AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3

Question 1.
Find the remainder when
x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each
Solution:
Let f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is f (- 1)
f (- 1) = (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

ii) \(x-\frac{1}{2}\)
f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is \(\mathrm{f}\left(\frac{1}{2}\right)\)
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 1

iii) x
Solution:
f(x) = x3 + 3x2 + 3x + 1
The remainder is f(0)
∴ f(0) = 03 + 3(0)2 + 3(0) + 1 = 1

iv) x + π
Solution:
f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is f(- π)
f(- π) = (- π)3 + 3(-π)2 + 3 (-π) + 1 .
= – π3 + 3π2 – 3π + 1

v) 5 + 2x
f(x) = x3 + 3x2 + 3x + 1
The remainder is \(\mathrm{f}\left(\frac{-5}{2}\right)\)
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 2

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

Question 2.
Find the remainder when x3 – px2 + 6x – p is divided by x – p.
Solution:
Let f(x) = x3 – px2 + 6x – p
(x – a) = x – p)
By Remainder theorem, the remainder is f(p)
∴ f(P) = P3 – P(P)2 + 6p – p
= p3 – p3 + 5p = 5p

Question 3.
Find the remainder when 2x2 – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.
Solution:
Let f(x) = 2x2 – 3x + 5 and
x – a = 2x – 3 = x – \(\frac{3}{2}\)
By Remainder theorem f(x) when divided by (x – \(\frac{3}{2}\) ) leaves a remainder \(\mathrm{f}\left(\frac{3}{2}\right)\)
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 3
As the remainder is 5 we say that (2x – 3) is not a factor of f(x).

Question 4.
Find the remainder when 9x3 – 3x2 + x – 5 is divided by x – \(\frac{2}{3}\)
Solution:
Let f(x) = 9x3 – 3x2 + x – 5
x-a = x – \(\frac{2}{3}\)
Remainder theorem the remainder is \(\mathrm{f}\left(\frac{2}{3}\right)\)
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 4

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

Question 5.
If the polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x + a leave the same remainder, when divided by x – 2, find the value of a.
Solution:
Let f(x) = 2x3 + ax2 + 3x – 5
g(x) = x3 + x2 – 4x + a
Given that f(x) and g(x) divided x – 2
give same remainder.
i e., f(2) = g(2)
By Remainder theorem.
But f(2) = 2(2)3 + a(2)2 + 3(2) – 5
= 2 x 8 + 4a + 6 – 5
= 17 +4a
g(2) = 23 + 22 – 4(2) + a .
= 8 + 4 – 8 + a = 4 + a
i.e., 4 + a = 17 + 4a
∴ a – 4a = 17 – 4
– 3a = 13
a = -13/3

Question 6.
If the polynomials x3 + ax2 + 5 and x3 – 2x2 + a are divided by (x + 2) leave the same remainder, find the value of a.
Solution:
Let f(x) = x3 + ax2 + 5
g(x) = x3 – 2x2 + a
Given that when f(x) and g(x) divided by (x + 2) leaves the same remainder.
i.e.,f(-2) = g(-2)
By Remainder theorem
f(- 2) = (- 2)3 + a(- 2)2 + 5
= -8 + 4a + 5 = 4a – 3
g(- 2) = (- 2)3 – 2(- 2)2 + a
= -8 – 8 + a = a – 16
By problem,
4a – 3 = a – 16
4a – a = – 16 + 3
⇒ 3a = – 13 ⇒ a = -13/3

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

Question 7.
Find the remainder when f(x) = x4 – 3x2 + 4 is divided by g(x) = x – 2 and verify the result by actual division.
Solution:
Given f(x) = x4 – 3x2 + 4
g(x) = x – 2
The remainder when f(x) is divided by g(x) is f(2).
f(2) = 24 – 3(2)2 + 4
= 16 – 12 + 4
= 8
Actual division
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 5
∴ The remainder either by Remainder theorem or by actual division is the same.

Question 8.
Find the remainder when p(x) = x3 – 6x2 + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.
Solution:
Given p(x) = x3 – 6x2 + 14x – 3
g(x) = 1 – 2x
By Remainder theorem when p(x) is divided by g(x) is p(1/2).
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 6

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

Question 9.
When a polynomial 2x3 + 3x2 + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.
Solution:Let f(x) = 2x3 + 3x2 + ax + b
The remainder when f(x) is divided by (x – 2) is 2.
i.e., f(2) = 2
⇒ 2(2)3 + 3(2)2+ a(2) + b = 2
⇒ 16 + 12 + 2a +b = 2
⇒ 2a + b = – 26 …………………..(1)

Also the remainder when f(x) is divided by (x + 2) is – 2.
i.e., f(- 2) = – 2
⇒ 2(- 2)3 + 3(- 2)2 + a (- 2) + b = – 2
⇒ -16 + 12 – 2a + b = – 2
– 2a + b = 2 ………………..(2)
Solving (1) and (2),
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 7
b = – 12
and 2a – 12 = – 26
2a = -26+ 12
a = -14/2 = -7,
a = -7, b = – 12

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.3

Question 1.
It is given that l // m; to prove ∠1 is supplement to ∠8. Write reasons for the Statements.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 1

Solution:

StatementReasons
i) l //m∠1 + ∠8 = 180° (exterior angles on the same side of the transversal)
ii) ∠1 = ∠5corresponding angles
iii) ∠5 + ∠8 = 180°linear pair of angles
iv) ∠1 + ∠8 = 180°exterior angles on the same side of the transversal.
v) ∠1 is supplement is ∠8exterior angles on the same side of the transversal.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 2.
In the given figure AB || CD; CD || EF and y: z = 3:7 find x.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 2
Solution:
Given that AB//CD; CD//EF.
⇒ AB // EF
Also y : z = 3 : 7
From the figure x + y = 180 …………. (1)
[∵ interior angles on the same side of the transversal]
Also y + z = 180 ………….. (2)
Sum of the terms of the ratio y : z
= 3 + 7 = 10
∴ y = \(\frac{3}{10}\) x 180° = 54°
y = \(\frac{7}{10}\) x 180° = 126°
From (1) and (2)
x + y = y + z
⇒ x = z = 126°

Question 3.
In the given figure AB//CD; EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 3
Solution:
Given that EF ⊥ CD; ∠GED = 126°
i. e., ∠FED = 90° and
∠GEF = ∠GED – ∠FED
∠GEF = 126° – 90° = 36°
In ∆GFE
∠GEF + ∠FGE + ∠EFG = 180°
36 + ∠FGE + 90° = 180°
∠FGE = 180° – 126° = 54°
∠AGE = ∠GFE + ∠GEF
(exterior angle in ∆GFE)
= 90°+ 36°= 126°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 4.
In the given figure PQ//ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint : Draw a line parallel to ST through point R.]
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 4
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 5
Given PQ // ST
Draw a lipe ‘l’ parallel to ST through R.
From the figure
a + 110° = 180° and c + 130 = 180°
[ ∵ Interior angles on the same side of the transversal]
∴ a = 180° -110° = 70°
c = 180° – 130° = 50°
Also a + b + c = 180° (angles at a point on a line)
70° + b + 50° = 180°
b = 180° – 120° = 60°
∴ ∠QRS = 60°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 5.
In the given figure m // n. A, B are any two points on in and n respectively. Let C be an interior point between the lines m and n. Find ∠ACB.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 6
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 7
Draw a line ‘l’ parallel to m and n through C.
From the figure
x = a [ ∵ alternate interior angles for l, m]
y = b [ ∵ alternate interior angles for l, n]
∴ z = a + b = x + y

Question 6.
Find the values of a and b, given that p // q and r // s.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 8
Solution:
Given that p // q and r // s.
∴ From the figure
2a = 80° (∵ corresponding angles)
a = \(\frac { 80° }{ 2 }\) = 40°
Also 80° + b = 180° ( ∵ interior angles on the same side of the transversal)
∴ b = 180° – 80° = 100°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 7.
If in the figure a // b and c // d, then name the angles that are congruent to (i) ∠1 and (ii) ∠2.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 9
Solution:
Given that a // b and c // d.
∠1 = ∠3 (∵ vertically opposite angles)
∠1 = ∠5 (∵ corresponding angles)
∠1 = ∠9 (∵ corresponding angles)
Also ∠1 = ∠3 = ∠5 = ∠7 ;
∠1 = ∠11 = ∠9 = ∠13 = ∠15
Similarly ∠2 = ∠4 = ∠6 = ∠8
Also ∠2 = ∠10 = ∠12 = ∠14 = ∠16

