Let E_i denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Then
Probability of the outcome 1 = \(\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{\text { Frequency of } 1}{\text { Total number of times the die is thrown }}\)
\(=\frac{179}{1000}=0.179\)
Similarly, \(\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{150}{1000}=0.15, \quad \mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{157}{1000}=0.157,\)
\(\mathrm{P}\left(\mathrm{E}_{4}\right)=\frac{149}{1000}=0.149, \quad \mathrm{P}\left(\mathrm{E}_{5}\right)=\frac{175}{1000}=0.175\)
\(\mathrm{P}\left(\mathrm{E}_{6}\right)=\frac{190}{1000}=0.19\)

Area of a parallelogram is divided into four equal parts by the diagonals.
\(\Rightarrow \ ar(AOB)=\frac { 1 }{ 4 } ar(ABCD)\)
\(ar(\Delta BOP)=\frac { 1 }{ 2 } ar(AOB)=\frac { 1 }{ 8 } ar\ (ABCD)\)
\(=\frac { 1 }{ 8 } \times 80=10\quad { cm }^{ 2 }\)

\(\text { Mean }=\frac{5000+5000+5000+5000+15000}{5}=\frac{35000}{5}=7000\)
So, the mean salary is Rs. 7000 per month.
To obtain the median, we arrange the salaries in ascending order:
5000, 5000, 5000, 5000, 15000
Since the number of employees in the factory is 5, the median is given by the \(\left(\frac{5+1}{2}\right) \mathrm{th}=\frac{6}{2} \mathrm{th}=3 \mathrm{rd}\) 3rd observation. Therefore, the median is Rs. 5000 per month.
To find the mode of the salaries, i.e., the modal salary, we see that 5000 occurs the maximum number of times in the data 5000, 5000, 5000, 5000, 15000. So, the modal salary is Rs. 5000 per month.

Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2\(\pi\)r.
So, the radius of the dome \(=17.6 \times \frac{7}{2 \times 22} \mathrm{~m}=2.8 \mathrm{~m}\)
The curved surface area of the dome = 2\(\pi\)r²
\(=2 \times \frac{22}{7} \times 2.8 \times 2.8 \mathrm{~m}^{2}\)
= 49.28 m²
Now, cost of painting 100 cm² is RS. 5.
So, cost of painting 1 m² = RS. 500
Therefore, cost of painting the whole dome
= RS. 500 x 49.28
= RS. 24640

Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of D ABD is given by
\(s=\frac{100+100+160}{2} \mathrm{~m}=180 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ABD}=\sqrt{180(180-100)(180-100)(180-160)}\)
\(=\sqrt{180 \times 80 \times 80 \times 20} \mathrm{~m}^{2}=4800 \mathrm{~m}^{2}\)
Therefore, each of them will get an area of 4800 m².

\(36°\),\(144°\),\(36°\), \(144°\)

x=50, \(92^{ 0 }\) and \(88^{ 0 }\)

Take any line l and a point P not on l. Then, by Playfair’s axiom, which is equivalent to the fifth postulate, we know that there is a unique line m through P which is parallel to l.

0x+5y-2=0

\(8x^3-y^3+27z^3+18xyz\)

7.87

\(2-\sqrt { 3 } \)

Required probability=\(\frac { 2+1 }{ 30 } =\frac { 1 }{ 10 } \)

Frequency of 4 = 3
Frequency of  9 = 3
For mode to be 9, x -10 = 9 \(\Rightarrow\) x =19
so that the observation 9 has the maximum frequency.

