l = 6 m, b = 4 m, h = 3 m.
Area of 4 walls = 2(l + b)h
= 2(6 + 4)3
= 60 m²
Area of door = 2.5\(\times\)1
= 2.5 m²
Net area to be white washed=Area of four walls-Area of door.
= 60-2.5
= 57.5 m²
Cost of white washing = 57.5 \(\times\)12
= Rs 690

In \(\Delta\)GHK; F and E are the mid-points of HG and GK respectively.
\(\therefore\) By mid-point theorem,
\(FE=\frac { 1 }{ 2 } KH\quad and\quad FE||KH\)  ...........(i)
In quadrilateral EFHK,
EF || HK        (by (i))
\(\therefore\) EFHK is a trapezium.
Also,
ar(EFHK) = ar(FHD) + ar(DEF) + ar(DEK)   .............(ii)
We have,    FE || HD and FE = HD
\(\therefore\) FEDH is a ||^gm,
So, ar(FHD) = ar(DEF)   ..........(iii)
Similarly, DFGE is a ||^gm, 
\(\therefore\) ar(DEF) = ar(GEF)   ...........(iv)
Also, DFEK is a ||^gm,
\(\therefore\) ar(DEF) = ar(DEK)    ...........(v)
Using (iii), (iv) & (v) we get,
ar(GEF) = ar(FHD)
= ar(DEK) = ar(DEF)
\(=\frac { 1 }{ 4 } \) ar(GHK)                ...........(vi)
Using (vi) and (ii), we get
ar(EFHK) = \(\frac { 3 }{ 4 } ar(GHK)\)
Hence proved

\(\angle ABC=\angle GBD=70^o\) (Vertically opp. angle)
\(x=\angle ABC+\angle CAB\) (Exterior angle)
=50^o+70^o=120^o
\(y=\angle GBD=70^o\) (Alternate angles)
\(z=180^o-\angle EAD-y\) (Angle sum property)
=180^o-50^o-70^o
=60^o

\({ x }^{ 2 }-{ y }^{ 2 }={ \left( \frac { \sqrt { 3 } +\sqrt { 2 } }{ \sqrt { 3 } -\sqrt { 2 } } \right) }^{ 2 }-{ \left( \frac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } \right) }^{ 2 }+\left( \frac { \sqrt { 3 } +\sqrt { 2 } }{ \sqrt { 3 } -\sqrt { 2 } } \right) \times \left( \frac { \sqrt { 3 } -\sqrt { 2 } }{ \sqrt { 3 } +\sqrt { 2 } } \right) \)
\(=\frac { 5+2\sqrt { 6 } }{ 5-2\sqrt { 6 } } -\frac { 5-2\sqrt { 6 } }{ 5+2\sqrt { 6 } } +1\)
\(=\frac { { (5+2\sqrt { 6 } ) }^{ 2 }-({ 5-2\sqrt { 6 } ) }^{ 2 } }{ (5-2\sqrt { 6 } )(5+2\sqrt { 6 } ) } +1\)
\(=\frac { 25+24+20\sqrt { 6 } -25-24+20\sqrt { 6 } }{ 25-24 } +1\)
\(=40\sqrt { 6 } +1\)
\(=40\times 2.4+1=96+1=97\)

(-1,-5) lies on the graph of
3x=ay+7
3(-1)=aX(-5)+7
\(\Rightarrow\)-3=-5a+7\(\Rightarrow\) 5a=10
\(\Rightarrow\)a=2

(ii) The two conclusions we can arrive at from the graph are as follows:
(a) The numbers of girls to the nearest ten per thousand boys is maximum in Scheduled Tribe section of the society and minimum in Urban section of the society.
(b) The number of girls to the nearest ten per thousand boys is the same for 'Non Sc/ST' and 'Non-backward Districts' sections of the society.

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.
∵ If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

The decimal expansion is terminating.
0.3796 is a rational number.

(i) 4 students were born in the month of November.
(ii) Maximum number of students were born in the month of August.

192.5 cm²

\(25\sqrt { 3 } \) cm²

\(140^{ 0 },140^{ 0 },140^{ 0 }\)

(i)(-5,-4)
(ii)(3,0)

IV

\(2(2x-y)(4x^2+2xy+y^2)\)

15

\(\angle ACD=\angle ABC=x\)
\(\angle BCP=\angle DCP=y\)
Ext \(\angle APC=x+y\)
\(\angle ACP=x+y\)
\(\angle ACP=\angle APC\)

For room
l = 12 m,
b = 9 m,
h = 8 m
\(\therefore\)   Length of the longest rod that can be placed in the room
= Length of the diagonal 
\(=\sqrt { { l }^{ 2 }+{ b }^{ 2 }+{ h }^{ 2 } } \) 
\(\\ =\sqrt { { \left( 12 \right) }^{ 2 }+{ \left( 9 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2 } } \)
\(\\ =\sqrt { 144+81+64 } \)
\(\\ =\sqrt { 289 } \)
\(\\ =17m.\)

(i) x = 2 x 35° = 70°
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(ii) \(x=\frac { 1 }{ 2 } \times 110°=55°\)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iii) \(x=\frac { 1 }{ 2 } \times 70°=35°\)
I ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle
(iv) x = 180° - (90° + 55°) I ∵ Angle in a semi-circle is 90°
= 180° - 145° = 35°
(v) \(x=\frac { 1 }{ 2 } \times \left( 180°-120° \right) \)
| ∵ Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point of the remaining part of the circle \(=\frac { 1 }{ 2 } \left( 60° \right) =30°\). 

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.

\(p(z)={ 3z }^{ 2 }=4z+\sqrt { 17 } \)
\(\therefore p(3)=3{ (3) }^{ 2 }-4(3)+\sqrt { 17 }\) 
\(=15+\sqrt { 17 }\)

\(=\frac { 40 }{ 2\sqrt { 10 } +\sqrt { 2\times 2\times 5 } +\sqrt { 2\times 2\times 10 } -2\sqrt { 5 } } \)
\(\\ =\frac { 40 }{ 2\sqrt { 10 } +\sqrt { 20 } +\sqrt { 40 } -2\sqrt { 5 } } \)
\(\\ =\frac { 40 }{ 4\sqrt { 10 } } =\frac { 10 }{ \sqrt { 10 } } \)
\(\\ \sqrt { 10 } =3.162\)

Proof: In \(\triangle\) PQR,
PQ =PR
\(\angle\) PQR = \(\angle\)PRQ
(Angles opp. to equal sides are equal) ...(i) 
In \(\triangle\)PQS, \(\angle\)PQR > \(\angle\)PSQ 
(Ext. angle of a 6 is greater than each of interior opp. angle)
\(\angle\)PRQ > \(\angle\)PSQ, using (i)
\(\Rightarrow\) \(\angle\)PRS >\(\angle\)PSR \(\Rightarrow\) PS > PR 
PS>PQ (\(\because\) PR = PQ)
(Side opp. to greater angle is larger)

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)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.