Question 8.
In the figure the arrow head segments are parallel, find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 10
Solution:
From the figure
y = 59° ( ∵ alternate interior angles)
x = 60° ( ∵ corresponding angles)

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 9.
In the figure the arrow head segments are parallel then find the values of x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 11
Solution:
From the figure 35° + 105° + y = 180°
∴ y = 180° – 140°
= 40°
∴ x = 40° (∵ x, y are corresponding angles)

Question 10.
Find the values of x and y from the figure.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 12
Solution:
From the figure 120° + x = 180°
(∵ exterior angles on the same side of the transversal)
∴ x = 180° – 120°
x = 60°
Also x = (3y + 6)
(∵ corresponding angles)
3y + 6 = 60°
3y = 60° – 6° = 54°
y = \(\frac { 54 }{ 3 }\) = 18°
∴ x = 60°; y = 18°

Question 11.
From the figure find x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 13
Solution:
From the figure
52° + 90° + (3y + 5)° = 180°
(∵ interior angles of a triangle)
∴ 3y + 147 = 180°
⇒ 3y = 33°
⇒ y = \(\frac { 33 }{ 3 }\) = 11°
Also x + 65° + 52° = 180°
(∵ interior angles on the same side of the transversal)
∴ x = 180° -117° = 63°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 12.
Draw figures for the following statement.
“If the two arms of one angle are respectively perpendicular to the two arms of another angle then the two angles are either equal or supplementary
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 14
AO ⊥ PQ, OB ⊥ QR
Angles are supplementary.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 15
AO ⊥ PQ, OB ⊥ QR
Angles are equal.

Question 13.
In the given figure, if AB // CD; ∠APQ = 50° and ∠PRD = 127°, find x and y.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 16
Solution:
Given that AB // CD.
∠PRD = 127°
From the figure x = 50°
(∵ alternate interior angles)
Also y + 50 = 127°
(∵ alternate interior angles)
∴ y = 127-50 = 77°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 14.
In the given figure PQ and RS are two mirrors placed parallel to each other.
An incident ray \(\overline{\mathrm{AB}}\) strikes the mirror PQ at B, the reflected ray moves along the path \(\overline{\mathrm{BC}}\) and strikes the mirror RS at C and again reflected back along CD. Prove that AB // CD. [Hint : Perpendiculars drawn to parallel lines are also parallel]
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 17
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 18
Draw the normals at B and C.
then ∠x = ∠y (angle of incidence angle of reflection are equal)
∠y = ∠w (alternate interior angles)
∠w = ∠z (angles of reflection and incidence)
∴ x + y = y + z (these are alternate interior angles to \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{CD}}\))
Hence AB // CD.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 15.
In the figures given below AB // CD. EF is the transversal intersecting AB and CD at G and H respectively. Find the values of x and y. Give reasons.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 19
Solution:
For fig(i)
3x = y (∵ alternate interior angles)
2x + y = 180° (∵ linear pair of angles)
∴ 2x + 3x = 180°
5x= 180°
x = \(\frac { 180 }{ 5 }\) = 36°
and y = 3x = 3 x 36 = 108°

For fig (ii)
2x + 15 = 3x- 20°
(∵ corresponding angles)
2x-3x = -20-15
– x = – 35
x = 35°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

For fig (iii)
(4x – 23) + 3x = 180° ,
( ∵ interior angles on the same side of the transversal)
7x- 23 = 180°
7x = 203
x = \(\frac { 203 }{ 7 }\) = 29°

Question 16.
In the given figure AB // CD, ‘t’ is a transversal intersecting E and F re-spectively. If ∠2 : ∠1 = 5 : 4, find the measure of each marked angles.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 20
Solution:
Given that AB // CD and ∠2 : ∠1 = 5 : 4 ∠1 + ∠2 = 180° (-. linear pair of angles) Sum of the terms of the ratio ∠2 :∠1 = 5 + 4 = 9
∴ ∠1 = \(\frac { 4 }{ 9 }\) x 180° = 80°
∠2= \(\frac { 5 }{9 }\) x 180° = 100°
Also ∠1, ∠3, ∠5, ∠7 are all equal to 80°. Similarly ∠2, ∠4, ∠6, ∠8 are all equal to 100°.

Question 17.
In the given figure AB//CD. Find the values of x, y and z.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 21
Solution:
Given that AB // CD.
From the figure (2x + 3x) + 80° = 180°
(∵ interior angles on the same side of the transversal)
∴ 5x = 180° – 80°
x = \(\frac{100}{5}\) = 20°
Now 3x = y (∵ alternate interior angles)
y = 3 x 20° = 60° .
And y + z = 180°
(∵ linear pair of angles)
∴ z = 180°-60° = lg0°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

Question 18.
In the given figure AB // CD. Find the values of x, y and z.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 22
Solution:
Given that AB // CD.
From the figure x° + 70° + x° = 180°
(∵ The angles at a point on the line)
∴ 2x = 180° – 70°
x = \(\frac { 110° }{ 2 }\) = 55-
Also 90° + x° + y° = 180°
[∵ interior angles of a triangle]
90° + 55° + y = 180°
y = 180° – 145° = 35°
And x° + z° = 180°
[∵ interior angles on the same side of a transversal]
55° + z = 180°
z = 180°-55° = 125°

Question 19.
In each of the following figures AB // CD. Find the values of x in each case.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 23
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 24
In each case draw a line ‘l’ parallel to AB and CD through F.
fig (i)
a + 104° = 180° ⇒ a = 180° – 104° = 76°
b+ 116° = 180° ⇒ -b = 180°- 116° = 64°
[∵ interior angles on the same side]
∴ a + b = x = 76° + 64° = 140°

fig-(ii)
a = 35°, b = 65° [∵ alt. int. angles]
x = a + b = 35° + 65° = 100°

fig- (iii)
a + 35° = 180° ⇒ a = 145°
b + 75° = 180° ⇒ b = 105°
[ ∵ interior angles on the same side]
∴ x = a + b = 145° + 105° = 250°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2

Question 1.
Find the value of the polynomial 4x2 – 5x + 3, when
(i) x = 0
Solution:
The value at x = 0 is
4(0)2 – 5(0) + 3
= 3

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

(ii) x = – 1
Solution:
The value at x = – 1 is
4 (- 1)2 – 5 (- 1) + 3
= 4 + 5 + 3
= 12

iii) x = 2
Solution:
The value at x = 2 is
4(2)2 – 5(2) + 3
= 16 -10 + 3
= 9

iv) x = \(\frac{1}{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 1

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials.
i) p(x) = x2 – x + 1
Solution:
p(0) = 02 – 0 + 1 = 1
p(1) = 12 – 1 + 1 = 1
p(2) = 22 – 2 + 1 = 3

ii) P(y) = 2 + y + 2y2 – y3
Solution:
p(0) = 2 + 0 + 2(0)2 – 03 = 2
p(1) = 2+ 1 + 2(1)2 – 13 = 4
p(2) = 2 + 2 + 2(2)2 -23 = 4 + 8- 8 = 4

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

iii) P(z) = z3
Solution:
p(0) = 03 = 0
p(1) = 13 = 1
p(2) = 23 = 8

iv) p(t) = (t – 1)(t + 1) = t2 – 1
Solution:
p(0) = (0 – 1) (0 + 1) = – 1
p(1) = t2 – 1 = 12 – 1 = 0
p(2) = 22 – 1 = 4 – 1 = 3

v) p(x) = x2 – 3x + 2
Solution:
p(0) = 02 – 3(0) + 2 = 2
p(1) = 12 – 3(1) + 2 = 1 – 3 + 2 = 0
p(2) = 22 – 3(2) + 2 = 4- 6 + 2 = 0

Question 3.
Verify whether the values of x given in each case are the zeroes of the polynomial or not ?
i) p(x) = 2x + 1; x = \(\frac{-1}{2}\)
Solution:
The value of p(x) at x = \(\frac{-1}{2}\) is
\(\mathrm{p}\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)+1\)
= -1 + 1 = 0
∴ x = \(\frac{-1}{2}\) is a zero of p(x).

