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

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°

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

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%

-2, 2, -2, 2, -2
t₂ - t₁ = 2-(-2) = 4
t₃ - t₂ = -2 -2 = -4
t₄ - t₃ = 2 - (-2) = 4
It is not an A.P.

f(3) =. 2x - 1
= 2(3) - 1 = 6 - 1 = 5

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

(A U B) x  C = (A x C) U (B x C)
A = {0,1} , B = {2,3,4} , C = {3,5}
\(A\cup B\) = {0,1,2,3,4}
\((A\cup B)\times C\) = {0,1,2,3,4} x {3,5}
= {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}             ...(1)
A x C = {0,1} x {3,5}
= {(0,3),(0,5),(1,3),(1,5)}
B x C = {2,3,4} x {3,5}
= {(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}
\((A\times C)\cup (B\times C)\) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}                .....(2)
From (1) and (2), it is clear that
(A U B) x  C = (A x C) U (B x C)
Hence verified.

Number of non defective bulbs = 20.
Let x be the number of defective bulbs
Then total number of bulbs n(s) = 20 + x
Let 'A' be the event of getting defective bulbs
\(\mathrm{P}(\mathrm{A})=\frac{x}{20+x}=\frac{3}{8}\)
8x = 3 (20 + x)
8x = 60 + 3x
8x - 3x = 60
5x = 60
\(x=\frac{60}{5}\)
x = 12
Number of defective bulbs = 12

Let h₁ and h₂ be the heights of a cone and cylinder respectively.
Also, let r be the radius of the cone.
Given that, height of the cone h₁ = 24 cm; radius of the cone and cylinder r = 6 cm
Since, Volume of cylinder = Volume of cone
\({ \pi r }^{ 2 }=\frac { 1 }{ 3 } { \pi r }^{ 2 }{ h }_{ 1 }\)
\({ h }_{ 2 }=\frac { 1 }{ 3 } \times { h }_{ 1 }\quad gives\quad { h }_{ 2 }=\frac { 1 }{ 3 } \times 24=8\)
Therefore, height of cylinder is 8 cm

Given vertices are A (6, 2), B (- 5, - 1) and C (1, 9)
Median through A :
Let D be the mid point of BC
Mid point of BC \(=\mathrm{D}\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\(=\mathrm{D}\left(\frac{-5+1}{2}, \frac{-1+9}{2}\right) \)
= D (-2 , 4)
Now AD is the median.
Equation of AD \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
\(\frac{y-2}{4-2}=\frac{x-6}{-2-6} \)
\(\frac{y-2}{2}=\frac{x-6}{-8} \)
- 4y + 8 = x - 6
x + 4y - 14 = 0
Altitude through A

Altitude is passing through 'A' and perpendicular to BC.
Now,
Slope of BC = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{-1-9}{-5-1}=\frac{-10}{-6}=\frac{5}{3}\)
Slope of Altitude \(=-\frac{3}{5}\)
Equation of the altitude which is passing through A (6,2)and having slope \(-\frac{3}{5}\) is
y - y₁ = m(x - x₁)
\(y-2=-\frac{3}{5}(x-6)\)
5y- 10 = -3x + 18
3x + 5y - 28 = 0

LHS = sec⁶\(\theta \)
= (sec²θ)³
= (1 + tan²θ)³
\( =1^{3}+\left(\tan ^{2} \theta\right)^{3}+3(1)\left(\tan ^{2} \theta\right)\left(1+\tan ^{2} \theta\right) \)
\({\left[\because(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]} \)
\(=1+\tan ^{6} \theta+3 \tan ^{2} \theta\left(\sec ^{2} \theta\right)\)
= tan⁶θ + 3tan²θ sec²θ + 1
= RHS

Let h, r and R be the height, top and bottom radii of the frustum.
Given that, h = 45 cm, R = 28 cm, r = 7 cm
Now, Volume \(=\frac { 1 }{ 3 } \pi \left[ { R }^{ 2 }+Rr+{ r }^{ 2 } \right] h\quad cu.units\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times \left[ { 28 }^{ 2 }+(28\times 7)+{ 7 }^{ 2 } \right] \times 45\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 1029\times 45=48510\)
Therefore, volume of the frustum is 48510 cm³

Let the weight of applies be a kg.
Let the weight of bananas be b kg.
a+b = 50
ax = Rs. 1800...(1)
by = Rs. 600 ..(2)
x = 2y ....(3)
Use (3) in (1) ⇒
∵ 3b = 2a ....(5)
∵ a + b = 50
\(a+\frac { 2a }{ 3 } =50\Rightarrow \frac { 5a }{ 3 } =50\)
\(\Rightarrow a=50\times \frac { 3 }{ 5 } \)
= 30
∴ b = 20
∴ Iniya bought 30kg of applies and 20 kg of bananas.

