\(\angle QPR=75^o\) (Vertically opposite angles)
Again, \(\angle PQR+\angle QPR=105^o\) (Exterior angle)
\(\Rightarrow \angle PQR+75^o=105^o\)
\(\Rightarrow PQR=30^o\)

In \(\triangle\)PQS and \(\triangle\)PRS,
PQ = PR  (Given)
PS = PS   (Common)
\(\angle\)PSQ = \(\angle\)PSR = 90°
(PS is altitude)
By R.H.S. rule,
\(\triangle PQS\cong \triangle PRS\)
\(\angle\)QPS = \(\angle\)RPS (By c.p.c.t.) 
Hence, PS bisects \(\angle\)P.

60.6

1728 m³

21 cm²

\(\angle\)AOD + \(\angle\)BOD = 180° (linear pair)
\(\angle\)BOD = 180°- \(\angle\)AOD
\(\angle\)BOD = 180°- 75° = 105°
\(\angle\)CED =90° (angle in semi-circle)
\(\angle\)CDE =90° - \(\angle\)OCE \(\Rightarrow\) 90° - 40° = 50°
\(\angle\)OBE = \(\angle\)OBD
\(\angle\)OBD = 180°- (105° + 50°)
(In \(\Delta\)DBO, Angle sum property of \(\Delta\))
\(\angle\)OBE = \(\angle\)OBD = 25°

x=1

(a)

(b) Rhombus, 16 sq.units

1

63³

Caring, gender equality, concern 8 triangles
Now \(\angle 2=90^o-36^o\)
=54^o
\(\angle 2=180^o-88^o\)
=92^o (Linear pair)
\(\angle AOD, \angle 1+\angle2+\angle3=180^o\)[Angle sum property of triangle]
\(\Rightarrow\angle1= \angle 180^o-54^o-92^o\)
\(\Rightarrow \angle 1=34^o\)

For roller
\(r=\frac { 1.5 }{ 2 } m=0.75m\)
\(\\ h=84cm=0.84m\)
\(\therefore \)  Curved surface area = \(2\pi rh\)  
\(=2\times \frac { 22 }{ 7 } \times 0.75\times 0.84\)
\(\\ =3.96{ m }^{ 2 }\)
\(\therefore \)  Area of the ground levelled in 1 revolution 
= 3.96 m²
\(\therefore \)  Area of the ground levelled in 100 revolutions 
= 3.96  100 m² = 396 m²
\(\therefore \)  Cost of levelling
= Rs \(396\times \frac { 50 }{ 100 } =\) Rs 198

Given: OD is perpendicular to chord AB of a circle where centre is O. BC is a diameter of the circle.
To Prove: CA=20D
Proof: OD丄AB
∴ D is the mid-point of AB
 |The perpendicular drawn from the centre of a circle to a chord bisects the chord.
In \(\Delta BAC\),
∵ \(OD\parallel AC\) I By mid-point theorem
and \(OD=\frac { 1 }{ 2 } AC\)
⇒ CA=2 OD

Let us recall what a rectangle is.
A rectangle is a parallelogram in which one angle is a right angle.
 
Let ABCD be a rectangle in which \(\angle\) A = 90°.
We have to show that \(\angle\) B = Ð C = \(\angle\) D = 90°
We have, AD || BC and AB is a transversal
(see Fig.).
So, \(\angle\) A + \(\angle\) B = 180° (Interior angles on the same
side of the transversal)
But, \(\angle\) A = 90°
So, \(\angle\) B = 180° – \(\angle\) A = 180° – 90° = 90°
Now, \(\angle\) C = Ð A and \(\angle\) D = \(\angle\) B
(Opposite angles of the parallellogram)
So, \(\angle\) C = 90° and \(\angle\) D = 90°.
Therefore, each of the angles of a rectangle is a right angle.