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

(ii) p(x) = 5x – π ; x = \(\frac{-3}{2}\)
Solution:
The value of p(x) at x = \(\frac{-3}{2}\) is
\(\mathrm{p}\left(\frac{-3}{2}\right)=5\left(\frac{-3}{2}\right)-\pi=\frac{-15}{2}-\pi \neq 0\)
∴ x = \(\frac{-3}{2}\) is not a zero of p(x).

iii) p(x) = x2 – 1; x = ±1
Solution:
The value of p(x) at x = 1 and – 1 is
p(1) = 12 – 1 = 0
p(-1) = (-1)2 -1 = 0
∴ x = ±1 is a zero of p(x).

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2
Solution:
The value of p(x) at x = – 1 is
p(-1) = (-1 – 1) (-1 + 2)
=-2 x 1 =-2 ≠ 0
Hence x = – 1 is not a zero of p(x).
And the value of p(x) at x = – 2 is
p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0
Hence, x = – 2 is a zero of p(x).

v) p(y) = y2; y = o
Solution:
The value of p(y) at y = 0 is p(0) = 02 = 0
Hence y = 0 is a zero of p(y).

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

vi) p(x) = ax + b ; x = \(\frac{-\mathbf{b}}{\mathbf{a}}\)
Solution:
The value of p(x) at x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is
\(\mathrm{p}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)=\mathrm{a}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)+\mathrm{b}\)
= -b + b = 0
∴ x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is a zero of p(x).

vii) f(x) = 3x2 – 1; x = \(\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 2

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

viii) f(x) = 2x – 1; x = \(\frac{1}{2} ;-\frac{1}{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 3

Question 4.
Find the zero of the polynomial in each of the following cases.
i) f(x) = x + 2
Solution:
x + 2 = 0
x = – 2

ii) f(x) = x – 2
Solution:
x – 2 = 0
x = 2

iii) f(x) = 2x + 3
Solution:
2x + 3 = 0
2x = – 3
x = \(\frac{-3}{2}\)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

iv) f(x) = 2x – 3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)

v) f(x) = x2
Solution:
x2 = 0
x = 0

vi) f(x) = px, p ≠ 0
Solutin:
px = 0
x = 0

vii) f(x) = px + q; p ≠ 0; p, q are real numbers.
Solution:
px + q = 0
px = -q
x = \(\frac{-\mathrm{q}}{\mathrm{p}}\)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

Question 5.
If 2 is a zero of the polynomial p(x) = 2x2 – 3x + 7a, find the value of
a.
Solution:
Given that 2 is a zero of p(x) = 2x2 – 3x + 7a
(i.e.) p(2) = 0
⇒ 2(2)2 – 3(2) + 7a = 0
⇒ 8 – 6 + 7a = 0
⇒ 2 + 7a = 0
⇒ 7a = – 2
⇒ a = \(\frac{-2}{7}\)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

Question 6.
If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the values of a and b.
Solution:
Given that f(0) = 0; f(1) = 0 and
f(x) = 2x3 – 3x2 + ax + b
∴ f(0) = 2(0)3 – 3(0)2 + a(0) + b
⇒ 0 = b
Also f(1) = 0
⇒ 2(1)3 – 3(1)2 + a(1) + 0 = 0
⇒ 2 – 3 + a = 0 .
⇒ a = 1
Hence a = 1; b = 0

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.2

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 1.
In the given figure three lines \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{EF}}\) intersecting at ‘O’. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 1
Solution:
From the figure
2x + 2y + 2z = 360°
⇒ x + y + z = 180°
x : y : z = 2 : 3 : 5
Sum of the terms of the ratio
= 2 + 3 + 5 = 10
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 2

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 2.
Find the value of x in the following figures.
i)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 3
Solution:
From the figure
3x + 18= 180° – 93° (∵ linear pair )
3x + 18 = 87
3x = 87- 18 = 69
∴ x = \(\frac{69}{3}\) = 23

ii)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 4
Solution:
From the figure
(x – 24)° + 29° + 296° = 360“
(∵ complete angle)
x + 301° = 360°
∴ x = 360° – 301° = 59°

iii)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 5
Solution:
From the figure
2 + 3x = 62
(∵ vertically opposite angles)
3x = 62 – 2
∴ 3x = 60° ⇒ x = \(\frac{60}{3}\)
∴ x = 20°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

iv)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 6
Solution:
From the figure
40 + (6x + 2) = 90°
(∵ complementary angles)
6x = 90° – 42°
6x = 48
x = \(\frac{48}{6}\) = 8°

Question 3.
In the given figure lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) intersect at ’O’. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 7
Solution:
Given that ∠AOC +∠BOE = 70°
∠BOD = 40°
∠AOC = 40°
(∵ ∠AOC, ∠BOD are vertically opposite angles)
∴ 40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Also ∠AOC + ∠COE +∠BOE = 180°
( ∵ AB is a line)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 180° -70° = 110°
∴ Reflex ∠COE = 110°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 4.
In the given figure lines \(\overline{\mathrm{XY}}\) and \(\overline{\mathrm{MN}}\) . intersect at O. If ∠POY = 90° and a : b = 2:3, find c.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 8
Solution:
Given that XY and MN are lines.
∠POY = 90°
a : b = 2 : 3
From the figure a + b = 90°
Sum of the terms of the ratio a : b
= 2 + 3 = 5
∴ b = \(\frac{3}{5}\) x 90° = 54°
From the figure b + c = 180°
(∵ linear pair of angles)
54° + c = 180°
c = 180°-54° = 126°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 5.
In the given figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 9
Solution:
Given that ∠PQR = ∠PRQ
From the figure
∠PQR + ∠PQS = 180° ………….. (1)
∠PRQ + ∠PRT = 180° …………..(2)
From (1) and (2)
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But ∠PQR = ∠PRQ
So ∠PQS = ∠PRT
Hence proved.

Question 6.
In the given figure, if x + y = w + z, then prove that AOB is a line.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 10
Solution:
Given that x + y = w + z = k say
From the figure
x + y + z + w = 360°
(∵ Angle around a point)
Also x + y = z + w
∴ x + y = z + w = \(\frac{360^{\circ}}{2}\)
∴ x + y = z + w = 180°

OR

k + k = 360°
2k = 360°
k = \(\frac{360^{\circ}}{2}\)

(i.e.) (x,y) and (z, w) are pairs of adjacent angles whose sum is 180°.
In other words (x, y) and (z, w) are linear pair of angles ⇒ AOB is a line.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 7.
In the given figure \(\overline{\mathrm{PQ}}\) is a line. Ray \(\overline{\mathrm{OR}}\) is perpendicular to line \(\overline{\mathrm{PQ}}\).\(\overline{\mathrm{OS}}\) os is another ray lying between rays \(\overline{\mathrm{OP}}\) and \(\overline{\mathrm{OR}}\) Prove that
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 11
Solution:
Given : OR ⊥ PQ ⇒ ∠ROQ = 90°
To prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Solution:
Proof: From the figure
∠ROS = ∠QOS – ∠QOR ……………(1)
∠ROS = ∠ROP – ∠POS ……………..(2)
Adding (1) and (2)
∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS [ ∵ ∠QOR = ∠ROP = 90° given]
⇒ 2∠ROS = ∠QOS – ∠POS
⇒ ∠ROS = \(\frac{1}{2}\) [∠QOS – ∠POS]
Hence proved.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2

Question 8.
It is given that ∠XYZ = 64° and XY is produced to point P. A ray \(\overline{\mathrm{YQ}}\) bisects ∠ZYP. Dräw a figure from the given Information. Find ∠XYQ and reflex ∠QYP.
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 12
∠XYQ = 32°
∠QYP = 32°

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.1

Question 1.
Find the degree of each of the polynomials given below,
i) x5 – x4 + 3
Solution:
Degree is 5.

ii) x2 + x – 5
Solution:
Degree is 2.

iii) 5
Solution:
Degree is 0.

iv) 3x6 + 6y3 – 7
Solution:
Degree is 6.

v) 4 – y2
Solution:
Degree is 2.

vi) 5t – √3
Solution:
Degree is 1.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