Length of the rod \(=\sqrt { { \left( 10 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 } } \)

\(\angle POR=2\angle PSR=2(180°-\angle PQR)\) 
\( =2(180°-100°)=160°\)
OP=OR
∴ \(\angle OPR=\angle ORP\)

A rectangle is essentially a parallelogram. A square is essentially a parallelogram. Parallelograms on the same base and between the same parallels are equal in area

\(\angle 1\) and \(\angle 2\) are simply adjacent angles

\(ax+b(0)+c=0 \ \ \ \Rightarrow\ \ \ x=-{c\over a}\)
\(a(0)+by+c=0\ \ \ \Rightarrow\ \ \ y=-{c\over b}\)

Evident

Steps of construction:
i) Draw a ray BX and cut off line segment BC = 4cm
ii) Construct ㄥXBY = 90^o
iii) From BY cut off line segment BD = 8 cm,
iv) Join CD.
v) Draw the ⊥ bisector of CD, intersecting BD at A
vi) Join AC, ABC is the required triangle

ar(\(\Delta\)DRC) = ar(\(\Delta\)DPC),   (Given)
But they are on the same base DC.
Therefore, \(\Delta\)DRC and \(\Delta\)DPC must lie between the same parallels.
So, DC || RP
i.e., one pair of opposite sides of quadrilateral DCPR is parallel.
therefore, DCPR is a trapezium.
Also, ar(\(\Delta\)BDP) = ar(\(\Delta\)ARC)    (Given)      ........(i)
and ar (\(\Delta\)DPC) = ar(\(\Delta\)DRC) (Given) .........(ii)
Subtracting (ii) from (i), we get
ar(\(\Delta\) BDP) - ar(\(\Delta\)DPC) = ar(\(\Delta\)ARC) - ar(\(\Delta\)DRC) ar(\(\Delta\)BDC) = ar(\(\Delta\)ADC)

But they are on the same base DC.
Therefore, \(\Delta\)BDC and \(\Delta\)ADC must lie between the same parallels.
So, AB || DC
i.e., One pair of opposite sides of quadrilateral ABCD is parallels.
So, AB || DC
i.e., One pair of opposite sides of quadrilateral ABCD is parallel.
Therefor, ABCD is a trapezium.

EQ||FP and transversal cut them
\(\therefore \angle AQD=\angle APF\) (Conresponding angles)
\(\angle AQD=50^o\)
\(\therefore \angle DQB=180^o-50^o=130^o\)
AB||CD and transversal EQ cuts them
\(\therefore \angle EDG=\angle DQB=130^o\)
\(\therefore \angle EDG=130^o-40^o=90^o\)
FP||EQ and transversal PG cut them
\(\therefore \angle FPD=\angle EDG=90^o\)

In \(\triangle\)AFE and \(\triangle\)CDE,
\(\therefore\) AF = CD
\(\therefore\) \(\angle\)AFE = \(\angle\)CDE
\(\therefore\) \(\angle\)E = \(\angle\)E
\(\therefore\) \(\angle\)FAE = \(\angle\)DCE
\(\therefore\) \(\triangle AFE\cong \triangle CDE\)
\(\therefore\) EF =ED

9x²+y²+z²-6xy+2yz-6xz
=(-3x)²+(y)²+(z)²+2\(\times\)(-3x)(y)+2\(\times\)(y)(z)+2(-3x)(z)
=(-3x+y+z)²
If x=1, y=2, z=-1, then
(-3x+y+z)²=(-3\(\times\)2-1)²
=4.

\((i)\frac { 1 }{ 10 }\)  
\((ii)\frac { 1 }{ 50 }\) 
\((iii)\frac { 8 }{ 25 } \)

Inner diameter = 10.5 cm
\(\therefore\)  Inner radius (r) = \(\frac { 10.5 }{ 2 } cm=5.25cm\)
\(\therefore\)  Inner surface area = \(2\pi { r }^{ 2 }\)
\(=2\times \frac { 22 }{ 7 } \times { \left( 5.25 \right) }^{ 2 }=173.25{ cm }^{ 2 }\)
\(\therefore\)  Cost of tin-plating at the rate of Rs 16 per 100 cm²
= Rs \(\frac { 16 }{ 100 } \times 173.25\)= Rs 27.72.

Rs.5376

Given: ABCD is a trapezium whose nonparallel sides AD and BC are equal.
To Prove: Trapezium ABCD is cyclic.
Construction: Draw BE 11 AD.
Proof: ∵ AB || DE        I Given
and     AD || BE           I By construction
∴  Quadrilateral ABCD is a parallelogram.

