\(\angle PSR=\angle SPQ=43^{ 0 }\)
|Alternate Interior angles
\(27^{ 0 }+43^{ 0 }+m=180^{ 0 }\)
|Angle sum property

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

\(x=(180^{ 0 }-120^{ 0 })+(180^{ 0 }-110^{ 0 })\)

Remainder \(=f(-1)=(-1)^2+2(-1)+1=0\)
Aliter: \(x^2+2x+1=(x+1)^2\)
ஃ Remainder =0

Remainder theorem

Value=\(6(1)^2+7(1)-3\)
\(​​​​​​​=6+7-3=10\)

A polynomial of degree n has maximum number of terms as (n+1)

\(\therefore \) CD || EF
and a transversal DE intersect them
\(\therefore y+40^{ 0 }\)=\(180^{ 0 }\)
Sum of the consecutive interior on the same side of a traversal is \(180^{ 0 }\) 
\(\Rightarrow Y=180^{ 0 }-40^{ 0 }=140^{ 0 }\)
\(\therefore \) AB||CD and a traversal BD intersects them
\(\therefore \) c=y | corresponding angles
\(\Rightarrow x=140^{ 0 }\)
\(\therefore EA\quad \bot \quad AB\quad and\quad AB||EF\)
\(\therefore EA\quad \bot \quad EF\quad \)
If a line is perpendicular to a line then it is perpendicular to the parallel line also
\(\Rightarrow \angle AEF=90^{ 0 }\) 
\(\Rightarrow Z+40^{ 0 }=90^{ 0 }\) 
\(\Rightarrow Z+50^{ 0 }\)

Let \(f(x)=​​{ p }x^{ 2 }+5x+r\)
 If (x-2) is a factor of f(x), then by factor theorem
\(f(2)=0\ |x-2=0\) 
\(\Rightarrow \ x=2\) 
\(\Rightarrow { p }(2)^{ 2 }+5(2)+r=0\)
\(\Rightarrow \ 4p+r+10=0\ ......(1)\)
If \(\left( x-\frac { 1 }{ 2 } \right) \)is a factor of f(x), then by factor theorem,
\(f\left( \frac { 1 }{ 2 } \right) =0\)
\(|x-\frac { 1 }{ 2 } =0\ \Rightarrow x=\frac { 1 }{ 2 } \)
\(\Rightarrow \ p{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+5\left( \frac { 1 }{ 2 } \right) +r=0\)
\(\Rightarrow \ \frac { p }{ 4 } +\frac { 5 }{ 2 } +r=0\)
\(\Rightarrow p+4r+10=0\ \ ..........(2)\)
Subtracting (2) from (1), we get
\(3p-3r=0\)
\(\Rightarrow\)\(p=r\) 

\(\frac { 2 }{ \sqrt { 3 } +\sqrt { 5 } } +\frac { 5 }{ \sqrt { 3 } -\sqrt { 5 } } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ =\frac { 2\left( \sqrt { 3 } -\sqrt { 3 } \right) }{ \left( \sqrt { 3 } -\sqrt { 5 } \right) } +\frac { 5 }{ \sqrt { 3 } -\sqrt { 5 } } \frac { \sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +\sqrt { 5 } } \)
\(\\ =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ \frac { 2\left( \sqrt { 3 } -\sqrt { 5 } \right) }{ 3-5 } +\frac { 5\left( \sqrt { 3 } +\sqrt { 5 } \right) }{ 3-5 } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ -\left( \sqrt { 3 } -\sqrt { 5 } \right) -\frac { 5 }{ 2 } \left( \sqrt { 3 } +\sqrt { 5 } \right) =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ -\frac { 7 }{ 2 } \sqrt { 3 } -\frac { 3 }{ 2 } \sqrt { 5 } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ a=-\frac { 7 }{ 2 } ,\ b=-\frac { 3 }{ 2 } \)

Since, AB = BC
AX + BX = BY + CY
Since, BX = BY
AX + BX - BX = BY + CY - BY
AX = CY
Axiom : If equals are subtracted from the equals, the remainders are equal.

1000x = 237.\(\overline{237}\)
x = \(\frac{237}{999}\)

x⁴-625
=(x²)²-(25)²=(x²-25)(x²+25)  \(\left[ \because { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)
=[(x)²-(5)²](x²+25)
=(x+5)(x-5)(x²+25)   \(\left[ \because { a }^{ 2 }-{ b }^{ 2 }=(a+b)(a-b) \right] \)

30°, 60°, 90°.