Question 2.
Which of the following expressions are polynomials in one variable and which are not ? Give reasons for your answer.
i) 3x2 – 2x + 5
Solution:
Given expression is a polynomial in one variable.

ii) x2 + √2
Solution:
Given expression is a polynomial in one variable.

iii) p2 – 3p + q
Solution:
Given expression is not a polynomial in one variable. It involves two variables p and q.

iv) y + \(\frac{2}{\mathbf{y}}\)
Solution:
Given expression is not a polynomial. Since the second term contains the variable in its denominator.

v) \(5 \sqrt{x}+x \sqrt{5}\)
Solution:
Given expression is not a polynomial. Since the first term’s exponent is not an integer.

vi) x100 + y100
Solution:
Given expression has two variables. So it is not a polynomial in one variable.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

Question 3.
Write the coefficient of x3 in each of the following.
i) x3 + x + 1
ii) 2 – x3+ x2
iii) \(\sqrt{2} x^{3}+5\)
iv) 2x3 + 5
v) \(\frac{\pi}{2} x^{3}+x\)
vi) \(-\frac{2}{3} x^{3}\)
vii) 2x2 + 5
viii) 4
Solution:
i) x3 + x + 1 : co-efficient of x3 is 1.
ii) 2 – x3+ x2 : co-efficient of x3 is – 1.
iii) \(\sqrt{2} x^{3}+5\) co-efficient of x3 is √2
iv) 2x3 + 5 : co-efficient of x3 is 2.
v) \(\frac{\pi}{2} x^{3}+x\) co-efficient of x3 is \(\frac{\pi}{2}\)
vi) \(-\frac{2}{3} x^{3}\) co-efficient of x3 is \(-\frac{2}{3}\)
vii) 2x2 + 5 : co-efficient of x3 is ‘0’.
viii) 4 : co-efficient of x3 is ‘0’.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

Question 4.
Classify the following as linear, quadratic and cubic polynomials.
i) 5x2+ x – 7 : degree 2 hence quadratic polynomial.
ii) x – x3 , : degree 3 hence cubic polynomial.
iii) x2 + x + 4 : degree 2 hence quadratic polynomial.
iv) x – 1 : degree 1 hence linear polynomial.
v) 3p : degree 1 hence linear polynomial.
vi) πr2 : degree 2 hence quadratic polynomial.

Question 5.
Write whether the following statements are True or False. Justify your answer.
i) A binomial can have at the most two terms
ii) Every polynomial is a binomial
iii) A binomial may have degree 3
iv) Degree of zero polynomial is zero
v) The degree of x2 + 2xy + y2 is 2
vi) πr2 is monomial
Solution :
i) A binomial can have at the most two terms -True
ii) Every polynomial is a binomial – False
[∵ A polynomial can have more than two terms]
iii) A binomial may have degree 3 – True
iv) Degree of zero polynomial is zero – False
v) The degree of x2 + 2xy + y2 is 2 – True
vi) πr2 is monomial – True

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

Question 6.
Give one example each of a monomial and trinomial of degree 10.
Solution :
– 7x10 is a monomial of degree 10.
3x2y8 + 7xy – 8 is a trinomial of degree 10.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.2

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

Question 1.
Find three different solutions of the each of the following equations.
i) 3x + 4y = 7
Solution:
Given equation is 3x + 4y = 7

Choice of  value of x Or ySimplification for y or xSolution
x = 0

 

3 x 0 + 4y = \(\frac{7}{4}\)(0, \(\frac{7}{4}\) )

 

y = 0

 

3x + 4(0) = 7 ⇒ x = \(\frac{7}{3}\)(\(\frac{7}{3}\) ,0)
x = 1

 

3(1) + 4y = 7

⇒ y = \(\frac{7-3}{-4}\) = 1

(1, 1)

Choice of x or y Simplification for y or x
Solution

ii) y = 6x
Solution: Given equation is y = 6x ⇒ 6x – y = 0

Choice of  value of x Or ySimplification for y or xSolution
x = 06(0) – y = 0 ⇒ y = 0(0,0)
y = 06x – 0 = 0 ⇒ x = 0(0,0)
x = 16(1) – y = 0 ⇒ y = 6(1,6)
Y = 16x – 1 = 0 ⇒ 6x = 1 ⇒ x = \(\frac{1}{6}\)(\(\frac{1}{6}\),1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) 2x – y = 7
Solution:
Given equation is 2x – y = 7

Choice of  value of x Or ySimplification for y or xSolution
x = 02(0) – y = 7 ⇒ y = -7(0, -7)
y = 02x – 0 = 7 ⇒ x = \(\frac{7}{2}\)(\(\frac{7}{2}\) , 0)
x = 12(1) – y = 7 ⇒ -y = 7 – 2  ⇒ y = -5(1, -5)

iv) 13x – 12y = 25
Solution:
Given equation is 13x – 12y = 25

Choice of  value of x Or ySimplification for y or xSolution:
x = 0

 

13(0) – 12y = 25 ⇒ y = \(-\frac{25}{12}\)(0, \(-\frac{25}{12}\) )

 

y = 0

 

13x – 12(0) = 25 ⇒ y = \(\frac{25}{13}\)(\(\frac{25}{13}\) ,0)
x = 1

 

13(1) – 12y = 25
⇒ -12y = 25 – 13
y = \(\frac{12}{-12}\) = -1
(1, -1)

v) 10x + 11y = 21
Solution:
Given equation is 10x + 11y = 21

Choice of  value of x Or ySimplification for y or xSolution
x = 010(0) + 11y = 21 ⇒ y = \(\frac{21}{11}\)(0, \(\frac{21}{11}\))
y = 010x +11(0) = 21 ⇒ x = \(\frac{21}{10}\)(\(\frac{21}{10}\) , 0)
x = 110(1) + 11y = 21 ⇒ 11y = 21 – 10  ⇒ y = \(\frac{11}{11}\) = 1(1, 1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

vi) x + y = 0
Solution:
Given equation is x + y = 0

Choice of a value of x or ySimplificationSolution
x = 00 + y = 0 ⇒ y = 0(0, 0)
x = 11 + y = 0 ⇒ y = -1(1,-1)
y = 1x + 1 = 0 ⇒ x = -1(- 1, 1)

Question 2.
If (0, a) and (b, 0) are the solutions of the following linear equations. Find a and b.
8x – y = 34
Solution:
Given that (0, a) and (b, 0) are the solutions of 8x – y = 34
∴ 8(0) – a = 34 and 8(b) – 0 = 34
⇒ a = -34 and b = \(\frac{34}{8}=\frac{17}{4}\)

ii) 3x = 7y – 21
Solution:
Given that (0, a) and (b, 0) are the solutions of 3x = 7y – 21
⇒ 3x – 7y = – 21
∴ 3(0) – 7(a) = -21 and 3(b) – 7(0) = -21
⇒ a = \(\frac{-21}{-7}\) and b = \(\frac{-21}{3}\)
⇒ a = 3 and b = -7

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) 5x – 2y + 3 = 0
Given that (0, a) and (b, 0) are the solutions of 5x – 2y = – 3
i. e., 5(0) – 2a = – 3 and 5 (b) – 2(0) = – 3
⇒ -2a = -3 and 5b = -3
⇒ a = \(\frac{3}{2}\) and b = \(\frac{-3}{5}\)

Question 3.
Check which of the following is solution of the equation 2x – 5y = 10.
(i) (0, 2) (ii) (0,-2) (iii)(5, 0) (iv) (2√3, -√3) (v) (\(\frac{1}{2}\) , 2)
Solution:
i) (0, 2)
The given equation is 2x – 5y = 10
On substituting (0, 2), the L.H.S becomes
2(0)-5(2) = 0-10 = -10
R.H.S = 10
L.H.S ≠ R.H.S
∴ (0, 2) is not a solution.

ii) (0. – 2)
Substituting (0, – 2) in the L.H.S of 2x – 5y = 10, we get
2(0) – 5 (- 2) = 0 + 10 = 10 = R.H.S
∴ (0, – 2) is a solution.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

iii) (5, 0)
Substituting (5, 0) in the L.H.S of 2x – 5y = 10, we get
2(5)-5(0) = 10-0 = 10 = R.H.S
∴ (5, 0) is a solution.

iv) 2√3, -√3
On substituting (2√3, -√3 ), the L.H.S becomes
2(2√3) – 5(-√3) = 4√3 + 5√3 = 9√3 ≠ R.H.S
∴ (2√3, -√3) is not a solution.

v) (\(\frac{1}{2}\), 2)
Given equation is 2x – 5y = 10
Put x = \(\frac{1}{2}\) and y = 2 in the given equation.
Then 2(\(\frac{1}{2}\)) – 5(2) =10
1-10 = 10
-9 = 10 false
∴ (\(\frac{1}{2}\) , 2) is not a solution.

