n = 16,
a_n= a + (n -1) d = 20 + (16 - 1) ( - 1)
= 20 - 15 = 5
When, n = 25,
a_n = a + (n - 1) d = 20 + (25-1) (-1)
= 20 - 24 = - 4
[\(\because\) number of logs cannot be negative]
Hence, the number of rows is 16 and number of logs in the top row is 5.
(ii) The pattern of logs show space saving creativity, reasoning and balancing.

(i) (c) In quadrilateral PQOR, we have

\(\begin{array}{r} \angle Q P R+\angle P R O+\angle P Q O+\angle R O Q=360^{\circ} \\ \end{array}\)
\(\Rightarrow \begin{array}{r} 30^{\circ}+90^{\circ}+90^{\circ}+\angle R O Q=360^{\circ} \end{array}\)
[\(\because\) radius is always perpendicular to the tangent at point of contact]
\(\Rightarrow \quad \angle R O Q=360^{\circ}-210^{\circ}=150^{\circ} \)
\((ii) (a) In \quad \triangle Q O R, O Q=O R\)     [radii]
\(\begin{array}{ll} \therefore & \angle O R Q=\angle O Q R \\ \end{array}\)
\(\begin{array}{ll} \text { Now, } & \angle R O Q+\angle O R Q+\angle O Q R=180^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=180^{\circ}-150^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & 2 \angle O Q R=30^{\circ} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \angle O Q R=15^{\circ} \end{array}\)
Again, \(\begin{array}{ll} \angle O Q R=90^{\circ} \end{array}\)    \([\because O Q \perp Q P]\)
\(\begin{aligned} \Rightarrow \angle O Q R+\angle R Q P=90 \\ \end{aligned}\)^\({\circ}\)
\(\Rightarrow \quad \angle R Q P=90^{\circ}-15^{\circ}=75^{\circ}\)
(iii) (b) We know that angle subtended by an are at centre is double the angle subtended by it at any other part of the circle.
\(\begin{aligned} 2 \angle R S Q & =\angle R O Q \\ \end{aligned}\)
\(\begin{aligned} \angle R S Q & =\frac{1}{2} \times 150^{\circ}=75^{\circ} \end{aligned}\)
(iv) (a) \(\angle\)ORP = 90^\(\circ\) as radius is always perpendicular to the tangent at the point of contact.

(i) (d): Let h be the height of the pole.
In \(\Delta\)ABC, 

\( \frac{h}{15}=\sin 45^{\circ} \Rightarrow \frac{h}{15}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad h =\frac{15}{\sqrt{2}} \mathrm{~m}\)
(ii) (a): Let x be the required distance.
In \(\Delta\)ABC,

\(\frac{x}{15}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad x=\frac{15}{\sqrt{2}} \mathrm{~m}\)
(iii) (c) : Let h be the height of the pole.
In right triangle ABC,

\(\frac{h}{15}=\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow \quad h=\frac{15}{2}=7.5 \mathrm{~m}\)
(iv) (d): If 3 m rope is broken, then the length of the rope is 12 m.

\(\text { In } \Delta A B C, \frac{h}{12}=\sin 30^{\circ}=\frac{1}{2} \)
\(\Rightarrow \quad h=\frac{12}{2}=6 \mathrm{~m}\)
(v) (c)

Here the savings form an A.P. i.e., Rs 2.75, Rs 3, Rs 3.25, ...
So, a = 2.75, d = 3 - 2.75 = 0.25
(i) (b): Amount saved by Anuj on 14^th day
= t₁₄ = a + 13d = 2.75 + 13(0.25) = ₹ 6
(ii) (d): Total amount saved by Anuj in 8 days
\(=S_{8}=\frac{8}{2}[2(2.75)+7(0.25)]=₹ 29\)
(iii) (a): Amount saved by Anuj on 30^th day
= t₃₀ = a + 29d = 2.75 + 29(0.25) = ₹ 10
(iv) (b): Number of days in June = 30
\(\therefore S_{30}=\frac{30}{2}[2(2.75)+29(0.25)]=₹ 191.25\)
(v) (d): Let on nth day, he saves 10 times as he saves on 1^st day.
t_n = 10(2.75) \(\Rightarrow\) a + (n - 1)d = 27.5 \(\Rightarrow\) n = 100.