∴ \(\angle BAD=\angle BED\)                        ....(1)
| Opp.\(\angle \) s of a || gm are equal
and    AD = BE                                 ...(2)
Opp. sides of a || gm are equal
But     AD = BC            ...(3) I Given
From (2) and (3),
BE = BC
∴ \(\angle BEC=\angle BCE\)        ....(4)
| Angles opposite to equal sides of a triangle are equal
\(\angle BEC+\angle BED=180°\)       | Linear Pair Axiom
⇒ \(\angle BCE+\angle BAD=180°\) I From (4) and (1)
⇒ Trapezium ABCD is cyclic.
I ∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic

If a straight line I falls on two straight lines m and n such that sum of the interior angles on one side of I is two right angles, then by Euclid's fifth postulate the lines m and n will not meet on this side of I. Next, we know that the sum of the interior angles on the other side of line I will also be two right angles. Therefore, they will not meet on the other side also. So, the lines m and n never meet and are, therefore arallel. 

y = 3x
y =3x
Table of solution

We plot the points (0, 0) and (1, 3) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 3x.

(i) The point (-2,4) lies in the II quadrant.
(ii) The point (3, -1) lies in the IV quadrant.
(iii) The point (- 1,0) lies on the negative x-axis.
(iv) The point (1, 2) lies in the I quadrant.
(v) The point (- 3, - 5) lies in the Ill quadrant.

\({ 125 }^{ 1/3 }={ \left( 5^{ 3 } \right) }^{ 1/3 }\)
\(\\ ={ 5^{ 3\times 1/3 } }={ 5 }^{ 1 }=5\)

OP=OR
= 10 cm (radii)
PQ = 16 cm,
RS = 12 cm
Let OL \(\bot\) PQ and OM\(\bot\)RS
Since perpendicular from the centre bisects the chord
\(\therefore\) PL = LQ = \(\frac { 1 }{ 2 } \)PQ = 8 cm
RM =MS=\(\frac { 1 }{ 2 } \)RS= 6cm
In right triangle OLP
OP² = OL² + PL²
(ByPythagoras theorem)
100 = OL² + 64
OL=\(\sqrt { 100-64 } =\sqrt { 36 } \)
OL =6cm
In right triangle OMR (By Pythagoras theorem)
OR² = OM² + RM²
100 = OM² + 36
OM=8cm
(i) If PQ and RS lie on the same side of centre O

The distance between PQ and RS
=LM= OM-OL
=8-6 = 2 cm
(ii) PQ and RS lie on opposite sides of centre O

The distance between PQ and RS
=LM
=OL+OM
=6+8 cm
= 14 cm

Let, Side of new cube = x
\(\therefore\) (12)³ = 8x³
or, \(\frac { { (12) }^{ 3 } }{ 8 } ={ x }^{ 3 }\)
\(\Rightarrow \ { \left( \frac { 12 }{ 2 } \right) }^{ 3 }={ x }^{ 3 }\)
or 6 = x
\(\therefore\) Side of new cube = 6 cm.
Ratio of surface areas of 8 new cubes to original cube
\(=\frac { 8\times { 6x }^{ 2 } }{ { 6(12 })^{ 2 } } \)
\(=\frac { 8\times 6(6)^{ 2 } }{ 6\times 12\times 12 } \)
\(=\frac { 8\times 6\times 6\times 6 }{ 6\times 12\times 12 } =\frac { 2 }{ 1 } \)
\(\therefore\) Ratio = 2:1.

