We are given a circle with centre O and a tangent XY to the circle at a point P. We need to prove that OP is perpendicular to XY.

Take a point Q on XY other than P and join OQ see  Fig.The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle, XY will become a secant and not a tangent to the circle). Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY.

Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere
= 2\(\pi\)rh + 2\(\pi\)r² = 2\(\pi\)r(h + r)
\(=2 \times \frac{22}{7} \times 30(145+30) \mathrm{cm}^2\)
= 33000 cm² = 3.3 m²

Given, P(E) = 0.05
we know that P(E) + P(\(\bar{E}\)) = 1
\(\therefore\) P(\(\bar{E}\)) = 1 - P(E)  \(\Rightarrow\) P(\(\bar{E}\)) = 1 - 0.05 = 0.95

We know that area of sector of a circle =\(\frac{\theta}{360^{\circ}} \times \pi r^2\)
Given, radius of circle, r = 6 cm
and angle of sector, \(\theta\)= 60°
\(\therefore\) Area of sector of a circle \(=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^2=\frac{132}{7} \mathrm{~cm}^2\)

Given, AP is 2, 7, 12, ..., to 10 terms.
Here, a = 2, d = 7 - 2 = 5 and n = 10
\(\because\) Sum of first n terms of an AP,
\(S_n=\frac{n}{2}[2 a+(n-1) d]\)
\(\therefore\) On putting a = 2, d = 5 and n = 10, we get
\(\begin{aligned} S_{10} & =\frac{10}{2}[2 \times 2+(10-1) 5] \\ \end{aligned}\)
\(\begin{aligned} & =5(4+9 \times 5)=5(4+45)=5 \times 49=245 \end{aligned}\)

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.

(i) The cumulative frequency table of given frequency distribution is

Here, n = 68
\(\therefore \quad \frac{n}{2}=34\)
Since, the cumulative frequency just greater than 34 is 42 and the corresponding class is 125- 145. So, the median class is 125-145.
Now, l = 125, f = 20, cf = 22 and h = 20
\(\begin{aligned} & \therefore \text { Median }=l+\left\{\frac{\frac{n}{2}-c f}{f}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} \quad=125+\left\{\frac{34-22}{20}\right\} \times 20=125+12=137 \text { units } \end{aligned}\)
(ii) Let the assumed mean, a = 135
and width of the class, h = 20
Table for the given data is

We have, N = 68, and \(\sum\)f_iu_i = 7
By step deviation method,
Mean = \(\begin{aligned} \text { Mean } & =a+b \times \frac{1}{N} \times \Sigma f_i u_i=135+20 \times \frac{1}{68} \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =135+\frac{35}{17}=135+2.05=137.05 \text { units } \end{aligned}\)
(iii) Here, the modal class is 125-145 having maximum frequency, f₁ = 20.
Now, f₀ = 13, f₂ = 14, l = 125 and h = 20
\(\begin{aligned} \therefore \text { Mode } & =l+\left\{\frac{f_1-f_0}{2 f_1-f_0-f_2}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} & =125+\left\{\frac{20-13}{40-13-14}\right\} \times 20 \\ \end{aligned}\)
\(\begin{aligned} & =125+\frac{7 \times 20}{13}=125+\frac{140}{13} \end{aligned}\)
= 125 + 10.77 = 135.77 units
Hence, median = 137 units, mean = 137.05 units and mode = 135.77 units
So, we conclude that three measures are approximately the same.
(iv) Electricity consumption can be reduced by
(a) switching off electric appliances when not in use.
(b) using star rated AC's, refrigerators.

