\(\angle ABC=\angle GBD=70^o\) (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

In \(\triangle ABC,\)
\(\angle A+\angle B+\angle C=180^o\) (Angle sum prop. of \(\triangle)\)
\(\angle A+\angle C=180^o-90^o=90^o\)
\(\frac{1}{2}(\angle A+\angle C)=45^o\)
In \(\triangle AOC,\)
\(\frac{1}{2}\angle A+\frac{1}{2}\angle C+\angle AOC=180^o\)
\(\angle AOC=180^o-45^o=135^o\)

\(\frac { 5+\sqrt { 3 } }{ 7-4\sqrt { 3 } } \times \frac { 7+4\sqrt { 3 } }{ 7+4\sqrt { 3 } } =\frac { 35+20\sqrt { 3 } +7\sqrt { 3 } +12 }{ 49-48 } \)
\(=\frac { 47+27\sqrt { 3 } }{ 1 } \)
and \(\frac { 5+\sqrt { 3 } }{ 4+4\sqrt { 3 } } =\frac { \left( 5+\sqrt { 3 } \right) \left( 7-4\sqrt { 3 } \right) }{ \left( 7+4\sqrt { 3 } \right) \left( 7-4\sqrt { 3 } \right) } \)
\(=\frac { 23-13\sqrt { 3 } }{ 1 } \)
ஃ \(\frac { 5+\sqrt { 3 } }{ 7-4\sqrt { 3 } } -\frac { 5+\sqrt { 3 } }{ 4+4\sqrt { 3 } } =(47+27\sqrt { 3 } )-(23-13\sqrt { 3 } )\)
= 24 + 40√3
= 8(3 + 5√3)

Let the given field be in the shape of a trapezium ABCD in which AB=25 m, CD= 10 m, BC = 13 m and AD = 14 m.
From D, draw DE 11 BC meeting AB at E. Also, draw DF \(\bot \) AB.
\(\therefore \) DE = BC =13 m
AE = AB - EB = AB - DC
= 25 - 10 = 15 m

For AED
a = 14 m, b = 13 m, c=  15 m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 14+13+15 }{ 2 } =\frac { 42 }{ 2 } =21\) m
\(\therefore \)    Area of the \(\Delta \)AED  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-14)(21-13)(21-15) }\)
\( \\ =\sqrt { 21(7)(8)(6) } =\sqrt { \left( 7\times 3 \right) (7)\left( 4\times 2 \right) \left( 2\times 3 \right) } \)
\(=7\times 3\times 2\times 2=84\) m²
\(\Rightarrow \frac { 1 }{ 2 } \times \)AE\(\times \)DE = 84
\(\Rightarrow \ \frac { 1 }{ 2 } \times \)15\(\times \)DF = 84
\(\Rightarrow \) DF = \(\frac { 84\times 2 }{ 15 } \)
\(\Rightarrow \) DF=  \(\frac { 56 }{ 5 } \) m = 11.2 m
\(\Rightarrow\) Height of the trapezium is 11.2 m.
\(\therefore \) Area of parallelogram EBCD = Base \(\times \) Height
= EB\(\times \) DF = 10 \(\times \)\(\frac { 56 }{ 5 } \) = 112 m²
\(\therefore \) Area of the field = Area of AED + Area of parallelogram EBCD = 84 m² + 112 m² = 196 m².                 

3=2x+y
⇒ y=3-2x
Table of solutions

We plot the points (0, 3) and (1, 1) on the graph paper and join the.same by a ruler to get the line which is the graph of the equation 3 = 2x + y.

(a) (0,5)
(b) (0,-3)
(c) (5,0)
(d) (-2,0)

(i) \(B\rightarrow (-5,2)\)
(ii) \(C\rightarrow \left( 5,-5 \right) \)
(iii) E
(iv) G
(v) 6
(vi) -3
(vii) \(L\rightarrow (0,5)\)
(viii) \(M\rightarrow (-3,0)\)

We know that 
\(x^3-y^3=(x+y)(x^2+xy+y^2)\)
Using Identity VII
\(​​\Rightarrow\ x^3-y^3=(x-y)^3+3xy(x-y)\)
\(\Rightarrow\ x^3-y^3=(x-y){(x-y^2+3xy)}\)
\(​​\Rightarrow\ x^3-y^3=(x-y)(x^2-2xy+y^2+3xy)\) Using Identity IV
\(​​\Rightarrow x^3-y^3=(x-y)(x-y)(x^2+xy+y^2)\) 

This expression is not a polynomial in one variable because in the expression, three variables (x, y and t) occur.

Yes! zero is a rational number.We can write zero in the form \(\frac { p }{ q } \),where p and q are integers and \(q\neq 0\)as follows:
  \(0=\frac { 0 }{ 1 } =\frac { 0 }{ 2 } =\frac { 0 }{ 3 } \)etc.

OA = OB (O is the mid-point of AB)
\(\angle\)AOC = \(\angle\)BOD (Vertically opposite angles)
OC = OD (O is the mid-point of CD)
\(\triangle AOC\cong \triangle BOD\)
\(\Rightarrow\) AC = BD. (By c.p.c.t) Proved.

