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AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles InText Questions

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Question 1.
There are some statements given below. Write whether they are true or false. [Page No. 150]
Solution:
i) Two circles are always congruent. – False
ii) Two line segments of same length are always congruent. – True
iii) Two right angle triangles are some times congruent. – True
iv) Two equilateral triangles with their sides equal are always congruent. – True

Question 2.
Which minimum measurements do you require to check if the given figures are congruent? [Page No. 150]

i) Two rectangles.
Solution:
Length and breadth are required.

ii) Two rhombuses.
Solution:
One side and one interior angle are required.

Do These

Question 1.
State whether the following triangles are congruent or not. Give reasons for your answer. [Page No. 153]

Solution:
i) ΔABC ≅ ΔDEF
∵ ∠B = ∠E (∵ Angle sum property Z E = 180° – (70° + 60°) = 50°)
BC = EF
∠C = ∠F
∴ By SAS congruence
ΔABC ≅ ΔDEF

ii) In ΔMNL and ΔTSR
MN = ST
NL = TR
∠ M = ∠ T (∵ ΔMNL is flipped to get ΔTSR)
∴ ΔMNL ≅ ΔTSR

Question 2.
In the given figure, the point P bisects AB and DC. Prove that ΔAPC ≅ ΔBPD. [Page No. 153]

Solution:
Given that P bisects AB and DC.
Now in ΔAPC and ΔBPD
AP = BP (∵ P bisects AB)
CP = DP (∵ P bisects CD)
∠ APC = ∠ BPD
∴ ΔAPC ≅ ΔRPP (∵ SAS congruence)

Activity

1. On a tracing paper draw a line segment BC of length 6 cm.
2. From vertices B and C draw rays with angle 600 each. Name the point A where they meet
3. Fold the paper so that B and C fit precisely on top of each other. What do you observe? Is AB = AC?
[Page No. 160]

Do This

Question 1.
In the figure given below ΔABC and ΔDBC are two triangles such that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A C}}=\overline{\mathbf{C D}}\) . Show that ΔABC ≅ ΔDBC [Page No. 164]

Solution:
Given that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\)

In ΔABC and ΔDBC
AB = BD (∵ given)
AC = DC (∵ given)
BC = BC (common side)
∴ ΔABC ≅ ΔDBC
( ∵ by SSS congruence)

Activity

Question 1.
Construct a right angled triangle with hypotenuse 5 cm. and one side 3 cm. long. How many different triangles can be constructed? Compare your triangle with those of the other members of your class. Are the triangles congruent? Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangle if necessary what do you observe? You will find that two right triangles are congruent, if side and hypotenuse of one triangle are respectively equal to the corresponding side and hypotenous of other triangle.
[Page No. 165]

Activity

Question 1.
Draw a triangle ABC mark a point A’ on CA produced (new position of it)
So A’C > AC (Comparing the lengths) Join A to B and complete the triangle ABC.

What can you say about ∠A’BC and ∠ABC ?
Compare them. What do you observe?

Clearly, ∠A’BC > ∠ABC Continue to mark more points on CA (extended) and draw the triangles with the side BC and the points marked. You will observe that as the length of the side AC is increases (by taking different positions of A), the angle opposite to it, that is ∠B also increases.
[Page No. 169]

Question 2.
Construct a scalene triangle ABC (that is a triangle in which all sides are of different lengths). Measure the lengths of the sides.

Now, measure the angles. What do you observe?
In ΔABC figure, BC is the longest side and AC is the shortest side. ‘
Also, ∠A is the largest and ∠B is the smallest.
Measure angles and sides of each of the above triangles, what is the rela tion between a side and its opposite angle when compared with another pair?

Activity

Question
Draw a line segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it.
Solution:

Join each of these points with A as well as with B (see figure). Observe that as we move from P to T, ∠A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠TAB > ∠SAB > ∠RAB > ∠QAB > ∠PAB
and TB > SB > RB > QB > PB.
Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see figure).

Observe that the side opposite to the largest angle is the longest. In figure, ZB is the largest angle and AC is the longest side
Repeat this activity for some more triangles and we see that the converse of the above Theorem is also true.

Measure angles and sides of each triangle given below. What relation you can visualize for a side and its opposite angle in each triangle.

In this way, we arrive at the following theorem. [Page No. 170]

Do This

Question
Draw a triangle ABC and measure its sides. Find the sum of the sides AB + BC, BC + AC; and AC + AB, compare it with the length of the third side. What do you observe ?
[Page No. 171]
Solution:

AB + BC = 4 + 3 = 7
= 7 > 4 = AC
BC + CA > AB;
3 + 4 > 4
CA + AB > BC;
4 + 4 > 3

DE + EF > DF
EF + DF > DE
FD + DE > EF
∴ Sum of any two sides of a triangle is greater than the third side.