Given, \(\begin{array}{ll} \tan A=\frac{1}{\sqrt{3}} \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \frac{P}{B}=\frac{1}{\sqrt{3}} \end{array}\)
Let, P = k and B = \(\sqrt{3}\)k, where k is any positive integer.
Draw a right angled \(\Delta\)ABC, right angled at B.
In right angled \(\Delta\)ABC,
H² = B² = P²  [by using Pythagoras theorem]
\(\Rightarrow \quad H^2=(\sqrt{3} k)^2+(k)^2=3 k^2+k^2=4 k^2\)
\(\therefore\) H = 2k [taking positive square root since, side cannot be negative]
With reference to \(\angle\)A, we have

Base = AB = \(\sqrt{3}\)k, perpendicular = BC = k
and hypotenuse = AC = 2k
With reference to \(\angle\)C, we have
Base = BC = k, perpendicular = AB = \(\sqrt{3}\)k
and hypotenuse = AC = 2k
sin A cos C + cos A sin C
\(\begin{aligned} & =\left(\frac{B C}{A C}\right)\left(\frac{B C}{A C}\right)+\left(\frac{A B}{A C}\right)\left(\frac{A B}{A C}\right) \\ \end{aligned}\)
\(\begin{aligned} {\left[\because \sin \theta=\frac{P}{H} \text { and } \cos \theta=\frac{B}{H}\right]} \end{aligned}\)
\(\begin{aligned} & =\left(\frac{k}{2 k}\right)\left(\frac{k}{2 k}\right)+\left(\frac{\sqrt{3} k}{2 k}\right)\left(\frac{\sqrt{3} k}{2 k}\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{k^2}{4 k^2}+\frac{3 k^2}{4 k^2}=\frac{4 k^2}{4 k^2}=1 \end{aligned}\)

(i) For this A.P.,
a = 2
a₃ = 26
We know that, a_n = a + (n − 1) d
a₃ = 2 + (3 − 1) d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 − 1) 12
= 14
Therefore, 14 is the missing term.
(ii) For this A.P.,
a₂ = 13 and
a₄ = 3
We know that, a_n = a + (n − 1) d
a₂ = a + (2 − 1) d
13 = a + d (I)
a₄ = a + (4 − 1) d
3 = a + 3d (II)
On subtracting (I) from (II), we obtain
−10 = 2d
d = −5
From equation (I), we obtain
13 = a + (−5)
a = 18
a₃ = 18 + (3 − 1) (−5)
= 18 + 2 (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For this A.P.,
a = 5 and
\(a_{4}=\frac{19}{2}\)
We know that, an = a + (n − 1) d
a₄ = a + (4 - 1) d
\(\frac{19}{2}=5+3 \mathrm{~d}\)
\(\frac{19}{2}-5=3 d\)
\(\frac{9}{2}=3 d\)
d = 3 / 2
a₂ = a + (2 - 1) d
a₂ = 5 + 3 / 2
a₂ = 13 / 2
a₃ = a + (3 - 1) d
a₃ = 5 + 2 x 3/2
a₃ = 8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For this A.P.,
a = −4 and
a₆ = 6
We know that,
a_n = a + (n − 1) d
a₆ = a + (6 − 1) d
6 = − 4 + 5d
10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2 (2) = 0
a₄ = a + 3d = − 4 + 3 (2) = 2
a₅ = a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) For this A.P.,
a₂ = 38
a₆ = −22
We know that
a_n = a + (n − 1) d
a₂ = a + (2 − 1) d
38 = a + d (1)
a₆ = a + (6 − 1) d
−22 = a + 5d (2)
On subtracting equation (1) from (2), we obtain
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a₂ − d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula : Median \(=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) h, \text { we get }\)
\(525=500+\left(\frac{50-36-x}{20}\right) \times 100\)
i.e. 525 - 500 = (14 - x) \(\times\) 5
i.e. 25 = 70 - 5 x
i.e. 5 x = 70 - 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e. y = 15

(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.