Question 4.
Find the value of k, if x = 2; y = 1 is a solution of the equation 2x + 3y = k. Find two more solutions of the resultant equation. £3 Each
Solution:
Given that x = 2 and y = 1 is a solution of 2x + 3y = k .
∴ 2(2) + 3(1) = k ⇒ 4 + 3 = k ⇒ k = .7 ’
∴ The equation becomes 2x + 3y = 7

x01
y2(0) + 3y = 7
3y = 7
y = \(\frac{7}{3}\)
2(1) + 3y = 7
3y = 7 – 2= y = \(\frac{5}{3}\)
(x, y)(0,\(\frac{7}{3}\))(1, \(\frac{5}{3}\))

Two more solutions are (0, \(\frac{7}{3}\)) and (1, \(\frac{5}{3}\))

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

Question 5.
If x = 2 – α and y = 2 + α is a solution of the equation 3x – 2y + 6 = 0, find the value of ‘α’. Find three more solutions of the resultant equation.
Solution:
Given that x = 2 – α and y = 2 + α is a solution of 3x – 2y + 6 = 0
Thus 3 (2 – α) – 2 (2 + α) + 6 = 0
⇒ 6 – 3α – 4 – 2α + 6 = 0
⇒ -5α + 8 = 0
⇒ -5α = -8
∴ α = \(\frac{8}{5}\)
Three more solutions are

x03x – 2y = -6
3x – 2(0) = -6
y = \(\frac{-6}{3}\) = -2
1
y3x – 2y = -6
3(0) – 2y = -6
y = 3
03x – 2y = -6
3(1) – 2y = -6
y = \(\frac{9}{2}\)
Solutions(0, 3)(-2, 0)(1, \(\frac{9}{2}\))

Question 6.
If x = 1;y = 1 is a solution of the equation 3x + ay = 6, find the value of ‘a’.
Solution:
Given that x = 1; y = 1 is a solution of 3x + ay = 6.
Thus 3(1) + a(1) = 6
⇒ 3 + a = 6 ⇒ a = 6 – 3 = 3

Question 7.
Write five different linear equations in two variables and find three solutions for
each of them.
Solution:
i) Let the equations are 2x – 4y = 10

x012
y2(0) – 4y = 10
y = \(\frac{-5}{2}\)
2x – 4y = 10
2(1)  – 4y =10y = -2
2x – 4y = 10
2(2) – 4y = 10
y = –\(\frac{-3}{2}\))
Solutions(0, –\(\frac{-5}{2}\))(1 , -2)(2, \(\frac{-3}{2}\))

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

ii) 5x + 6y = 15

x05x + 6(0) = 15
x = 3
1
y5(0) + 6y = 15
y = \(\frac{15}{6}\) = \(\frac{5}{2}\)
05(1) + 6y = 15
y = \(\frac{5}{3}\)
(x, y)(0, \(\frac{5}{2}\))(3, 0 )(1, \(\frac{5}{3}\))

iii) 3x-4y = 12

x041
y-30\(\frac{-9}{4}\)
(x, y)(0, -3)(4, 0 )(1, \(\frac{-9}{4}\))

iv) 2x – 7y = 9

x0\(\frac{9}{2}\)1
y\(\frac{-9}{7}\)0-1
(x, y)(0, \(\frac{-9}{7}\))( \(\frac{9}{2}\) , 0 )(1, -1)

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.2

v) 7x- 5y = 3

x0\(\frac{3}{7}\)1
y\(\frac{-3}{5}\)0\(\frac{4}{5}\)
(x, y)(0, \(\frac{-3}{5}\) )(\(\frac{3}{7}\), 0 )(1, \(\frac{4}{5}\))

 

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.1

Question 1.
In the given figure, name:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 1
i) Any six points
Solution:
A, B, C, D, P, Q, M, N etc.

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

ii) Any five line segments
Solution:
\(\overline{\mathrm{AX}}, \overline{\mathrm{XM}}, \overline{\mathrm{MP}}, \overline{\mathrm{PB}}, \overline{\mathrm{MN}}, \overline{\mathrm{PQ}}, \overline{\mathrm{AB}} \ldots \ldots\) etc.

iii) Any four rays
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\) etc.

iv) Any four lines
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\)

v) Any four collinear points
Solution:
A, X, M, P and B are collinear points on the line \(\overline{\mathrm{AB}}\).

Question 2.
Observe the following figures and identify the type of angles in them.
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 2
Solution:
∠A – reflex angle
∠B – right angle
∠C – acute angle

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

Question 3.
State whether the following state¬ments are true or false
i) A ray has no end point.
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\)
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\)
iv) A line has a definite length.
v) A plane, has length and breadth but no thickness.
vii) Two lines may intersect in two points.
viii) Two intersecting lines cannot both be parallel to the same line.
Solution:
i) A ray has no end point. – False
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\) – True
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\) – False
iv) A line has a definite length. – False
v) A plane, has length and breadth but no thickness. – True
vi) Two distinct points always determine a unique line. – True
vii) Two lines may intersect in two points. – False
viii) Two intersecting lines cannot both be parallel to the same line. – True

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

Question 4.
What is the angle between two hands of a clock when the lime in the clock
is
a) 9 ‘o clock
b) 6 ‘o clock
c) 7 ; 00 p.m.
Solution:
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 3
a) 12 hours = = 360°
1 hour = \(\frac{360^{\circ}}{12}\) = 30°
∴Angle between hands when the time is 9 o clock = 3 x 30 = 90

b)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 4
Angle between hands = 6 x 30° = 180°

AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1

c)
AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 5
Angle between hands = 7 x 30° = 210°

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry Exercise 3.1

Question 1.
Answer the following:
i) How many dimensions a solid has ?
Solution:
A solid has three dimensions namely length, breadth and height or depth.

ii) How many books are there in Euclid’s Elements ?
Solution:
There are 13 volumes in Euclid’s elements.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

iii) Write the number of faces of a cube and cuboid.
Solution:
Cube : 6 faces
Cuboid : 6 faces

iv) What is the sum of interior angles of a triangle ?
Solution:
The sum of interior angles of a triangle is 180°.

v) Write three undefined terms of geometry.
Solution:
Point, line and plane are three undefined terms in geometry.

Question 2.
State whether the following statements are true or false. Also give reasons for your answers.
a) Only one line can pass through a given point
b) All right angles are equal
c) Circles with same radii are equal
d) A finite line can be extended on its both sides endlessly to get a straight line
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 1
e) From figure AB > AC
Solution:
a) Only one line can pass through a given point – False.
Reason : (Since, infinitely many lines can pass through a given point)
b) All right angles are equal – True.
c) Circles with same radii are equal – True.
d) A finite line can be extended on its both sides endlessly to get a straight line – True.
e) From figure AB > AC – True.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 3.
In the figure given below, show that the length AH > AB + BC + CD.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 2
Solution:
Given a line \(\stackrel{\leftrightarrow}{\mathrm{AH}}\)
To prove AH > AB + BC + CD
From the figure AB + BC + CD = AD
AD is a part of whole AH.
From Euclid’s axiom whole is greater than part.
∴ AH > AD
⇒ AH > AB + BC + CD

Question 4.
If a point Q lies between two points P and R such PQ = QR, prove that PQ = \(\frac{1}{2}\)PR.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 3
Let PR be a given line.
Given that PQ = QR
i. e., Q is a point on PR.
⇒ PQ + QR = PR
⇒ PQ + PQ = PR [∵ PQ = QR]
⇒ 2PQ = PR
⇒ PQ = \(\frac{1}{2}\) PR
Hence proved.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 5.
Draw an equilateral triangle whose sides are 5.2 cm
Soluton:
Step – 1 : Draw a line segment AB of length 5.2 cm. *
Step – 2 : Draw an arc of radius 5.2 cm with centre A.
Step – 3 : Draw an arc of radius 5.2 cm with centre B.
Step – 4 : Two arcs intersect at C; join C to A and B.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 4
Δ ABC is the required triangle.

