Given PQ||RS
Given \(\angle 1=\angle 2\ and \angle 3 =\angle 4\)
But, \(\angle 1=\angle 3 \) (Corres angles)
\(2\angle 1=2\angle3\)
\(\angle MPB=\angle PRD\)
Hence, AB||CD.

y=180^o-(30^o+20^o)=130^o
l||m \(\Rightarrow\)x+100^o=180^o\(\Rightarrow\)x=80^o
x+y=210^o,y-x=50^o
(x+y):(y-x)=21:5

PQRS is a square. (given)

(i) SRT is an equilateral triangle. (given)
\(\therefore\) \(\angle\)PSR = 90°, \(\angle\)TSR = 60° 
\(\Rightarrow\) \(\angle\)PSR + \(\angle\)TSR = 150°.
Similarly, \(\angle\)QRT = 150°
In \(\triangle\)PST and \(\triangle\)QRT, we have PS = QR (S)
\(\angle\)PST =\(\angle\)QRT = 150° (A)
and ST = RT (5) Y,
By SAS, \(\triangle PST\cong \triangle QRT\)
\(\Rightarrow\) PT = QT (By c.p.c.t.) Proved.
(ii) In \(\triangle\)TQR, QR = RT
(Square and equilateral 6 on same base) Yz
\(\Rightarrow\) \(\angle\)TQR = \(\angle\)QTR = x 
\(\therefore\) x + x + \(\angle\)QRT = 180°
\(\Rightarrow\) 2x + 150° = 180° ~ 2x = 30°
\(\therefore\) x = 15° Proved.

Given: Circles are drawn with sides AB and AC of a triangle ABC as diameters. They intersect at a point D.
To Prove: D lies on the third side BC of \(\Delta \)ABC
Construction: Join AD

Proof: ∵ Circle drawn on AB as diameter intersects BC in D.
∴  \(\angle ADB=90°\)
| Angle in a semi-circle
But \(\angle ADB+\angle ADC=180°\)
Linear Pair Axiom
∴ \(\angle ADC=90°\)
Hence, the circle described on AC as diameter must pass through D.
Thus, the two circles intersect in D.
Now, \(\angle ADB+\angle ADC=180°\).
∴ Points B, D, C are collinear.
∴ D lies on BC.

Given: In parallelogram ABCD, AC = BD   To Prove: ||gm ABCD is a rectangle.

Proof: In \(\Delta\) ACB and \(\Delta\)BDA,
AC = BD                                 I Given
AB = BA                                 I Common
BC = AD                                 I Opposite sides of II gm ABCD
\(\therefore\) \(\Delta\)ACB =\(\Delta\)BDA               | SSS Congruence Rule
\(\therefore\)   \(\angle ABC=\angle BAD\quad \quad \quad ......(1)\quad C.P.C.T\) 
AD || BC                    | Opp. sides of II gm ABCD
and transversal AB intersects them.
 \(\angle BAD+\angle ABC=180°\quad \quad \quad ....(2)\)
|Sum of consecutive interior angles on the same side of a transversal is 180
From (1) and (2),
 \(\angle BAD=\angle ABC=90°\)
\(\therefore\) ​​​​​|| gm ABCD is a rectangle.

There can be infinitely many rational numbers between 3 and 4.
\(\frac { 3+4 }{ 2 } =\frac { 7 }{ 2 } \)
\(\\ \frac { 3+\frac { 7 }{ 2 } }{ 2 } =\frac { 13 }{ 4 } \)
\(\\ \frac { 3+\frac { 13 }{ 4 } }{ 2 } =\frac { 25 }{ 8 }\)
\( \\ \frac { 3+\frac { 25 }{ 8 } }{ 2 } =\frac { 49 }{ 16 } =\frac { 3+\frac { 49 }{ 16 } }{ 2 } =\frac { 97 }{ 32 } =\frac { 3+\frac { 97 }{ 32 } }{ 2 } =\frac { 193 }{ 64 } \)
Thus, six rational numbers between 3 and 4
\(\frac { 193 }{ 64 } ,\frac { 97 }{ 32 } ,\frac { 49 }{ 16 } ,\frac { 25 }{ 8 } ,\frac { 13 }{ 4 } \)and \(\frac { 7 }{ 2 } \)
Aliter
\(3=\frac { 3 }{ 1 } =\frac { 3\times 7 }{ 1\times 7 } =\frac { 21 }{ 7 } \)
\(\\ 4=\frac { 4 }{ 1 } =\frac { 4\times 7 }{ 1\times 7 } =\frac { 28 }{ 7 } \)
6 + 1 = 7
the six rational numbers between 3 and 4 can be taken as 
\(\frac { 22 }{ 7 } ,\frac { 23 }{ 7 } ,\frac { 24 }{ 7 } ,\frac { 25 }{ 7 } ,\frac { 26 }{ 7 } \) and \(\frac { 27 }{ 7 } \)

Given, r = 3.5 m, h = 12 m
Capacity V=\(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(V=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 3.5\times 3.5\times 12\)
= 154 m³   [1 m³ = 1000 litre]
= 154000 litre.

\(\frac{1}{\sqrt{2}+1}\)=\(\frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } \) = √2-1
= 1.414-1
= 0.414

\(\frac { (3-\sqrt { 5 } )107 }{ 120 } \)

3

1

28

\(93-24\sqrt { 15 } \)

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.

Proof: In \(\triangle\)ADE, we have
AD = AE
\(\angle\)ADE = \(\angle\)AED
180°- \(\angle\)ADE = 180°- \(\angle\)AED
\(\Rightarrow\) \(\angle\)ADB = \(\angle\)AEC
Consider \(\triangle\)ABD and \(\triangle\)ACE
AD =AE
\(\angle\)ADB = \(\angle\)AEC
BD = EC
By SAS congruence,
\(\triangle ADB\cong \triangle AEC\)
By c.p.c.t., AB = AC
\(\therefore\) \(\triangle\)ABC is an isosceles triangle.

\(\angle\)1 + \(\angle\)5 = 180⁰ = \(\angle\)2 + \(\angle\)6
(linear pair)
\(\Rightarrow\) \(\angle\)1 =\(\angle\)2 (\(\because\) \(\angle\)5 = \(\angle\)6)
In \(\triangle\)CAP and \(\triangle\)BAP,
\(\angle\)1 = \(\angle\)2 (proved)
\(\angle\)3 = \(\angle\)4
(AD is the bisector of \(\angle\)BAC)
AP =AP
\(\triangle CAP\cong \triangle BAP\) (By SAS) 
\(\Rightarrow\) CP = BP (By c.p.c.t.) Proved

\(\frac { 1 }{ 1+{ x }^{ a-b } } +\frac { 1 }{ 1+{ x }^{ b-a } } =\frac { { x }^{ b } }{ { x }^{ b }+{ x }^{ a } } +\frac { { x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)
\(=\frac { { x }^{ b }+{ x }^{ a } }{ { x }^{ a }+{ x }^{ b } } \)