Height (h₁) of larger cylinder = 220 cm
Radius (r₁) of larger cylinder = \(\frac{24}{2}\)= 12 cm
Height (h₂) of smaller cylinder = 60 cm
Radius (r₂) of smaller cylinder = 8 cm
Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder
\(\begin{aligned} & =\pi r_1^2 h_1+\pi r_2^2 h_2 \\ \end{aligned}\)
\(\begin{aligned} & =\left(\pi(12)^2 \times 220\right)+\left(\pi(8)^2 \times 60\right) \\ \end{aligned}\)
\(\begin{aligned} & =\pi[144 \times 220+64 \times 60] \\ \end{aligned}\)
\(\begin{aligned} & =3.14[31,680+3,840] \\ \end{aligned}\)
\(\begin{aligned} & =3.14 \times 35520=111,532.8 \mathrm{~cm}^3 \end{aligned}\)
Mass of 1 cm³ iron = 8 g
Mass of 111532.8 cm³ iron = 11532.8 \(\times\)8 = 892262.4 g

(i) Total number of discs in a box = 90
\(\therefore\) Number of all possible outcomes = 90
Let E₁ = Event of getting a disc bearing a two-digit number
Here, two-digit numbers are 10, 11, .., 90
\(\therefore\)Number of outcomes favourable to E₁ =81
Hence, probability of getting a disc bearing a two-digit number, \(P\left(E_1\right)=\frac{81}{90}=\frac{9}{10}\)
(ii) Let E₂ = Event of getting a disc bearing a perfect square number
Here, perfect square number are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
\(\therefore\) Number of outcomes favourable to E₂ =9
Hence, probability of getting a disc bearing a perfect square number, \(P\left(E_2\right)=\frac{9}{90}=\frac{1}{10}\)
(iii) Let E₃ = Event of getting a disc bearing a number divisible by 5
Here, the numbers divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90
\(\therefore\) Number of outcomes favourable to E₃ = 18
Hence, required probability = \(P\left(E_3\right)=\frac{18}{90}=\frac{1}{5}\)

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
OV/OS = cos 60º
OV/12 = 1/2
OV = 6 cm
\(S \frac{V}{S} O=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
\(\frac{S V}{12}=\frac{\sqrt{3}}{2} \)
\(S V=6 \sqrt{3} \mathrm{~cm} \)
\(S T=2 S V=2 \times 6 \sqrt{3}=12 \sqrt{3} \mathrm{~cm}\)
Area of ΔOST = 1/2 x ST x OV
\(\frac{1}{2} \times 12 \sqrt{3} \times 6 \)
\(=36 \sqrt{3}=36 \times 1.73=62.28 \mathrm{~cm}^{2}\)
Area of sector OSUT \(=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}\)
Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm²

Let CD = h m be the height of the tower. At point D of the tower, a man is standing and observes the car at an angle of depression of 30°. After six seconds, the angle of depression of the car is 60°.
i.e. \(\angle\)ODA = 30° and \(\angle\)ODB = 60°
\(\Rightarrow\) \(\angle\)DAC = \(\angle\)ODA = 30°  [alternate angles]
and \(\angle\)DBC = \(\angle\)ODB = 60°  [alternate angles]
Let AB = y m and BC = x m
In right angled \(\Delta\)BCD,

\(\begin{array}{rlrl} \tan 60^{\circ} & =\frac{P}{B}=\frac{C D}{B C} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \sqrt{3} & =\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad h & =\sqrt{3} x & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \end{array}\)
\(\Rightarrow \quad h = \sqrt3 x\)....(i)
In right angled \(\Delta\)ACD,
\(\begin{array}{rlrl} \tan 30^{\circ} & =\frac{C D}{A C}=\frac{C D}{A B+B C} & & {[\because A C=A B+B C]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{h}{x+y} & & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad x+y & =h \sqrt{3} & \end{array}\)
\(\Rightarrow \quad x + y=\sqrt3 x(\sqrt3)\) [from Eq. (i)]
\(\Rightarrow\) x + y = 3x  .....(ii)
It is given that a car moves from point A to B in six seconds. Let its speed be k km/s.
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
\(\Rightarrow \quad 6=\frac{y}{k} \Rightarrow y=6 k\)
On putting y = 6k in Eq. (ii), we get
x + 6k = 3x \(\Rightarrow\) 6k - 2x \(\Rightarrow\) x = 3k
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}=\frac{x}{k}=\frac{3 k}{k}=3 \mathrm{~s}\)
Hence, the car moves from point B to point C in 3s.