Question 6.
What is a conjecture? Give an example for it.
Solution:
Mathematical statements which are neither proved nor disproved are called conjectures. Mathematical discoveries often start out as conjectures. This may be an educated guess based on observations.
Eg : Every even number greater than 4 can be written as sum of two primes. This example is called Gold Bach Conjecture

Question 7.
Mark two points P and Q. Draw a line through P and Q. Now how many lines are parallel to PQ, can you draw ?
Solution:
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 5
Infinitely many lines parallel to PQ can be drawn.

Question 8.
In the figure given below, a line n falls on lines / and m such that the sum of the interior angles 1 and 2 is less than 180°, then what can you say about lines l and m ?
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 6
Solution:
Given : l, m and n are lines, n is a transversal.
∠1 < 90°
∠2 < 90°
If the lines l and m are produced on the side where angles 1 and 2 are formed, they intersect at one point.

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 9.
In the figure given below, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4 write the rela-tion between ∠1 and ∠2 using Euclid’s postulate.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 7
Solution :
Given : ∠1 = ∠3
∠3 = ∠4
∠2 = ∠4
∴∠1 = ∠2

∵Both ∠1 and ∠2 are equal to ∠4. (By Euclid’s axiom things which are equal to same things are equal to one another).

AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1

Question 10.
In the figure given below, we have BX = \(\frac{1}{2}\) AB, BY= \(\frac{1}{2}\) BC and AB = BC. Show that BX = BY.
AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 8
Solution:
Given : BX = \(\frac{1}{2}\) AB
BY = \(\frac{1}{2}\)BC
AB = BC
To prove : BX = BY
Proof: Given AB = BC [ ∵ By Euclid’s axiom things which are halves of the same things are equal to one another]
\(\frac{1}{2}\) AB = \(\frac{1}{2}\) BC
BX = BY
Hence proved.

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables Exercise 6.1

Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the values of a, b and c in each case.
i) 8x + 5y – 3 = 0
Solution:
8x + 5y – 3 = 0
⇒ 8x + 5y + (- 3) = 0
Here a = 8, b = 5 and c = – 3

ii) 28x – 35y = – 7
Solution:
28x – 35y = – 7
⇒ 28x + (- 35) y + 7 = 0
Here a = 28, b = – 35 and c = 7

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iii) 93x = 12- 15y
Solution:
93x = 12 – 15y
⇒ 93x + 15y -12 = 0
⇒ 93x + 15y + (- 12) = 0
Here a = 93, b = 15 and c = – 12

iv) 2x = – 5y
Solution:
2x = – 5y
⇒ 2x + 5y = 0
Here a = 2, b = 5 and c = 0

v) \(\frac{x}{3}+\frac{y}{4}=7\)
Solution:
\(\frac{x}{3}+\frac{y}{4}=7\)
⇒ \(\frac{x}{3}+\frac{y}{4}-7=0\)
⇒\(\frac{4 x+3 y-84}{12}=0\)
⇒ 4x + 3y – 84 = 0
Here a = 4, b = 3 and c = – 84

vi) y = \(-\frac{3}{2} x\)
Solution:
y = \(-\frac{3}{2} x\)
⇒ 2y = -3x
⇒ 3x + 2y = 0
Here a = 3, b = 2 and c = 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

vii) 3x + 5y = 12
Solution:
3x + 5y = 12
⇒ 3x + 5y + (- 12) = 0
Here a = 3, b = 5 and c = – 12

Question 2.
Write each of the following in the form of ax + by + c = 0 and find the values of a, b and c.
i) 2x = 5
Solution:
2x – 5 = 0
a = 2
b = 0
c = -5

ii) y – 2 = 0
Solution:
y – 2 = 0
a = 0
b = 1
c = – 2

iii) \(\frac{y}{7}\) = 3
Solution:
\(\frac{y}{7}\) = 3
y = 21
y – 21 = 0
a = 0
b = 1
c = -21

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iv) x = \(-\frac{14}{13}\)
x = \(-\frac{14}{3}\)
⇒ 13x = – 14
⇒ 13x + 14 = 0
a = 13
b = 0
c = 14

Question 3.
Express the following statements as a linear equation in two variables,
i)The sum of two numbers is 34.
Solution:
x + y = 34; x, y are any two numbers ⇒ x + y – 34 = 0

ii) The cost of a ball pen is ?5 less than half the cost of a fountain pen.
Solution:
Let the cost of a fountain pen = x
Let the cost of ball pen = y
Then y = x – 5 or x – y – 5 = 0

AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables Ex 6.1

iii) Bhargavi got 10 more marks than double of the marks of Sindhu. |l M)
Solution:
Let Sindhu’s marks = x
Bhargavi’s marks = y
Then by problem y = 2x + 10 or 2x – y + 10 = 0

iv) The cost of a pencil is ₹2 and one ball point pen costs ₹15. Sheela pays ₹100 for the pencils and pens she purchased.
Solution:
Giver: that cost of a pencil = ₹2
Cost of a ball point pen = ₹15
Let the number of pencils purchased = x
Let the number of pens purchased = y
Then the total cost of x – pencils = 2x
Then the total cost of y – pens = 15y
By problem 2x + 15y = 100

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.
Use suitable identities to find the following products.
i) (x + 5) (x + 2)
Solution:
(x + 5) (x + 2)
= x2 + (5 + 2)x + 5 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= x2 + 7x + 10

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) (x – 5) (x – 5)
Solution:
(x – 5) (x – 5)
= (x – 5)2 = x2 – 2(x) (5) + 52
[ ∵(x – y)2 = x2 – 2xy + y2]
= x2 – 10x + 25

iii) (3x + 2) (3x – 2)
Solution:
(3x + 2) (3x – 2) = (3x)2 – (2)2
[∵ (x + y) (x – y) =x2 – y2]
= 9x2 – 4

iv) \(\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 1(i)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

v) (1 + x) (1 + x)
Solution:
(1 + x) (1 + x)
= (1 + x)2 = 12 + 2 (1) (x) + x2
[∵(x + y)2 = x2 + 2xy + y2]
= 1 + 2x + x2

Question 2.
Evaluate the following products with¬out actual multiplication.
i) 101 x 99
Solution:
101 x 99
= (100 + 1) (100 – 1)
= 1002 – 12
= 10000 – 1
= 9999

ii) 999 x 999
Solution:
999 x 999
= 9992
= (1000 – 1)2
= 10002 – 2 x (1000) x 1 + 12
= 1000000-2000 + 1
= 998001

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(50 \frac{1}{2} \times 49 \frac{1}{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 1

iv) 501 x 501
Solution:
501 x 501
= (500 + 1) (500 + 1)
= (500 + 1)2
= 5002 + 2 x (500) x 1 + 12
= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)
= 302 – (0.5)2
= 900 – 0.25
= 899.75

Question 3.
Factorise the following using appro-priate identities.
i) 16x2 + 24xy + 9y2
Solution:
16x2 + 24xy + 9y2
= (4x)2 + 2 (4x) (3y) + (3y)2
= (4x + 3y)2 = (4x + 3y) (4x + 3y)
[ ∵ (x + y)2 = x2 + 2xy + y2]

ii) 4y2 – 4y + 1
Solution:
4y2 – 4y + 1
= (2y)2 – 2 (2y) (1) + (1)2
[ ∵ (x -y)2 = x2 – 2xy + y2]
= (2y -1)2 = (2y – 1) (2y-1)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(4 x^{2}-\frac{y^{2}}{25}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 2

iv) 18a2 – 50
Solution:
18a2 – 50 = 2 (9a2 – 25)
= 2[(3a)2 – (5)2]
[ ∵ x2 – y2 = (x + y) (x – y)]
= 2 (3a + 5) (3a – 5)

v) x2 + 5x + 6
Solution:
x2 + 5x + 6 = x2 + (3 + 2) x + 3 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + a . b]
= (x + 3) (x + 2)

vi) 3p2 – 24p + 36
Solution:
3p2 – 24p + 36
= 3[p2 – 8p + 12]
= 3[p2 + (- 6 – 2)p + (- 6) (- 2)]
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= 3 (p – 6) (p – 2)