\((x^2-4x)(x^2-4x-1)-20\)
\(=y(y-1)-20\quad\quad y=x^2-4x\)
\(=y^2-y-20\)
\(=y^2-5y+4y-20\)
\(=y(y-5)+4(y-5)\)
\(=(y-5)(y+5)\)
\(=(x^2-5x+x-5)(x-2)^2\)
Using Identity II
\(=\left\{ x(x-5)+1(x-5) \right\} { (x-2) }^{ 2 }\)
\(=(x-5)(x+1){ (x-2) }^{ 2 }\) 

\(\frac { 1 }{ 3 } \)=0.3333...
\(\frac { 1 }{ 2 } \)=0.5
Hence two irrational numbers between \(\frac { 1 }{ 3 } \) and \(\frac { 1 }{ 2 } \) can be taken as 0.343443444... and 0.353553555...

In \(\triangle BDE,\)
\(\angle B+\angle D+\angle DEB=180^o\)( Angle sum property of a triangle)
\(\Rightarrow\) 40^o+x+90^o=180^o
\(\Rightarrow\)x=50^o
In \(\triangle DCF,\)
\(\angle D+\angle FCD=\angle AFD\) (Exterior angle is the sum of the two interior opposite angles)
\(\Rightarrow \)50^o+y=110^o
y=60^o

(i) Diameter = 14 cm
Radius (r) = \(\frac { 14 }{ 2 } cm=7cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 7 \right) }^{ 2 }=616{ cm }^{ 2 }.\)
(ii) Diameter = 21 cm
Radius (r)   = \(\frac { 21 }{ 2 } cm\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
 \(=4\times \frac { 22 }{ 7 } \times { \left( \frac { 21 }{ 2 } \right) }^{ 2 }=1386{ cm }^{ 2 }.\)
(iii) Diameter = 3.5 m
Radius (r)   = \(\frac { 3.5 }{ 2 } m=1.75m\)
\(\therefore\)  Surface area = \(4\pi { r }^{ 2 }\)
\(=4\times \frac { 22 }{ 7 } \times { \left( 1.75 \right) }^{ 2 }\)
\(\\ =38.5{ m }^{ 2 }.\)

ar(parallelogram ABCD) = AB x AE
= 16 x 8 cm²
= 128 cm²             ...(1)
ar(parallelogram ABCD) = AD x CF
= AD x 10 cm²   .......(2)
From (1) and (2), we get
AD x 10 = 128
AD = \(\frac { 128 }{ 10 } \)
AD = 12.8 cm.

Given: In \(\triangle ABC\) , AD is the perpendicular bisector of BC
To Prove:  \(\triangle ABC\) is isosceles triangle in which AB = AC.
Proof: In \(\angle ADB=\angle ADC\)
DB = DC | AD is the perpendicular bisector of BC
AD = AD
\(\triangle ADB\cong \triangle ADC\) |By SAS rule
AB = AC  | C.P.C.T
\(\triangle ABC\)  is an isosceles triangle in which
AB = AC 

(a)x+y=50

0

In \(\triangle\)ABC, as AB is the greatest side
\(\Rightarrow\) AB > BC \(\Rightarrow\) \(\angle\)C > \(\angle\)A
AB > AC \(\Rightarrow\) \(\angle\)C > \(\angle\)B
On adding (1) and (2), we get
2\(\angle\)C > \(\angle\)A + \(\angle\)B
\(\Rightarrow\) 2\(\angle\)C + \(\angle\)C > \(\angle\)A + \(\angle\)B + \(\angle\)C
\(\Rightarrow\) 3\(\angle\)C > 180°
\(\therefore\) \(\angle\)C > 60°.

\(={ \left( { x }^{ { a }^{ 2 }-{ b }^{ 2 } } \right) }^{ \frac { 1 }{ a+b } }.{ \left( { x }^{ { b }^{ 2 }-{ c }^{ 2 } } \right) }^{ \frac { 1 }{ b+c } }.{ \left( { x }^{ { c }^{ 2 }-{ a }^{ 2 } } \right) }^{ \frac { 1 }{ c+a } }\)
\(={ x }^{ \frac { { a }^{ 2 }-{ b }^{ 2 } }{ a+b } }.{ x }^{ \frac { { b }^{ 2 }-{ c }^{ 2 } }{ a+b } }.{ x }^{ \frac { { c }^{ 2 }-{ a }^{ 2 } }{ c+a } }\)
\(={ x }^{ a-b }.{ x }^{ b-c }.{ x }^{ c-a }\)
\(={ x }^{ 0 }\)