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

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables InText Questions

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Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the value of a, b, c in each case. [Page No. 128]
i) 3x + 2y = 9
Solution:
3x + 2y = 9
⇒ 3x + 2y – 9 = 0
Here a = 3, b = 2, c = -9

ii) – 2x + 3y = 6
Solution:
– 2x + 3y = 6
⇒ – 2x + 3y – 6 = 0
Here a = -2, b = 3, c = -6

iii) 9x – 5y = 10
Solution:
9x- 5y = 10
⇒ 9x- 5y – 10 = 0
Here a = 9, b = -5, c = -10

iv) \(\frac{x}{2}-\frac{y}{3}-5=0\)
⇒ \(\frac{3 x-2 y-30}{6}=0\)
⇒ 3x – 2y – 30 = 0
Here a = 3, b = – 2 and c = – 30

v) 2x = y
Solution:
2x = y
⇒ 2x – y = 0
Here a = 2, b = – 1 and c = 0

Try These

Take a graph paper, plot the point (2, 4), and draw a line passing through it. Now answer the following questions. [Page No. 135]

Question 1.
Can you draw another line that passes through the point (2, 4) ?
Solution:
Yes, We can draw another line passing through (2, 4).

Question 2.
How many such lines can be drawn ?
Solution:
Infinitely many lines can be drawn.

Question 3.
How many linear equations in two variables exist for which (2, 4) is a solution ? Solution:
Infinitely many linear equations in two variables exist.
Graph is given in next page.

Do This

Question 1.
i) Draw the graph of following equations. [Page No. 145]
a) x = 2 b) x = -2 c) x = 4 d) x = -4

ii) Are the graphs of all these equations parallel to Y-axis? .
Solution:
Yes; all these lines are parallel to Y-axis.

iii) Find the distance between the graph and the Y-axis in each case.
Solution:

Graph	DIstance	Right side/left side of the origin
a) x = 2	2 units	Right side
b) x = -2	2 units	Left side
c) x = 4	4 units	Right side
d) x = -4	4 units	Left side

Question 2.
i) Draw the graph of the following equations. [Page No. 145]
a) y = 2 b) y = -2 c) y = 3 d) y = – 3
Solution:

ii) Are all these parallel to the X-axis?
Solution:
Yes, these are all parallel to X-axis.

iii) Find the distance between the graph and the X-axis in each case.

Graph	DIstance	side of the origin above/below
a) y = 2	2 units	above origin
b) y = -2	2 units	below origin
c) y = 3	3 units	above origin
d) y = -3	3 units	below origin

Try These

Question
Find the equation of the Y – axis. [Page No. 146]
Solution:
All points on Y – axis have their x – co-ordinates as zero.
∴ Equation of Y- axis is x = O.

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

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry InText Questions

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Question
Describe the seating position of any five students in your classroom. [Page No. 108]
Solution:
Oral answer – depend on the classroom.

Activity (Ring Game)

Question
Have you seen ‘Ring game’ in exhibitions ? We throw rings on the objects arranged in rows and columns. Observe the following picture. [Page No. 108]
Complete the following table.
Solution:

Question
Is the object 3rd column and 4th row is same as 4th column 3rd row ?
Solution:
No

Do This

Question
Among the points given below, some of the points lie on X – axis. Identify them.
i) (0, 5) ii) (0, 0) iii) (3, 0) iv) (- 5, 0) v) (- 2, – 3) vi) (- 6,0) vii)(0,6) viii)(0, a) ix) (b, 0) [Page No. 114]
Solution:
The points (ii) (0, 0), (iii) (3, 0) (iv) (- 5, 0) (vi) (- 6, 0) and (ix) (b, 0) lie on X – axis.

Try These

Question 1.
Which axis the points such as (0, x), (0, y), (0, 2) and (0, – 5) lie on ? Why ? [Page No. 114]
Solution:
All the above points lie on Y – axis, since the X – coordinate of all these points is zero.

Question 2.
What is the general form of the points which lie on X – axis. [Page No. 114]
Solution:
The general form of points lying on X – axis is (x, 0).

Do This

Question
Plot the following points on a Cartesian plane.
1. B (- 2, 3) 2. L (5, – 8) 3. U (6, 4) 4. E (-3, – 3) [Page No. 120]
Solution:

Do This

Question
i) Write the co-ordinates of the points A, B, C, D and E. [Page No. 121]
ii) Write the co-ordinates of F, G, H, I and J.

