We have DB = 3 CD.
BC = BD + DC
BC = 3CD + CD
BC = 4CD
\(CD=\cfrac { 1 }{ 4 } BC\)
\(CD=\cfrac { 1 }{ 4 } BC\)
\(BD=3cD=\cfrac { 3 }{ 4 } BC\)
Since  \(\Delta ABD\) is a right triangle (i) right angled at D.
AB² = AD² + BD²
By \(\Delta ACD\) is a right triangle right angled at D 
AC² = AD² + CD²
Subtracting equation (iii) from equation (ii), we got
AB² - AC² = BD² - CD²
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\left( \cfrac { 3 }{ 4 } BC \right) ^{ 2 }-\left( \cfrac { 1 }{ 4 } BC \right) ^{ 2 }\)
\((from \ CD=\cfrac { 1 }{ 4 } BC,BD=\cfrac { 3 }{ 4 } BC)\)
(i) \(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 9 }{ 16 } { BC }^{ 2 }-\cfrac { 1 }{ 16 } { BC }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } { BC }^{ 2 }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } BC{ 2 }^{ 2 }\)
\(\Rightarrow 2({ AB }^{ 2 }-{ AC }^{ 2 })={ BC }^{ 2 }\)
\(\Rightarrow { 2AB }^{ 2 }=2{ AC }^{ 2 }+{ BC }^{ 2 }\)

\(\Delta ACD\sim \Delta ABC\)
So,
\(\cfrac { AC }{ AB } =\cfrac { AD }{ AC } \)
AC²  = AB ·AD
Similarly \(\Delta BCD\sim \Delta BAC\)
So,
\(\cfrac { BC }{ BA } =\cfrac { BD }{ BC } \)
BC²  = BA·BD
From (1) and (2)
\(\cfrac { { BC }^{ 2 } }{ AC^{ 2 } } =\cfrac { BA.BD }{ AB.AD } =\cfrac { BD }{ AD } \)

Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB² = BP² + OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC²  = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

Let AB be the ladder and CA be the wall with . the window at A.
Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem
AB² =  BC²+ CA²
= (2.5)² + (6)²
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.

We are given a right triangle ABC right angled at B.
We need to prove that AC² = AB² + BC²
Let us draw \(BD\bot AC\)
Now,\(\Delta ADB\sim \Delta ABC\)
\(\cfrac { AD }{ DB } =\cfrac { BC }{ AC } \)
(sides are proportional)
Also,
\(\Delta BDC\sim \Delta ABC\)
\(\cfrac { CD }{ BC } =\cfrac { BC }{ AC } \)
CD·AC = BC²     ..(2)
Adding (1) and (2)
AD .AC + CD . AC = AB²+ BC²
AC(AD + CD) = AB² + BC²
AC.AC = AB² + BC²
AC = AB² + BC²

We know:
cot A = tan(90^o - A)
So,
cot 25° = tan (90° - 25°) = tan 65°
\(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = \(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = 1

Since, sin(A - B) = \(\frac12\), ∴ A-B = 30°       ..... (1)
Also, since cos (A + B) = \(\frac12\)
∴ A + B = 60°  ...(2)
Solving (1) and (2)
A - B + A + B = 30^o + 60^o
2A = 90^o
A = 45^o
We get,
A = 45° and B = 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

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Total number of cards = 52
Event of selecting a spade card = A
Event of selecting a heart card = B
n(A) = 13, n(B) = 13
P(A) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \), P(B) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \).
A ⋂ B refers to a card is both spade and heart. It is not possible

A, B are exclusive events.
P(A,B) = P(A) + P(B) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

\(\bar { { x }_{ 1 } } \) =\(\frac { 140 }{ 5 } \) =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.

σ  = 21.2, \(\bar { x } \)  = 36.6
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 21.2 }{ 36.6 } \) x 100 = 57.92%

No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)

= 450

Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V₁ : V₂ : V₃

\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r)    (where 2r = h)
∴  V₁ : V₂ : V₃ = 3: 1 : 2

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

Mean \(\bar { x } \)  = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \)  = 17

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

Σ(x -10)² = 79 = Σx²- 20x + 100 = 79
= Σx²- 20Σx + 100 x 9 = 79
= Σx²- 20 x 99 + 900 = 79
Σx² = 79 + 1980 - 900 = 1159
Σ(x - \(\bar { x } \))² = Σ(x - 11)² = Σ(x² - 22x + 121)
= Σx² - 22Σx + 121 x 9
= 1159 - 22 x 99 + 1089 = 70
∴ Σx² = 1159, Σ(x - \(\bar { x } \))² = 70

Let the tens digits be x. Then the uints digits=\(\frac{18}{x}\)
∴ Number =10x+\(\frac{18}{x}\)
and number obtained by interchanging the digits =10x+\(\frac{18}{x}\)
\(\\ \therefore \left( 10x+\frac { 18 }{ x } \right) -\left( 10\times \frac { 18 }{ x } +x \right) =63\)
\(\Rightarrow 10x+\frac { 18 }{ x } -\frac { 180 }{ x } -x=0\)
\(\Rightarrow 9x-\frac { 162 }{ x } -63=0\)
⇒ 9x²-63x-162=0
⇒ x²-7x-18=0
⇒ (x-9)(x+2)=0⇒ x=9,-2
But a digit can never be (-ve), so x = 9.
So, the required number \(=10\times 9+\frac { 18 }{ 9 } =92\)

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 length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

Seven years ago, let Swathi's age be x years .
Seven years ago, let Varun's age was 5x² years.
Swathi's present age = x + 7 years
Varun's present age = (5x² + 7) years
3 years hence, we have
Swathi's age = x + 7 + 3 years
=x + 10 years
Varun's age = 5x² + 7 + 3 years
= 5x² + 10 years
It is given that 3 years hence Swathi's age will
be \(\frac{2}{5}\) of Varun's age.
∴ x+10=\(\frac{2}{5}\)(5x²+10)
⇒ x+10=2x²+4
⇒ 2x²-x-6=0
⇒ 2x(x-2)+3(x-2)=0
⇒(2x+3)(x-2)=0
⇒ x-2=0
⇒ x=2(∵2x+3≠0 as x>0)
Hence Swathi's present age = (2 + 7) years
= 9 years
Varun's present age = (5 x 2² + 7) years
= 27 years

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

a_n = 3 +2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11,.........
Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an A.P. with common difference d = 2.
To find S₂₄ we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac{24}{2}\) [2 x 5 + (24 - 1) x ]
= 12 [10 + 46] = 672.
So, sum of first 24 terms of the list of numbers is 672.

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.

We have
a₃ = a + (3 - 1)d = a + 2d = 5  (1)
a₇ = a + (7 - 1)d = a + 6d = 9 (2)
(1) - (2) ⇒ -4d -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

F(5) = 3x² - 10
= 3(5)²- 10 = 75 - 10 = 65

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

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

f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 x 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

4, 10, 16,22, ...
We have a² - a² = 10 - 4 = 6
a³ - a² = 16 -10 = 6
ஃ It is an A.P. with common difference 6
ஃ The next two terms are, 

The area of the triangle formed by the vertices A(1, -1), B(-4, 6) and C(-3, -5), by using the formula above, is given by
= \(\frac { 1 }{ 2 } \)[1(6 + 5) +(-4) (-5 + 1) + (-3)(-1 - 6)]
= \(\frac { 1 }{ 2 } \)[11 + 16 + 21] = 24 square units.

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).