Let x=3k, y=2k
Then, x+y=3k+2k=180^o
(Angles on the same side of transversal)
5k=180^o
k=36⁰
x=3k=108^o
y=2k=72^o
Thus, \(\angle z=\angle x=108^o\) (Alternate interior angles)

∠APR=ㄥDRP
or ㄥ1=ㄥ2
But these are alternate interior angles
SP II RQ, SR II PQ
PQRS is a parallelogram
∠APR+ㄥBPR=180°,(linear pair)
⇒ \(\frac{1}{2}\) ∠APR+\(\frac{1}{2}\)ㄥBPR=\(\frac{1}{2}\)x180°
⇒ ∠1+ㄥ3=90°
⇒ ∠SPQ=90°

PQRS is a rectangle

PQ = QR
QX = QY

If equals are subtracted from equals, the remainders are also equal.
We have PQ - QX = QR - QY
PX = RY

\((i)\frac { 53 }{ 100 }\) 
\((ii)\frac { 3 }{ 25 } \)

84 cm²

36, 60

x=4

Comparing the given expression with (x + y + z)², we find that
x = 3a, y = 4b and z = 5c.
Therefore, using Identity V, we have
(3a + 4b + 5c)² = (3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ac

\(\quad a=-\frac { 179 }{ 171 } ,b=\frac { -20 }{ 171 } \)

P(E)+P(F)=1

\(\overset{-}{x}=\frac {3+4+5+6+7}{5}=5.\)

Volume = 15 \(\times\) 10 \(\times\) 8 = 1200 cm³

Smallest side = \(\frac { 12 }{ 25+17+12 } \times 540\) = 120 m.

\(\angle CBD=\angle CAD=50°\)

A rhombus is a parallelogram and in a parallelogram, opposite angles are equal.

↔ denotes correspondence

(a, 3a) satisfies y=3x

\(f(x)=x^3+2x^2-3ax-8\)
\(\Rightarrow \ f(4)=0\)
\(\Rightarrow \ 4^3+2 (4)^2-3a(4)-8=0\)
\(\Rightarrow \ 64+32-12a-8=0\)
\(\Rightarrow \ 12a=88\)
\(\Rightarrow \ a=\frac{22}{3}\)

\(OP\bot AB\)
\(\Rightarrow \angle POA=90^o\)
Let \(\angle POQ=a\)
\(\therefore \angle QOR=3a\)
\(\angle ROA=5a\)
\(\Rightarrow\) a+3a+5a=90^o
\(\Rightarrow\) 9a=90^o
\(\Rightarrow\)a=10^o
\(\therefore\)x=10^o
and y=3x10^o=30^o
z=5x10^o=50^o

Here, \(\angle1=\angle3, \angle2=\angle4\ and\ \angle3=\angle4\), Euclid's first axiom says, the things which are equal to same things are equal to one another.

So, \(\angle 1=\angle 2\)

\(b=\frac { 1 }{ 2+\sqrt { 5 } } \)
\(=\frac { 1 }{ 2+\sqrt { 5 } } \times \frac { 2-\sqrt { 5 } }{ 2-\sqrt { 5 } } \)
\(=\frac { 2-\sqrt { 5 } }{ -1 } =-2+\sqrt { 5 } \)
\({ a }^{ 2 }={ \left( 2+\sqrt { 5 } \right) }^{ 2 }=9+4\sqrt { 5 } \)
\({ b }^{ 2 }={ \left( -2+\sqrt { 5 } \right) }^{ 2 }=9-4\sqrt { 5 } \)
\({ a }^{ 2 }+{ b }^{ 2 }=9+4\sqrt { 5 } +9-4\sqrt { 5 } \)
= 18

\((i)\frac { 21 }{ 40 }\) 
\((ii)\frac { 29 }{ 40 }\) 
\((iii)\frac { 29 }{ 40 } \)

Modified Continues Distribution

(ii) Frequency Polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the original interval 145-153 (or modified interval 144.5-153.5).