Given A quadrilateral ABCD, circumscribing a circde.
To prove AB + CD = AD + BC
Proof We know that the lengths of tangents drawn from an external point to a circde are equal.
\(\therefore\)   AP = AS               ...(i)
[\(\because\) both are tangents to a circle from point A]
Similarly, BP = BQ,   ...(ii)
CR = CQ                  ....(iii)
and DR = DS           .....(iv)
On adding Eqs. (i), (ii), (ii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
\(\Rightarrow\)   AB + CD = (AS + DS) + (BQ + CQ)
\(\Rightarrow\)  AB + CD = AD + BC
Hence Proved.

Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them.
Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken out is blue’, and R be the event ‘the ball taken out is red’. Now, the number of possible outcomes = 3.
(i) The number of outcomes favourable to the event Y = 1.
So, P(Y) = \(\frac{1}{3}\)
Similarly, (ii) P(R) = \(\frac{1}{3}\)and
(iii) P(B) =\(\frac{1}{3}\)

\(\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}\)
\(\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\)
\(=\frac{5\left(\frac{1}{4}\right)+\left(\frac{16}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}}\)
\(=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}=\frac{67}{12}\) 

We are given two concentric circles C₁ and C₂ with centre O and a chord AB of the larger circle C₁ which touches the smaller circle C₂ at the point P (see Fig). We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C₂ at P and OP is its radius. Therefore, by Theorem.
OP \(\perp\) AB
Now AB is a chord of the circle C₁ and OP \(\perp\) AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,
i.e., AP = BP

Here, S₁₄ = 1050, n = 14, a = 10.
As  \(\begin{aligned} \mathrm{S}_n & =\frac{n}{2}[2 a+(n-1) d] \\ \end{aligned}\)
So, \(\begin{aligned} 1050 & =\frac{14}{2}[20+13 d]=140+91 d \end{aligned}\)
i.e., 910 = 91d
or, d = 10
Therefore, a₂₀ = 10 + (20 - 1) \(\times\) 10 = 200, i.e. 20th term is 200.

We know that the formula to calculate simple interest is given by
Simple Interest  \(=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\)
So, the interest at the end of the 1st year = \(Rs \frac{1000 \times 8 \times 1}{100}=Rs 80\)
The interest at the end of the 2nd year = \(Rs \frac{1000 \times 8 \times 2}{100}=Rs 160\)
The interest at the end of the 3rd year = \(RS \frac{1000 \times 8 \times 3}{100}=Rs 240\)
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240, .
It is an AP as the difference between the consecutive terms in the list is 80, i.e., d = 80. Also, a = 80
So, to find the interest at the end of 30 years, we shall find a₃₀
Now, a₃₀ = a + (30 – 1) d = 80 + 29 x 80 = 2400
So, the interest at the end of 30 years will be Rs. 2400

Given, diameter of circle, d = 35 mm
\(\therefore\) Circumference of circle = \(\pi\)d      [\(\because\) d = 2r]
\(=\frac{22}{7} \times 35=110 \mathrm{~mm}^2\)
Now, length of 5 diameters = 5 \(\times\) 35 = 175 mm
(i) Total length of the silver wire = \(\pi\)d + 5d
= 110 + 175 = 285 mm²
(ii) Here, we see that total circle is divided into 10 sectors.
\(\therefore\) Angle of each sector = \(\frac{360^{\circ}}{10}=36^{\circ}\)
Then, area of each sector ofthe brooch = \(=\frac{\theta}{360^{\circ}} \times \pi r^2\)
\(\begin{aligned} & =\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2 \quad\left[\because r=\frac{d}{2}=\frac{35}{2} \mathrm{~mm}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}=\frac{11 \times 35}{2 \times 2}=\frac{385}{4} \mathrm{~mm}^2 \end{aligned}\)