To prove that: Ar(||^gm DLOP) = Ar(||^gm BMOQ)
Proof: Give ABCD is a ||^gm and AC is one if its digonal
\(\therefore\) ar(\(\Delta\)ABC) = ar(\(\Delta\)ADC)     .......(i)
[\(\because\) The diagonal of a ||^gm divides it into two equal triangles]
Now, LM is parallel to AD
(\(\because\) AD=BC and AD||BC as ABCD is a ||^gm)
\(\therefore\) LM||BC \(\Rightarrow\) LO||CQ
Similarly PQ||DC \(\Rightarrow\) OQ||LC
\(\therefore\)   CLOQ is also a ||^gm, as its opposite sides on parallel to each other.
In ||^gm CLOG, CO is the diagonal
\(\therefore\)   ar(\(\Delta\)COQ) = ar(\(\Delta\)CLO)      ..........(ii)
[\(\because\) The diagonal of a ||^gm divides it into two equal triangles]
Again, AD || LM \(\Rightarrow\) AP||OM
and AB || PQ \(\Rightarrow\) AM||PO
\(\therefore\) APOM is also a ||^gm, as its opposite sides one parallel to each other.
In ||^gm AOPM, AO is the diagonal
\(\therefore\) ar(\(\Delta\) AOP) = ar(\(\Delta\)AOM)    .........(iii)
[\(\because\) The diagonal of a ||^gm divides it into two equal triangles]
Now using (i) ar(\(\Delta\)ABC) = ar(\(\Delta\)ADC)
\(\Rightarrow\)  ar(\(\Delta\)COQ) + ar(\(\Delta\)AOM) + ar(BMOQ) = ar(\(\Delta\)CLO) + ar(AOP)+ar(DLOP)
using (ii) and (iii), we get
ar(BMOQ) = ar(DLOP)
Now, BMOQ is a ||^gm as
PQ||AB \(\Rightarrow\) OQ||BM
and BC || AD \(\Rightarrow\) OM||BQ
Similarly, DLOP is also || gm as
AD||LM \(\Rightarrow\) DP||LO
and PQ||DC \(\Rightarrow\) PO||DL
\(\therefore\) ar(||^gm BMOQ) = ar(||^gm DLOP)
Hence Proved.

Since \(\angle1=\angle3 \ and \ \angle2=\angle4,\) therefore adding both equation
\(\angle1+\angle2=\angle3+\angle4\)
\(\Rightarrow \angle BAD=\angle BCD\)
\(\Rightarrow \angle A=\angle C\)
If equals are added to equal, the wholes are equal
Two more axioms:
Things which are equal to the same thing are equal to one another
e.g., if \(\overline {AB}=\overline{ PQ} \ and\ \overline {PQ}=\overline {XY}, then\ \angle{AB}=\angle{XY}\)
If equals are subtracted from equals, the remainders are equal.
e.g., if \(m\angle1=m\angle2\) then
\(m\angle3=m\angle3\)
\(=m\angle2-m\angle3\)

Let p(x)=bx³+3x²-3
Put x-4 = 0 or x = 4 in p(x),
We get p(4) = b(4)³+3(4)²- 3
=R₁
64b + 48-3 = R₁
64b+45=R₁
Let q(x) = 2x³-5x+b
Again put x-4=0 or x=4 in q(x),
We get q(4) = 2(4)³-5(4)+b = R₂
= 128 - 20 + b = R₂
108+b = R₂
Given that 2R₁-R₂=0
\(\Rightarrow\) 2(64b+45)-(108+b) = 0
\(\Rightarrow\) 128b + 90 - 108 -b = 0
\(\Rightarrow\) 127b-18 = 0
\(\Rightarrow\) \(b=\frac { 18 }{ 127 } \)

Number of two wheelers = 57
Number of three wheelers = 33
Number of four wheelers = 30
Total number of vehicles = 57 + 33 + 30 = 120
Number of vehicles that is not a four-wheeler = 57 + 33 = 90
Probability that the vehicle chosen at random is not a four-wheeler
 \(\frac { 90 }{ 120 } =\frac { 3 }{ 4 } \)

Arranging the given data in the ascending order, we have
40,50,65,70,75,75,95,100
Number of observations (n) = 8 which is even, \(\therefore\) Median
\(=\frac { { \left( \frac { n }{ 2 } \right) }^{ th }observation+{ \left( \frac { n }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(=\frac { { \left( \frac { 8 }{ 2 } \right) }^{ th }observation+{ \left( \frac { 8 }{ 2 } +1 \right) }^{ th }observation }{ 2 } \)
\(=\frac { { 4 }^{ th }observation+{ 5 }^{ th }observation }{ 2 } \)
\(=\frac { 70+75 }{ 2 } \)
= 72.5