Let the radius of the base of the cylinder be r cm.
= 5 cm
Lateral surface =94.2 cm²
\(\Rightarrow 2\pi rh=94.2\)
\(\Rightarrow 2\times 3.14 \times r\times 5=94.2\)
\(\Rightarrow r=\frac{94.2}{2\times3.14\times5}\)
\(\Rightarrow r=\frac{94.2}{31.4}\)
\(\Rightarrow r=3cm\)
(ii) r = 3cm,  h = 5 cm
ஃ Volume of the cylinder =\(\pi r^2h\)
= 3.14 x (3)² x 5 =141.3 cm³

Area of paper of shade I \(2\times \left( \frac { 1 }{ 2 } \times 16\times 16 \right) =256\)cm²
Similarly, Area of paper of shade II = 256 cm²
For area of paper of shade III
a = 8 cm, b = 6 cm, c = 6cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 8+6+6 }{ 2 } =10\) cm
\(\therefore \) Area of paper of shade III = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-8)(10-6)(10-6) } \)
 \(=\sqrt { (10)(2)(4)(4) } =8\sqrt { 5 } \) 
= 17.89 cm²

Construction: Join OA and OB.
∵ OA = OB = AB              I Given
∴ \(\Delta AOB\) is equilateral.
∴ \(\angle AOB=60°\)

\(\angle ACB=\frac { 1 }{ 2 } \angle AOB\)
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
\(=\frac { 1 }{ 2 } \times 60°=30°\)
Now, ∵ ADBC is a cyclic quadrilateral.
∴ \(\angle ADB+\angle ACB=180°\)
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°
\(\Rightarrow \ \angle ADB+30°=180°\)
\(\\ \Rightarrow \angle ADB=180°-30°\)
\(\\ \Rightarrow angle ADB=150°.\) 

x=4y
\(\Rightarrow\ \ y={x\over 4}\)
Put x=0, we get \(y={0\over4}=0\)
Put x=4, we get \(y={4\over4}=1\)
Put x=-4, we get \(y={-4\over 4}=-1\)
Put x=2, we get \(y={2\over4}={1\over 2}\)
Four solutions are (0,0), (4, 1), (-4, -1) and \(\left(2,{1\over 2}\right)\)

(0,0); (-p,0); (5,0); (0,-q)

Term with the highest power of x = 5x³
Exponent of x in this term           = 3
Therefore Degree of this polynomial = 3

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA = 90°
⇒ ㄥLON = 90°
Similarly, ㄥOLM =ㄥLMN =ㄥMNO = 90°

ஃ A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is rectangle.

In \(\triangle\)AMB and \(\triangle\)PNQ,
AB = PQ   (Given)
AM = PN   (Given)
\(\angle\)1 = \(\angle\)2 = 90°
(AM\(\bot \)BC & PN \(\bot \)QR)
\(\Rightarrow\) \( \triangle AMP\cong \triangle PNQ\) (By RHS)
\(\Rightarrow\) \(\angle\)3 =\(\angle\)4 (By c.p.c.t.) 
Similarly, \(\triangle AMC\cong \triangle PNR\)
\(\Rightarrow\) \(\angle\)5 = \(\angle\)6
(In congruent triangles, corresponding angles are equal)
\(\therefore\) In \(\triangle\)ABC and \(\triangle\)PQR,
AB = PQ  (Given)
AC = PR (Given)
\(\angle\)A =\(\angle\)P
(\(\angle\)3 + \(\angle\)5 = \(\angle\)4 + \(\angle\)6) (proved)
\(\triangle ABC\cong \triangle PQR\)     (By SAS)

\(\frac { 1 }{ 3+\sqrt { 7 } } =\frac { 1 }{ 3+\sqrt { 7 } } \times \frac { 3-\sqrt { 7 } }{ 3-\sqrt { 7 } } =\frac { 3-\sqrt { 7 } }{ 9-7 } =\frac { 3-\sqrt { 7 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } =\frac { 1 }{ \sqrt { 7 } +\sqrt { 5 } } \times \frac { \sqrt { 7 } -\sqrt { 5 } }{ \sqrt { 7 } -\sqrt { 5 } } =\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { 1 }{ \sqrt { 5 } +\sqrt { 3 } } \times \frac { \sqrt { 5 } -\sqrt { 3 } }{ \sqrt { 5 } -\sqrt { 3 } } =\frac { \sqrt { 5 } -\sqrt { 3 } }{ 5-3 } \)
\(=\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } \)
\(\frac { 1 }{ \sqrt { 3 } +1 } =\frac { 1 }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } =\frac { \sqrt { 3 } -1 }{ 3-1 } =\frac { \sqrt { 3 } -1 }{ 2 } \)
\(LHS=\frac { 3-\sqrt { 7 } }{ 2 } +\frac { \sqrt { 7 } -\sqrt { 5 } }{ 2 } +\frac { \sqrt { 5 } -\sqrt { 3 } }{ 2 } +\frac { \sqrt { 3 } -1 }{ 2 } \)
\(=\frac { 3-1 }{ 2 } =\frac { 2 }{ 2 } =1=RHS\)