On x-axis, y co-ordinates is zero,
So put y=0 in 3x-2y=6, we get
3x-0=6
\(\Rightarrow x=\frac{6}{3}=2\)
3x-2y=6 meets the x-axis at (2,0)

Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cm
So, 2s = 32, i.e., s = 16 cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11) cm = 5 cm,
s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16 \times 8 \times 5 \times 3} \mathrm{~cm}^{2}=8 \sqrt{30} \mathrm{~cm}^{2}\)

\(x=37^{ 0 },y=33^{ 0 }\)

\(z=29^{ 0 },\angle DNM=45^{ 0 },\angle CNM=135^{ 0 }\)

(4,0); (0,6)

(i)

(ii) Rectangle
(iii) 14 units

(x-1)(x-3)(2x-5)

Yes, No

2

-11/48, 7/96, 43/192

So, 2y-2 must be subtracted.

For \(\Delta \)ABD
a = 13 cm, b = 14 cm, c = 15 cm

\(\therefore s=\frac { a+b+c }{ 2 } s=\frac { 13+14+15 }{ 2 } \) = 21 cm
\(\therefore \)   Area = \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 21(21-13)(21-14)(21-15) } \)
\(\sqrt { 21\times 8\times 7\times 6 } =84\) cm²
For\(\Delta \) BCD
1 = 9 cm, b = 12 cm, c = 15 cm
\( \therefore s=\frac { a+b+c }{ 2 } s=\frac { a+12+15 }{ 2 } \)=18 cm
\(\therefore \)Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt {18(18-9)(18-12)(18-15) } \)
\(\sqrt { 18\times 9\times 6\times 3 } =54\) cm²
Now, area of quadrilateral ABCD = Area of \(\Delta \)ABD + Area of \(\Delta \)BCD
= 84 cm² + 54 cm² = 138 cm²

Given: In ΔABC, D is the mid-point of BC The perpendiculars from D to AB and AC are equal.
To Prove: ΔABC is isosceles.

Proof: In righ~ triangles DEC and DFB, Hyp. DC = Hyp. DB
| D is the mid-point of BC
Side DE = Side DF     IGiven
∴ ΔDEC ≌ ΔDFB I RHS congruence rule
∴ ㄥDCE = ㄥDBF
∴ ㄥBCA = ㄥCBA
∴ AB = AC
|Sides opposite to equal angles of a triangle are equal

x+2y=8
2y=8-x
\(y={8-x\over 2}\)
This expresses y in terms of x
This line will intersect x-axis at the point for while y=0.So, put y=0 in (1), we get
x+2(0)=8
x=8
Hence, line(1) intersects x-axis at the point (8, 0).
This line will cut y-axis at the point for which x=0.So, put x=0 in (1), we get
0+2y=8
2y=8
\(y={8\over2}=4\)
Hence, line(1) cuts y-axis at the point (0, 4).

(i) Coordinates of point P are (0,4)
(ii) Abscissae of point Q is 5.
(iii) The point identified by the coordinates (-4,4) is S.
(iv) The point identified by the coordinates (-3,-6) is R.

\(2(\sqrt { 10 } +\sqrt { 6 } -1)\)

Formula

ㄥB > ㄥC
∴ AC > AB

\(\angle BAC=20^{ 0 }\) 
\(\angle ABC+\angle BAC=100^{ 0 }\)

Required angle=\(180^{ 0 }\)-(\(90^{ 0 }\)+\(9^{ 0 }\)) = \(90^{ 0 }\)+\(9^{ 0 }\)

(1, 1) satisfies 3x=2y+1 and 2x=3y-1 both

x=2y+10 ⇒ x-2y-10=0

By convention

\((x^3+5)(4-x^5)\)\(=4x^3-x^8+20-5x^5\)

(i) AB and CD intersect atO0
\(\therefore\) \(\angle\)AOD = \(\angle\)BOC
(Vertically opp. angles) ...(i)
In \(\triangle\)AOD and \(\triangle\)BOC, we have 
\(\angle\)AOD = \(\angle\)BOC ...(ii)
\(\angle\)DAO = \(\angle\)CBO = 90° (Given)
and AD = BC (Given)
\(\triangle AOD\cong \triangle BOC\)
(By AAS congruence criterion)
\(\Rightarrow\) OA = OB (By c.p.c.t.)
i.e., O is the mid-point of AB
Hence, CD bisects AB.
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

(i) I and m are two parallel lines intersected byanother pair of parallel lines p and q.
AD II BC
and AB II CD.
\(\Rightarrow\) ABCD is a parallelogram.
i.e., AB = CD
and BC = AD
Now in \(\triangle\)ABC and \(\triangle\)CDA,we have
AB = CD (Prop. of II^gm)
BC =AD
and AC = AC (Common)
\(\therefore\) \(\triangle ABC\cong \triangle CDA\)
(By SSS criterion of congruence) 
(ii) Congruency of triangles.
(iii) Equality is the sign of democracy.

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.
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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°.

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)

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

In \(\triangle\)ABY and \(\triangle\)ACX,
AB = AC   (Given)
AY = AX     (Given)
\(\angle\)A =\(\angle\)A   (Common)
\(\therefore\) By SAS, \(\triangle ABY\cong \triangle ACX\)

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