The coordinates of A are (-4, 3).
The coordinates of B are (4, 2).
The coordinates of C are (-3, -2).
The coordinates of D are (4, -2).

\(\angle 1=\angle 2=x, \angle B=45^o\)
\(\angle A+\angle B+\angle, C=180^o \angle C=55^o\) (Angle sum prop of \(\triangle)\)
\(\Rightarrow\) 2x+45^o+55^o=180^o
2x=80^o
x=40^o
\(\angle ADB=\angle 2+\angle C\) (Exterior angle is the sum of the two interior opposite angles)
=40^o+55^o
=95^o
Similarly, \(\angle ADC=\angle 1+\angle B\)
=45^o+40^o
=85^o

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

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

(i) IV   (ii) II   (iii) III (iv) I

\(55^{ 0 }\)

\((6x+\frac{1}{5})(36x^2-\frac{6x}{5}+\frac{1}{25})\)

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

Let \(l\bot m\)
Then \(\angle 1=90^{ 0 }\)
\(P\bot 1\)
Then \(\angle 2=90^{ 0 }\)

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

\(x^2+2 x+3=0\)
\(\Rightarrow\quad x=\frac{-2\pm\sqrt{4-12}}{2}=\frac{-2\pm2\sqrt{2}i}{2}\)
\(=-1\pm\sqrt{2}i\)

Convention

\(\angle CBD=34^o+30^o\)
=64^o( Exterior \(\angle\)of \(\triangle ABC\))
\(x^o=\angle EBD+\angle EDB\)
=64^o+65^o
x^o=109^o

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

For Point P
(i) Abscissa of the point = 2
(ii) Ordinate of the point = 3
(iii) Coordinates of the point = (2, 3)
(iv) The point (2, 3) lies in the I quadrant.
For Point Q
(i) Abssissa of the point = - 2
(ii) Ordinate of the point = 4
(iii) Coordinates of the point = (-2, 4)
(iv) The point (-2, 4) lies in the II quadrant.
For Point R
(i) Abscissa of the point = - 5
(ii) Ordinate of the point = - 3
(iii) Coordinates of the point = (-5, - 3)
(iv) The point (-5, -3) lies in the III quadrant.
For Point S
(i) Abscissa of the point = 5
(il; Ordinate of the point = -1
(iii) Coordinates of the point = (5, -1)
(iv) The point (5, -1) lies in the IV quadrant.

(i) False. This can be seen visually.
(ii) False. This contradicts the Axiom.
[Given two distinct points, there is a unique line that passes through them.]
(iii) True by Euclid's Postulate 
[A terminated line can be produced indefinitely.]
(iv) True. If we superimpose the region bounded by one circle on the other, then they coincide. So, their centres and boundaries coincide, therefore, their radii will coincide.
(v) True by the first Axiom of Euclid.
[Things which are equal to the same thing are equal to one another.]

This expression is not a polynomial because in the term \(\frac { 2 }{ y } \) , the exponent of y is (- 1) which is not a whole number.

We have AB ॥ CD [Given] and PQ is a transversal.
∴ Interior alternate angles are equal.
∴ ∠APQ = ∠PQR
or 50° = x [∵ APQ = 50° (Given)] ...(1)
Again, AB ॥ CD and PR is a transversal.
∴ ∠APR = ∠PRD [Interior alternate angles]
⇒ ∠APR = 127° [∵ It is given that ∠PRD = 127^o]
But ∠APR = ∠APQ + ∠QPR
∴ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [∵ It is given that ∠APQ = 50^о]
⇒ y = 127° - 50° = 77°
Thus, x = 50° and y = 77°

According to axiom 5, we have the whole is greater than the part, which is a universal truth. Let a line segment PQ = 8 cm. Consider a point R in its interior, such that PR = 5cm.
Clearly, PR is a part of the line segment PQ and R lies in its interior. So, PR is smaller than PQ. Hence, the whole is greater than its part and this is true for anything in any part of the world.