Question 4.
Expand each of the following, using suitable identities.
i) (x + 2y + 4z)2
(x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) (2a – 3b)3
Solution:
(2a – 3b)3 = (2a)3 – 3 (2a)2 (3b) + 3 (2a) (3b)2 – (3b)3
[ ∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 8a3 – 3(4a2) (3b) + 3 (2a) (9b2) – 27b3
= 8a3 – 36a2b + 54ab2-27b3
(or)
∵ (a – b)3 = a3 – b3– 3ab (a – b)]
= (2a)3 – (3b)3 – 3(2a) (3b) (2a – 3b)
= 8a3 – 27b3 – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)2
Solution:
(- 2a + 5b – 3c)2
= (- 2a)2 + (5b)2 + (- 3c)2 + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)
= 4a2 + 25b2 + 9c2 – 20ab – 30bc + 12ca
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz +2za]

iv) \(\left[\frac{a}{4}-\frac{b}{2}+1\right]^{2}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 3

v) (p + 1)3
Solution:
(p + 1)3
= (P)3 + 3 (p)2 (1) + 3 (p) (1)2 + (1)3
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= p3 + 3p2 + 3p + 1

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

vi) \(\left(x-\frac{2}{3} y\right)^{3}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 4

Question 5.
Factorise
i) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
Solution:
25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
= (5x)2 + (- 4y)2 + (- 2z)2 + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)
= (5x – 4y – 2z)2 = (- 5x + 4y +, 2z)2

ii) 9a2 + 4b2 + 16c2 + 12ab – 16bc – 24ca
Solution:
9a2 + 4b2 + 16c2 + 12ab – 16bc -24ca
= (3a)2 + (2b)2 + (- 4c)2+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)
= (3a + 2b – 4c)2

Question 6.
If a + b + c = 9 and ab + be + ca = 26, find a2 + b2 + c2.
Solution:
Given that a + b + c = 9
Squaring on both sides,
(a + b + c)2 = 92
⇒ a2+ b2 + c2+ 2 (ab + be + ca) = 81 ⇒ a2 + b2 + c2 = 81 – 2 (ab + be + ca)
(by problem)
= 81 – 2 x 26
= 81 – 52 = 29

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 7.
Evaluate the following by using suit¬able identities. m EachgM)
i) (99)3
Solution:
(99)2 = (100 – 1)3
= 1003 – 3 (100)2 (1) + 3 (100) (1)2 – 13
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 + y3]
= 1000000 – 30000 + 300 – 1
= 970299

ii) (102)3
Solution:
(102)3 = (100 + 2)3
= 1003 + 3 (100)2 (2) + 3 (100) (2)2 + 23
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= 1000000 + 60000 + 1200 + 8
= 1061208

iii) (998)3
Solution:
(998)3 =(1000 – 2)3
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 – y3] = 10003– 3(1000)2(2) + 3(1000)(2)2– 23
= 1000000000 – 6000000 + 12000 – 8
= 994011992

iv) (1001)3
Solution:
(1001)3 = (1000 + 1)3 .
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3] = 10003 + 3(1000)2(1) + 3(1000) (1)2 + 13
= 1000000000 + 3000000 + 3000 + 1
= 1003003001

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 8.
Factorise each of the following.
i) 8a3 + b3 + 12a2 b + 6ab2
Solution:
8a3 + b3 + 12a2 b + 6ab2
= (2a)3 + (b)3 + 3 (2a)2 (b) + 3 (2a) (b)2
= (2a + b)3

ii) 8a3 – b3 – 12a2 b + 6ab2
Solution:
8a3 – b3 – 12a2 b + 6ab2
= (2a)3 – (b)3 – 3 (2a)2 (b) + 3 (2a) (b)2
= (2a – b)3

iii) 1 – 64a3 -12a + 48a2
Solution:
1 – 64a3 – 12a + 48a2
= (1)3 – (4a)3 – 3(1)2 (4a) + 3(1) (4a)2
= (1 – 4a)3

iv) \(8 p^{3}-\frac{12}{5} p^{2}+\frac{6}{25} p-\frac{1}{125}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 5

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 9.
Verify i) x3 + y3 = (x + y) (x2 – xy + y2);
ii) x3 – y3 = (x – y) (x2 + xy + y2)
Using some non-zero positive integers and check by actual multiplication. Can you
call these as identities ?
i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
Given x3 + y3 = (x + y) (x2 – xy + y2)
L.H.S = x3 + y3
R.H.S = (x + y) (x2 – xy + y2)
= x (x2 – xy + y2) + y (x2 – xy + y2)
= x3 -x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S
∴ L.H.S = R.H.S

Take x = 3, y = 2
L.H.S = 33 + 23 = 27 + 8 = 35
R.H.S = (3 + 2) (32 – 3 x 2 + 22)
= 5 x (9 – 6 + 4)
= 5 x 7 = 35
∴ L.H.S = R.H.S

ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
Given that x3 – y3 = (x – y) (x2 + xy + y2)
L.H.S = x3 – y3
R.H.S = (x – y) (x2 + xy + y2)
= x (x2 + xy + y2) – y (x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3= L.H.S

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

L.H.S = 33 – 23 = 27 – 8 = 19
R.H.S = (3 – 2) (32 + 3 x 2 + 22)
= 1 x (9 + 6 + 4)
= 1 x 19 = 19
∴ L.H.S = R.H.S
We can call the above two expressions as identities

Question 10.
Factorise by using the above results (identities).
i) 27a3 + 64b3
Solution:
27a3+ 64b3 = (3a)3 + (4b)3
= (3a + 4b) {(3a)2 – (3a) (4b) + (4b)2}
= (3a + 4b) (9a2 – 12ab + 16b2)

ii) 343y3 – 1000
Solution:
343y3 – 1000 = (7y)3 – (10)3
= (7y – 10) [(7y)2 + (7y) (10) + (10)2]
= (7y – 10) (49y2 + 70y + 100)

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 11.
Factorise 27x3 + y3 + z3 – 9xyz using identity.
Solution:
Given 27x3 + y3 + z3 – 9xyz
= (3x)3 + (y)3 + (z)3 – 3 (3x) (y) (z)
= (3x + y + z)
[(3x)2 + y2 + z2 – (3x) (y) – (y) (z) – (z) (3x)]
[ ∵ (x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2– xy – yz – zx)
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

Question 12.
Verify that x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
(OR)
Verify that
p3 + q3 + r3 – 3pqr = 1/2 (p + q + r)
[(p – q)2 + (q – r)2 + (r – p)2]
Solution:
Given x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
R-H.S = 1/2 (x + y + z) [(x – y)2 + (y – z)2+ (z – x)2]
= 1/2 (x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2xz]
= 1/2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= 1/2 (x + y + z) (2) [x2 + y2 + z2 – xy – yz – zx]
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= L.H.S
Hence proved.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz
Solution:
Given x + y + z = 0
To prove x3 + y3 + z3 = 3xyz
We have an identity
(x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz
Substituting x + y + z = 0in the above equation, we get
0 x (x2 + y2 + z2 -xy-yz-zx)
= x3 + y3 + z3 – 3xyz
⇒ x3 + y3 + z3 – 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz

Question 14.
Without actual calculating the cubes, find the value of each of the following.
i) (- 10)3 + 73 + 33
Solution:
Given (-10)3 + 73 + 33
Sum of the bases = -10 + 7 + 3 = = 0
∴ (- 10)3 + 73 + 33
= 3 (- 10) x (7) x 3
= -630
[ ∵ x + y + z = 0 then x3 + y3 + z3 = 3xyz]

ii) (28)3 + (- 15)3 + (- 13)3
Solution:
Given (28)3 + (- 15)3+ (- 13)3
Sum of the bases = 28 + (- 15) + (- 13) = 0
∴ (28)3 + (- 15)3 + (- 13)3
= 3 x 28 x (- 15) x (- 13)
= 16380

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

iii) \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}\) read it as \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(\frac{-5}{6}\right)^{3}\)
Solution:
AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 6

iv) (0.2)3 – (0.3)3 + (0.1)3
Solution:
Given that (0.2)3 – (0.3)3 + (0.1)3
= (0.2)3 + (- 0.3)3 + (0.1)3
Sum of the bases = 0.2 – 0.3 + 0.1 = 0
∴ (0.2)3 + (-0.3)3 + (0.1)3
= 3 x (0.2) (- 0.3) (0.1)
= -0.018

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

Question 15.
Give possible expressions for the length and breadth of the rectangle whose area is given by
i) 4a2 + 4a – 3
Given that area = 4a2 + 4a – 3
= 4a2 + 6a – 2a – 3
= 2a (2a + 3) – 1 (2a + 3)
= (2a – 1) (2a + 3)
∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a2 – 35a + 12
Solution:
Given that area = 25a2 – 35a +12
= 25a2 – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?
i) 3x3 – 12x
Solution:
Volume = 3x3 – 12x
= 3x (x2 – 4)
= 3x (x + 2) (x – 2) are the dimensions.

AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

ii) 12y2 + 8y – 20
Solution:
Given that volume = 12y2 + 8y – 20
= 4 (3y2 + 2y – 5)
= 4 [3y2 + 5y – 3y – 5]
= 4 [y (3y + 5) – 1 (3y + 5)]
= 4 (3y + 5) (y – 1)
Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.
Show that if 2 (a2 + b2 ) = (a + b)2 then a = b
Solution:
Given that 2 (a2 + b2 ) = (a + b)2
To prove a = b
As 2 (a2 + b2 ) = (a + b)2
We have
2a2 + 2b2 = a2 + 2ab + b2
2a2 – a2 + 2b2 – b2 = 2ab
a2 + b2 = 2ab
This is possible only when a = b
∴ a = b

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.3

Question 1.
Plot the following points on the Cartesian plane whose x, y co-ordinates are given.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 1
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 1 (i)
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 2

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 2.
Are the positions of (5, -8) and (-8, 5) Is same ? JustIfy your answer.
Solution:
The positions of (5, -8) and (-8, 5) are not same. They are two distinct points. (5, -8) lies at a distance of 5 units from Y – axis and 8 units from X – axis on down side of the origin. So it lies in Q4. Where as (- 8, 5) lies in Q2. The point is at a distance of 8 units from Y – axis on left side of the origin and 5 units from X – axis.

Question 3.
What can you say about the position of the points (1, 2), (1, 3), (1, – 4), (1, 0) and (1, 8), Locate on a graph sheet.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 3
All the given points lie on a line parallel to Y-axis at a distance of 1 cm.

Question 4.
What can you say about the position of the points (5, 4), (8, 4), (3, 4), (0, 4), (-4, 4), (-2,4)? Locate the points on a graph sheet and justify your answer.
Solution:
The points (5, 4), (8, 4), (3, 4), (0, 4), (- 4, 4) and (- 2, 4) all lie on a line parallel to X-axis at a distance of 4-units from it.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 4

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 5.
Plot the points (0, 3), (4, 3), (3, 4) (4, 0) in graph sheet. Join the points with straight lines to make a rectangle. Find the area of the rectangle.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 5
From the graph, area of the rectangle =12 square units (OR)
Length = 4 units ; breadth = 3 units
A = l/b = 4 × 3 = 12 sq. units.

Question 6.
Plot the points (2, 3), (6, 3) and (4, 7) in a graph sheet. Join them to make it a triangle. Find the area of the triangle.
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 6
From the graph
Base of the triangle = 4 units
Height of the triangle = 4 units
∴ Area of the triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) × 4 × 4 = 8sq. units.

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 7.
Plot at least six points in a graph sheet, each having the sum of its co-ordinates equal to 5. [Hint: (- 2, 7), (1, 4)………….]
Solution:
Given that (x co-ordinate) + (y co-ordinate) = 5
Let the points be
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 7

Question 8.
Look at the graph. Write the co-ordinates of the points
A, B, C, D, E, F. G. H, 1, J, K, L, M, N, O, P and Q.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 8
Solution:
A(- 3, 4) ; B(0, 5) : C (3, 4) ; D (2, 4) ; E (2, 0) ;
F (3, 0) ; G (3, – 1) ; H (0, – 1) ; I (- 3, – 1) ;
J (- 3, 0) ; K (- 2, 0) ; L (- 2, 4) ; M (- 1, 0) ;
N (-1, 3); O (0, 0) ; P (1. 3) and Q (1, 0)

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3

Question 9.
In a graph sheet plot each pair of points, join them by line segments.
i) (2, 5), (4, 7)
ii) (-3, 5) (-1,7)
iii) (-3, -4), (2, -4)
iv) (-3, -5) (2, -5)
v) (4, -2), (4, -3)
vi) (-2, 4), (-2, 3)
vii) (-2, 1), (-2, 0)
Now join the following pairs of points by straight line segments, in the same graph.
viii) (-3, 5), (-3, 4)
ix) (2, 5), (2, -4)
x) (2, -4), (4, -2)
xi) (2, -4), (4, -3)
xii) (4, -2), (4, 7)
xiii) (4, 7), (-1, 7)
xiv) (-3, 2), (2, 2)
Now you will get a surprise figure. What is it ?
Solution:
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.3 9

AP Board 9th Class Maths Solutions

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry Exercise 5.2

Question 1.
Write the quadrant in which the following points lie.
i) (- 2, 3)
ii) (5, – 3)
iii) (4, 2)
iv) (- 7, – 6)
v) (0, 8)
vi) (3, 0)
vii) (-4,0)
viii) (0, – 6)
Solution:
i) (- 2, 3) – Q2 (second quadrant)
ii) (5, – 3) – Q4 (fourth quadrant)
iii) (4, 2) – Q1 (Iirst quadrant)
iv) (- 7, – 6) – Q3 (third quadrant)
v) (0, 8) – on Y-axis
vi) (3, 0) – on X-axis
vii) (-4,0) – on X’ – axis
viii) (0, – 6) – on Y’: axis

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 2.
Write the abscissae and ordinates of the following points.
i)(4,-8)
ii)(-5,3)
iii)(0,0)
iv)(5, 0)
v)(0, -8)
Note: Plural of abscissa is abscissae.
Solution:

Pointabscissaordinate
i) (4, – 8)4-8
ii)  (-5, 3)-53
iii) (0,0)00
iv) (5,0)50
v) (0,-8) .0-8

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 3.
Which of the following points lie on the axes ? Also name the axis.
i) (-5,-8)
ii) (0, 13)
iii) (4, – 2)
iv) (- 2,0)
v) (0, – 8)
vi) (7,0)
vii) (0,0)
Solution:
The points (ii) (0,13) ; (v) (0,- 8) lie on Y – axis.
The points (iv) (- 2, 0), (vi) (7, 0) lie on X – axis.
The point (vii) (0, 0) lie on both X – axis and Y – axis.
The points (i) (- 5, – 8); (iii) (4, – 2) do not lie on any axis.

Question 4.
Write the following based on the graph.
i) The ordinate of L
ii) The ordinate of Q
iii) The point denoted by (- 2,-2)
iv) The point denoted by (5, – 4)
v) The abscissa of N
vi) The abscissa of M
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 1
Solution:
i) – 7
ii) 7
iii) The point ’R’
iv) The point P
v) 4
vi) – 3

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 5.
State true or false and write correct statement.
i) In the Cartesian plane the horizon-tal line is called Y – axis, if) in the Cartesian plane, the vertical line is called Y – axis.
iii) The point which lies on both the axes is called origin.
iv) The point (2, – 3) lies in the third quadrant.
v) (-5, -8) lies in the fourth quadrant.
vi) The point (- x, – y) lies in the first quadrant where x < 0; y < 0.
Solution:
i) False
Correct statement: In the Cartesian plane the horizontal line is called X – axis.
ii) True
iii) True
iv) False
Correct statement: The point (2, -3) lies in the fourth quadrant.
v) False
Correct statement: (- 5, – 8) lies in the third quadrant.
vi) True

AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2

Question 6.
Plot the following ordered pairs on a graph sheet. What do you observe ?
i) (1, 0), (3, 0), (- 2, 0), (- 5, 0), (0, 0), (5, 0), (- 6, 0)
ii) (0, 1), (0, 3), (0, – 2), (0, – 5), (0, 0), (0,5), (0,-6)
Solution:
i) All points lie on X – axis,
ii) All points lie on Y – axis.
AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry Ex 5.2 2