Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

Given XY IIA C

So,

\(\therefore \Delta ABC\sim \Delta XbY\) (AAA similarity criterion)
So, \(\cfrac { ar(ABC) }{ ar(XBY) } =\left( \cfrac { AB }{ XB } \right) ^{ 2 }\)         ...(1)
ar(ABC) = 2ar(XBY)
\(\cfrac { ar(ABC) }{ ar(ABC) } =\cfrac { 2 }{ 1 } \)           ...(2)
From (1) and (2),
\(\left( \cfrac { AB }{ XB } \right) ^{ 2 }=\cfrac { 2 }{ 1 } i.e.,\cfrac { AB }{ XB } =\cfrac { \sqrt { 2 } }{ 1 } \)
\(\cfrac { XB }{ AB } =\cfrac { 1 }{ \sqrt { 2 } } \)
\(1-\frac{X B}{A B}=1-\frac{1}{\sqrt{2}}\)
\(\cfrac { AB-XB }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } \)
\(\cfrac { AX }{ AB } =\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } =\cfrac { 2-\sqrt { 2 } }{ 2 } \)

Let assumed mean A = 30
C = 5

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

D= b²-4ac
⇒ 4(ac + bd)² - 4(a² + b²)(c² + d²)
⇒ 4[(ac + bd)² - (a² + b²)(c² + d²)]
⇒ 4(a²c² + b²d² + 2acbd - a²c²b²c² - a²d² - b²d²]
⇒ 4[2acbd - a²d² - b²c²]
⇒ 4[a²d² + b²c² - 2adbc]
⇒-4[ ad - bc]²
We have ad≠ bc
∴ ad- be of 0
⇒ (ad - bc)² > 0
⇒ 4(ad - bc)² < 0 ⇒ D < 0
Hence the given equation has no real roots.

Since 12576 > 4052 we apply the division lemma to 12576 and 4052, to get
12576 = 4052 x 3 + 420.
Since the remainder 420 ≠ 0, we apply the division lemma to 4052
4052 = 420 x 9 + 272.
We consider the new divisor 420 and the new remainder 272 and apply the division lemma to get
420 = 272 x 1 + 148, 148 ≠ 0
∴ Again by division lemma
272 = 148 x 1 + 124, here 124 ≠ 0
∴ Again by division lemma
148 = 124 x 1 + 24, Here 24 ≠ 0
∴ Again by division lemma
124 = 24 x 5 + 4, Here 4 ≠ 0
∴ Again by division lemma
24 = 4 x 6 + 0.
The remainder has now become zero. So our procedure stops. Since the divisor at this stage is 4.
∴ The HCF of 12576 and 4052 is 4.

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

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) \(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 9 }{ 16 } { BC }^{ 2 }-\cfrac { 1 }{ 16 } { BC }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } { BC }^{ 2 }\)
\(\Rightarrow { AB }^{ 2 }-{ AC }^{ 2 }=\cfrac { 1 }{ 2 } BC{ 2 }^{ 2 }\)
\(\Rightarrow 2({ AB }^{ 2 }-{ AC }^{ 2 })={ BC }^{ 2 }\)
\(\Rightarrow { 2AB }^{ 2 }=2{ AC }^{ 2 }+{ BC }^{ 2 }\)

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

BL and CM are medians at the \(\triangle\)ABC in which
\(A=\angle { 90 }^{ 0 }\)
From \(\triangle\)ABC
BC² = AB² + AC²
(Pythagoras theorem)

From \(\Delta ABL\)
BL² = AL² + AB²
\({ BL }^{ 2 }=\left( \cfrac { { AC }^{ 2 } }{ 2 } \right) +{ AB }^{ 2 }\)
(L is the mid-point at AC)
\({ BL }^{ 2 }=\cfrac { { AC }^{ 2 } }{ 4 } +{ AB }^{ 2 }\)
4BL² = AC² + 4AB²
From \(\Delta CMA\)
CM² = AC² + AM²
\({ CM }^{ 2 }={ Ac }^{ 2 }+\left( \cfrac { AB }{ 2 } \right) ^{ 2 }\)
(M is the mid-point at AB)
\({ CM }^{ 2 }={ AC }^{ 2 }+\cfrac { { AB }^{ 2 } }{ 4 } \)
4CM² = 4AC²+ AB²
Adding (2) and (3), we have
4(BL² + CM²) = 5(AC² + AB²)
4(BL² + CM²) = 5BC²

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

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

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

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%

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.

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.

1,-1,-3, -5, ...
t₂ - t₁ = -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

A = {5, 6, 8, to}, B = {19, 15,9, 11}
f(x) = 2x - 1
f(5) = 2(5) - 1 = 9
f(8) = 2(5)-1 = 15
\(\therefore\) a = 9; b = 15

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

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

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).