The required number is the L.C.M. of first ten natural numbers
i.e., L.C.M. of (1,2,3,4,5,6,7,8,9,10)

L.C.M. is 5 x 2 x 3 x 2x7 x 2 x 3 = 2520
The least number divisible by first ten natural numbers is 2520.

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

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

PQ II RS


Also

\(\angle \therefore \Delta POQ\sim \Delta SOR\) (AAA similarity criterion)

\(\bar { x } =\frac { \Sigma x }{ n } =\frac { 75 }{ 5 } \)=15
d'=\(\frac { x-\bar { x } }{ c } =\frac { x-A }{ c } \)
A is assumed mean c is common factor.
Here A= 15, 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 c
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
If 3 is added to each value, we get 8, 13, 18,23, 28 as new values.

∴ 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\(\sqrt { 2 } \)
S.D. doesn't change when a number is added or subtracted to the values.

x + y + z = 6            ....(1)
2x + 3y + 4z = 20     ...(2)
3x + 2y + 5z = 22   ....(3)

Sub. z = 3 in (5) ⇒ y - 2(3) =-4
y=2
Sub. y = 2, z = 3 in (1), we get
x+2+3=6
x=1
x= 1,y = 2, z = 3

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

Let A(1, 7), B(4, 2), C(-1, -1) and D(-4, 4) be the given paints. One way at showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonals should be equal.
Now,
AB = \(\sqrt { (1-4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
BC = \(\\ \sqrt { (4+1)^{ 2 }+(2+1)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
CD = \(\sqrt { (-1+4)^{ 2 }+(-1-4)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
DA =\(\sqrt { (1+4)^{ 2 }+(7-4)^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
AC = \(\sqrt { (1+1)^{ 2 }+(7+1)^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
BD = \(\\ \sqrt { (4+4)^{ 2 }+(2-4)^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
Since, AB = BC = CD = DA and AC = BD, all the four sides at the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

 43.5, 13.6, 18.9,38.4,61.4,29.8
Largest value L= 61.4
Smallest value S = 13.6
R = L - S
= 61.4 - 13.6 = 4
Co-efficient of range = \(\frac { L-S }{ L+S } \)
= \(\frac { 47.8 }{ 75 } =0.64\)
Range = 47.8; co-efficient of range = 0.64.

\(\frac { { x }^{ 4 }{ b }^{ 2 } }{ x-1 } \times \frac { { x }^{ 2 }-1 }{ { a }^{ 4 }{ b }^{ 3 } } =\frac { { x }^{ 4 }\times { b }^{ 2 } }{ x-1 } \times \frac { \left( x+1 \right) \left( x-1 \right) }{ { a }^{ 4 }\times { b }^{ 3 } } =\frac { { x }^{ 4 }\left( x+1 \right) }{ { a }^{ 4 }b } \)

Let AD is the first building.
BC is the second building.
AD = BE = BC - CE
o
From the right triangle CED
\(\tan 45^{\circ} =\frac{C E}{D E} \)
\(1 =\frac{C E}{A B}=\frac{C E}{70 m} \)
CE = 70 m
BE = BC - EC
= 120 m - 70 m = 50 m
AD = 50 m
Height of the first building is 50 m.

Given the A.P. - 11, - 15, - 19,....
Here First term a = - 11
Common difference d = t₂ - t₁ = - 15 - ( - 11)
= -15 + 11
d = - 4
n^th term of an A.P. is t_n = a + (n - 1)d
19^th term (t₁₉) = -11 + (19 - 1) (-4)
= -11 + 18 (- 4)
= -11 + (- 72) = -83 
19th term of -11,-15, -19,...is - 83.

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

Given function f({1,2), (2,3), (3,2), (4,), (5,4)}
(i) An arrow diagram
 
(ii) a table form

(iii) A graph representation

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0

x² + 2x + 5 = 0
Let y = x² + 2x + 5
Step 1 Prepare a table of values for the equation y = x² + 2x + 5

Step 2: Plot the above ordered pairs(x, y) on the graph using suitable scale.

Step 3: Join the points by a free-hand smooth curve this smooth curve is the graph of  y = x² + 2x + 5
Step 4: The solutions of the given quadratic equation are the x coordinates of the intersecting points of the parabola the X axis.
Here the parabola doesn’t intersect or touch the X axis.
So, we conclude that there is no real root for the given quadratic equation.

-1

Diameter = 6 cm
Radius =\(\frac { 6 }{ 2 } =3cm\) 

Length of the tangents PA = PB = 4 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 3cm.
(2) Draw a line OP = 5 cm
(3) Draw a bisector of OP, which cuts OP and M
(4) With M as centre and MO as radius draw a circle which cuts previous circle at A and B
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 4 cm.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.