Consider line RS. Then, 4b + 75° + b = 180°
[sum of all angles on a straight line is 180°]
b = 21°
Consider line PQ. Then, 75° + b + a = 180° \(\Rightarrow\)a = 84°
[sum of all angles on a straight line is 180°]
Again, consider line RS, 2c+ a = 180° [linear pair]
c = 48°

(i)  \((7a+5b)^2\)
(ii) \((\frac{5}{2}x+\frac{y}{3})(\frac{5}{2}x-\frac{y}{3})\)

\(\sqrt[3]{44 x-7}\) = 5 ⇒ (44x - 7)^1/3 = 5
= [(44x - 7)^1/3]³ = [5]³
= 44x - 7 = 125 ⇒ x = \(\frac{132}{44}\) = 3

Let x = \(0 . \overline{38}\) = 0.3838...
∴ 100x = 100 x (0.3838...)
or 100x = 38.3838
Subtracting (1) from (2), we have
100x - x = (38.3838 ... ) - (0.3838 ... )
or 99x = 38
or \(x=\frac{38}{99}\)
Thus, \(0 . \overline{38}=\frac{38}{99}\).

y=88^o (Corresponding angles)
a=180^o-88^o=92^o (linear pair)
b=180^o-110^o=70^o (linear pair)
x=180^o-(92+70)^o (Angle sum property)
=180^o-162^o
=18^o

EQ||FP and transversal cut them
\(\therefore \angle AQD=\angle APF\) (Conresponding angles)
\(\angle AQD=50^o\)
\(\therefore \angle DQB=180^o-50^o=130^o\)
AB||CD and transversal EQ cuts them
\(\therefore \angle EDG=\angle DQB=130^o\)
\(\therefore \angle EDG=130^o-40^o=90^o\)
FP||EQ and transversal PG cut them
\(\therefore \angle FPD=\angle EDG=90^o\)

Euclid's axiom 2

\((4a-3b)^3\)

a=5, b=4

(i) For rectangle ABCD:
AB = 8 - 2 = 6 units ← Length
AD = 12 - 8 = 4 units ← Breadth
∴ Area of rectangle ABCD = AB x AD
= 6 units x 4 units
= 24 sq. units.
For triangle PQR:
QR = 12 - 4 = 8 units ← Base
SP = 6 - 0 = 6 units ← Height
∴ Area of ΔPQR = \(\frac{1}{2}\) base x height
= \(\frac{1}{2}\) x 8 x 6 sq. units
= 24 sq. units.
⇒ \(\left[\begin{array}{l} \text { area of } \\ \text { rect. } \mathrm{ABCD} \end{array}\right]=\left[\begin{array}{l} \text { area of } \\ \Delta \mathrm{PQR} \end{array}\right]\)
(ii) Co-ordinate Geometry.
(iii) Community or social upliftment.

From the figure, we have
(i) The co-ordinates of B are (-5, 2).
(ii) The co-ordinates of C are (5, -5).
(iii) The point E is identified by the co-ordinates (-3, -5)
(iv) The point G is identified by the co-ordinates (2, -4).
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point H is -3.
(vii) The co-ordinates of the point L are (0, 5).
(viii) The co-ordinates of the point M are (-3, 0).
Plotting a point in the plane when coordinates of the point are given
To plot a point in the coordinate plane we draw the coordinate axes and choose our units such that we can mark equal distances on the x-axis or on the y-axis or on both the axes. We take origin as zero for x-axis as well as on y-axis. The distances marked along OX and OY are taken as positive and those along OX and OY are taken as negative.

\(\frac { { 3 }^{ 30 }+{ 3 }^{ 29 }+{ 3 }^{ 28 } }{ { 3 }^{ 31 }+{ 3 }^{ 30 }-{ 3 }^{ 29 } } =\frac { { 3 }^{ 38 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }+1 \right) }{ { 3 }^{ 29 }\left( { 3 }^{ 2 }+{ 3 }^{ 1 }-1 \right) } \)
\(=\frac { \left( 9+3+1 \right) }{ 3\left( 9+3-1 \right) } \)
\(=\frac { 13 }{ 3\times 11 } =\frac { 13 }{ 33 } \)

322

(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

(a) (iii) A(3, 5); B(7, 9)
(b) (i) C(11, 5); D(7, 1)
(c) (iii) 8 units
(d) (iii) 8 units
(e) (i) (7, 5)

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 a point both of whose coordinates are negative lies in III quadrant
So, Reason is correct.
Hence, Point A(-2, -4) lies on III quadrant
So, Assertion is also correct and Reason explains Assertion
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).