Let the third side be x cm. Then, 12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm, b = 12 cms, c = 6 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 12+12+6 }{ 2 } \)
= 15 cm
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
 \(=9\sqrt { 15 } \) cm²

(i) Steps of Construction
1. Draw the base BC = 5 em.
2. At point B make an angle XBC = 45°.
3. CutthelinesegmentBD=AB-AC(=2.8 cm) from the ray BX.
4. Join DC.
5. Draw the perpendicular bisector, say PQ of DC.
6. Let it intersect BX at a point A.
7. Join AC.
Then, ABC is the required triangle.

(ii) By measurement, AB = 13cm
(iii) By measurement, AC = 10.2cm
(iv) AB - AC = 13 - 10 .2 = 2.8cm
(v) Yes! Hari is true as by measurement ㄥACB = 112°.
The value 'wise' is depicted by comment of Hari.

Given: ABC is an isosceles triangle in which AB = AC. AD bisects L PAC and CD IIAB.
To Prove:
(i)  \(\angle \)DAC =\(\angle \)BCA
(ii) ABCD is a parallelogram.
Proof:
(i) In \(\Delta \) ABC,
\(\because\) AB = AC
\(\therefore\) \(\angle \)B =\(\angle \)C               .......(1)      I Angles opposite to equal sides of a triangle are equal
Also, Ext.   \(\angle \)PAC =\(\angle \)B +\(\angle \)C     I An exterior angle of a triangle is equal to the sum of its two interior opposite angles
⇒    \(\angle \)PAC = \(\angle \)C+\(\angle \)C   | From (1)
⇒ 2\(\angle \)CAD = 2\(\angle \)C   | \(\because\) AD bisects \(\angle \)PAC 
⇒ \(\angle \)CAD =\(\angle \)C
⇒  \(\angle \) DAC =\(\angle \)BCA
(ii) But these angles form a pair of equal alternate interior angles
\(\therefore\)  AD II BC
Also, CD II AB         I Given
\(\therefore\)   ABCD is a parallelogram   IA quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel.

Given AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD 
To prove \(2\angle APD=\angle B+\angle C\)
We know that the sum of all the angles of a quadrilateral is \(360^{ 0 }\)
\(\Rightarrow \angle A+\angle D=360^{ 0 }-(\angle B+\angle C)\)
Now in \(\triangle \)PAD
\(\angle APD+\angle PAD+\angle PDA=180^{ 0 }\)
|Angle sum property of  a triangle  
\(\Rightarrow \angle APD+\frac { 1 }{ 2 } \angle A+\frac { 1 }{ 2 } \angle D=180^{ 0 }\)
|AP and DP are the bisectors of two adjacent angles And D of quadrilateral ABCD 
\(\Rightarrow 2\angle APD=360^{ 0 }-(\angle A+\angle D)\) 
\(\\ \Rightarrow 2\angle APD=\angle B+\angle C\) 

x=3y
x-3y=0
(1)x+(-3)y+(0)=0
Comparing with ax+by+c=0, we get
a=1
b=-3
c=0
Now x=3y
\(\Rightarrow\ \ \ y={x\over 3}\)
Put x=0, then \(y={0\over3}=0\)
Put x=3, then \(y={3\over 3}=1\)
Hence, (0, 0) and (3, 1) are the two solutions of the equation x=3y.

(i) The point A(-3, -4) lies in III quadrant.
(ii) The point B(-2, 0) lies on x-axis.
(iii) The point C(-1, 4) lies in II quadrant.
(iv) The point D(1, 0) lies on x-axis.

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.