Factor theorem - statement
Let p(x)=x³-3x²-x+3
the factors of the constant term 3 are \(\pm \)1, \(\pm \)3
p(1)=1³-3(1)²-1+3=0
(x-1) is a factor
p(-1)=(-1)³-3(-1)²-(-1)+3=0
(x+1) is a factor
p(3) = 3³-3(3)²-3+3 = 0
(x-3) is a factor
(x-1)(x+1)(x-3) are the factors of p(x)

Total number of subjects = 5
(i) Number of subjects in which the student gets a first class = 4
Probability that the students gets a first class \(=\frac { 4 }{ 5 } \)
(ii) Number of subjects in which the student gets a distinction = 2
Probability that the student gets a distinction  \(=\frac { 2 }{ 5 } \)
(iii) Number of subjects in which the student gets marks between 70% and 80% = 1
Probability that the students gets marks between 70% and 80% \(=\frac { 1 }{ 5 } \) 

Area =  \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)
 \(\Rightarrow 60=\frac { 24 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ (24) }^{ 2 } } \)
\(\Rightarrow 10=\sqrt { 4{ b }^{ 2 }-576 } \)
\(\Rightarrow 100=4{ b }^{ 2 }-576\)      | Squaring
\(\Rightarrow 4{ b }^{ 2 }=676\)
\(\\ \Rightarrow { b }^{ 2 }=\frac { 676 }{ 4 } =169\)
\(\Rightarrow b=\sqrt { 169 } \) = 13 cm
\(\therefore\) Perimeter = a + b + b
= 24 + 13 + 13 = 50 cm

Given: Diameter AB and a chord AC have a common end point A. AB = 20 cm and AC = 12 cm.
To determine: OD
Determination: ∵ OD丄AC
∴ \(AD=DC=\frac { 1 }{ 2 } AC=\frac { 1 }{ 2 } \times 12=6\quad cm\)
| ∵ The perpendicular drawn from the centre of a circle to a chord bisects the chord.
\(OA=OB=\frac { 1 }{ 2 } AB=\frac { 1 }{ 2 } \times 20=10\quad cm\)
In right triangle ODA,
OA² = OD² + AD² I By Pythagoras Theorem
⇒ (10)²=OD²+(6)²
⇒ OD=8 cm
Hence, AC is 8 cm far from the centre of the circle.

Given l,m,n are three lines such that l || m and  n \(\bot \)m

\(\therefore \)l || m and n is a transversal
\(\therefore \) \(\angle \)1=\(\angle \)2 |Corresponding angles
But \(\angle \)1=\(90^{ 0 }\)
\(\therefore \) \(\angle \) 2= \(90^{ 0 }\)
\(\Rightarrow \) n \(\bot \) m

We have
AB = BC
\(\Rightarrow \) AB - BX = BC - BX
|If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
AB - BX = BC - BY  \(|\)\( \because\)  BX = BY
AB - BX coincides with AX;
BC - BY coincides with CY
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

Let us represent time (in hour) by x and distance (in km) by y. Then, we have y =80x.
Table of solution

We plot the points (1, 80), (2, 160), (3, 240) and (4, 320) on a graph paper and join these points by a ruler to get the line which is the graph of the equation y = 80x.

The figure formed is a square.
Its all the sides are of equal length.

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) 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

\(x=3-2\sqrt { 2 } \Rightarrow \frac { 1 }{ x } =3+2\sqrt { 2 } \)
\({ \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) }^{ 2 }=8\)
\(\Rightarrow \sqrt { x } +\frac { 1 }{ \sqrt { x } } =\pm 2\sqrt { 2 } \)