Let \(\angle DAF=\angle 1\)
\(\angle CFA=\angle 2,\)
\(\angle BFE=\angle 3\)
\(\angle BEF=\angle 4\)
Since, AE||DC
\(\angle D+\angle 1=180^o\)
(Angles on the same side of transversal)
\(\angle 1=180^o-110^o=70^o\)
\(\angle 4=\angle 1=70^o\) (Alternate angle)
Again, \(97^o+\angle2=180^o\) (Angle on the same side of transversal)
\(\angle 2=180^o-97^o=83^o\)
\(\angle 3=\angle2=83^o\)
(Vertically opp.angles)

In \(\triangle BEF,\)
\(\angle 3+\angle 4+\angle EBF=180^o\)
(Angle sum property)
\(\Rightarrow 83^o+70^0+\angle EBF=180^o\)
\(\Rightarrow \angle EBF=180^o-153^o\)
\(\Rightarrow \angle EBF=27^o\)

p(x)=x⁴-2x³+3x²-px+3p-7
Put, x+1=0 or x=-1 in p(x), we get
p(-1)=(-1)⁴-2(-1)³+3(-1)²-p(-1)+3p-7=19
\(\Rightarrow\) 1+2+3+p+3p-7=19
\(\Rightarrow\) 4p-1=19
\(\Rightarrow\) 4p = 20
\(\therefore\) p = 5
\(\therefore\) The polynomial p(x)=x⁴-2x³+3x²-5x+15-7
=x⁴-2x³+3x²-5x+8
Put, x+2 = 0 or x = -2 in p(x)
p(-2)=(-2)⁴-2(-2)³+3(-2)²-5(-2)+8
= 16+ 16 + 12 + 10 + 8
= 62.

Plot the points (4,- 5), (5.5,- 3), (-2, 5), (-1, - 6), (0, 5) and (2.5, 0) on the graph paper.

Given, ㄥABC = ㄥACB   ......(i)
and     ㄥ4 = ㄥ3       ............ (ii)
According to Euclid's axiom 3, if equals are subtracted from equals, then reminders are also equal.
On subtracting Eq. (ii) from Eq. (i), we get
ㄥABC-ㄥ4 =ㄥACB-ㄥ3     ⇒  ㄥ1 =ㄥ2
Now, in ΔBDC, ㄥ1 = ㄥ2    ⇒ DC = BD
[since sides opposite to equal angles are equal]
ஃ BD = DC 
Hence proved.

We know that the sum of angles that are formed on a straight line is equal to 180°.
So, Reason is correct
We have : ∠AOC+∠BOC=180°   [Since AOB is a straight line ]
⇒3x + 10 + 4x − 26 = 180°
⇒7x = 196°
⇒x = 28°
∴∠BOC = [4 × 28 − 26]°
⇒∠BOC=86°.
So, Assertion (A) is also true.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that If two parallel lines are intersected by a transversal, then each pair of corresponding angles so formed is equal.
So, Reason is correct.
Also, we know that If a transversal intersects two parallel lines, then the sum of the interior angles on the same side of the transversal is 180^о.
From figure we have, 120^о – x + 5x = 180^о
⇒ 4x = 180^о – 120^о
⇒ 4x = 60^о
⇒ x = 15^о
So, Assertion is also correct.
But reason (R) is not the correct explanation of assertion (A). Correct option is (b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).

We know that the if the point lies on y-axis, its x-coordinate is 0.
So, Reason is correct.
The x co-ordinate of the point (0, 4) is zero.
So, Point (0, 4) lies on y -axis.
So, Assertion is also correct
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that the perpendicular distance of a point from y-axis is called its x-coordinate or abscissa.
So, Reason is correct.
The x co-ordinate of the point (5, 2) is 5.
So, Assertion is also correct
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

We know that a decimal in which a digit or a set of digits is repeated periodically, is called a repeating, or a recurring, decimal.
So, Reason is correct.
Also, we know that a decimal that ends after a finite number of digits is called a terminating decimal.
Hence Assertion is correct but reason is not the correct explanation of Assertion Correct option is (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

(a) (iv) C(10, 6)
(b) (iii) R(5, 6)
(c) (ii) \(\sqrt{34}\) units
PS² = AS² + AP² = 5² + 3²
= 25 + 9 = 34
⇒ PS = \(\sqrt{34}\)
(d) (iv) Rhombus
(e) (i) 30 sq. units
Area of rhombus = \(1 / 2\) x product of diagonals
= \(1 / 2\) x 6 x 10 
= 30 sq. units