(i) Let us assume that A, Band C are the position of Priyanka, Sania and David respectively on the boundary of circular park with centre O.
Draw AD\(\bot\) BC
Since the centre of the circle coincides with the centroid of the equilateral \(\triangle\) ABC
\(\therefore\) Radius of circumscribed circle = \(\frac{2}{3}\) AD
\(\Rightarrow 20=\frac { 2 }{ 3 } AD\)
\(\Rightarrow AD=20\times \frac { 2 }{ 3 } \)
\(\Rightarrow AD=30m\)
Now, AD\(\bot\)BC, and let AB=BC=CA=x
\(\Rightarrow BD=CD=\frac { 1 }{ 2 } BC=\frac { x }{ 2 } \)

In rt. \(\triangle\)BDA, D=90⁰
By Pythagoras Theorem, we have
AB²=BD²+AD²
\(\Rightarrow { x }^{ 2 }={ \left( \frac { x }{ 2 } \right) }^{ 2 }+{ (30) }^{ 2 }\)
\(\Rightarrow{ x }^{ 2 }-{ \frac { { x }^{ 2 } }{ 4 } }=90\)
\(\Rightarrow \frac { 3 }{ 4 } { x }^{ 2 }=90\)
\(\Rightarrow { x }^{ 2 }=900\times \frac { 4 }{ 3 } \)
\(\Rightarrow { x }^{ 2 }=1200\)
\(\therefore x=\sqrt { 1200 } =20\sqrt { 3 } \)
Hence, the distance between each of them is \(20\sqrt { 3 } .\)
(ii) Properties of the circle, equilateral triangle and Pythagoras theorem.
(iii) Live and let live.

Proof: QO is bisector of \(\angle\)PQR
\(\angle\)OQR = \(\frac{1}{2}\)\(\angle\)PQR = \(\frac{1}{2}\) =\(\angle\)Q
RO is bisector \(\angle\)ORQ
\(\therefore\) \(\angle\)ORQ =\(\frac{1}{2}\) \(\angle\)PRQ = \(\frac{1}{2}\) \(\angle\)R
In \(\angle\)OQR
\(\angle\)QOR + \(\angle\)OQR + \(\angle\)ORQ = 180°
(Angle sum property)
\(\angle\)QOR + \(\frac{1}{2}\) \(\angle\)Q + \(\frac{1}{2}\) \(\angle\)R = 180°
\(\angle\)QOR = 180°- \(\frac{1}{2}\)(\(\angle\)Q + \(\angle\)R)
But in \(\angle\)PQR
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180°
\(\angle\)Q + \(\angle\)R = 180°- \(\angle\)P
\(\angle\)QOR = 180°- \(\frac{1}{2}\) (180°- \(\angle\)P)
= 180°-90° + \(\frac{1}{2}\)\(\angle\)P
= 90° + \(\angle\)P Hence Proved.
These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio.

(i) Required number of students = 8 + 32 = 40
(ii) Here, We notice that classes are continuous but class-size is not the same for all the classes. We notice minimum class-size is of class 45-50, i.e., 5. We will first find proportionate length of rectangle (adjusted frequency) for each class.
Length of rectangle (adjusted frequency) =\(\frac { Frequency\ of\ Class }{ Width\ of\ class } \times Minimum\ class-size\)

Now, we construct rectangles with respective class-intervals as widths and adjusted frequencies as heights.
Histogram representing marks obtained by students in unit test of Mathematics.

(iii) Hardwork and Dilligence.

(i) As EB II DL and ED II BL.
Therefore EBLD is a parallelogram.
ஃ BL = ED
=\(\frac{1}{2}\)BC=CL ...(i)
Now in triangles DCL and FBL, we have
CL=BL from (i)
ㄥDLC= ㄥFLB 
ㄥCDL=ㄥBFL
ΔCDL≡ΔBFL
CD= BF
and DL= FL
Now, BF = DC = AB
⇒ 2AB = 2DC
⇒ AF = 2DC
(ii) ∵ DL=FL
⇒ DF = 2DL

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x³ is an irrational number but x⁵ is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x³ = (5√7)³ = (7)^3/5
is an irrational number.
But x⁵ = (\(\sqrt[5]{7}\))⁵=(7)^5/5  = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.