Solution”
i) A(2, 9) ; B(5, 9); C(2, 6) ; D(5, 3) ; E(2, 3)
ii) F(- 6, – 2) ; G(- 4, – 5) ; H(- 3, – 7) ; I(- 9, – 7) ; J(- 8, – 5)

Activity

Question
Study the positions of different cities like Hyderabad, New Delhi, Chennai and Vishakapatnam with respect to longitudes and latitudes on a globe. [Page No. 123]
Solution:
Student Activity

Creative Activity

Question
Take a graph sheet and plot the following pairs of points on the axes and join them with line segments. [Page No. 123]
(1, 0) (0, 9) ; (2, 0) (0, 8) ; (3, 0) (0, 7) ; (4, 0) (0, 6) ;
(5, 0) (0, 5) ; (6, 0) (0, 4) ; (7, 0) (0, 3) ; (8, 0) (0, 2) ; (9, 0) (0, 1).
Try to complete the picture by using above points. What did you observe ?
Solution:
Student Activity

AP Board 9th Class Maths Solutions

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

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles InText Questions

Question
Observe your surroundings carefully and write any three situations of your daily life where you can observe lines and angles. [Page No. 71]
Solution:
The edges of a blackboard; the edges of a ruler/scale and the edges of a table gives the idea of lines and their corners give the idea of angles.

Question
Draw the pictures in your notebook and collect some pictures. [Page No. 71]
Solution:

Think, Discuss and Write

What is the difference between inter-secting lines and concurrent lines ? [Page No. 74]
Solution:
If the number of lines meeting at a point are only two then they are said to be intersecting lines whereas if the lines are three or more than three meeting at a point then they are said to be concurrent lines.

Do These

Question 1.
Write the complementary, supplementary and conjugate angles for the following angles, a) 45° b) 75° c) 215° d) 30° e) 60° f) 90° g) 180° [Page No. 76]
Solution:

Question 2.
Which pairs of the following angles become complementary or supplementary. [Page No. 76]

Solution:
The angles formed by the figures (i) and (ii) are complementary.
The angles formed by the figures (ii) and (iii) are supplementary.

Try This

Question 1.
Find the pairs of adjacent and non- adjacent angles in the given figures. [Page No. 77]

Solution:
In figure (i) ∠1 and ∠2 are pair of adjacent angles.
In figure (ii), no adjacent angles.
In figure (iii) (∠1, ∠2), (∠2, ∠3) are pairs of adjacent angles.
In figure (iv) ∠1 and Z2 are adjacent angles.

ii) List the adjacent angles in the given figure. [Page No. 77]

Solution:
In the given figure
(∠1, ∠2), (∠3, ∠4), (∠4,∠5) and (∠3, ∠5) are the pairs of adjacent angles.

Think Discuss and Write

Question
Linear pair of angles are always supplementary. But supplementary angles need not form a linear pair. Why ? [Page No. 77]
Solution:
A pair of supplementary angles need not necessarily a linear pair because they may exists in separate figures.

Activity

Measure the angles in the following figure and complete the table.
[Page No. 78]

Solution:

Measure the four angles 1,2,3,4 in each of the above figure and complete the table:
[Page No. 79]

Solution:

Do This

Question 1.
Classify the given angles as pairs of complementary, linear pair, vertically opposite and adjacent angles. [Page No. 80]

Solution:
The pairs of angles a and b in the fig.(i) are linear pair of angles.
The pairs of angles a and b in the fig.(ii) are adjacent angles.
The pairs of angles a and b in the fig.(iii) are complementary angles*
The pairs of angles a and b in the fig.(iv) are vertically opposite angles.

Question 2.
Find the measure of angle ‘a’ in each figure. Give reasons in each case. [Page No. 81]
i)

a = 180° – 50° = 130°
(linear pair of angles)

ii)

a = 43°
( ∵ vertically opposite angles)

iii)

a = 360° – (209° + 96°)
= 360° – 305° = 55°
(∵ complete angle = 360°)

iv)

a = 90° – 63° = 27°
(pair of complementary angles )

Do These

Question 1.
Find the measure of each angle indi-cated in each figure where / and m are parallel lines intersected by a transversal n. [Page No. 87]

Solution:
x = 110° (alternate exterior angles)

Solution:
y = 84° (alternate interior angles)

Solution:
z = 180° – 100° = 80° .
(interior angles on the same side of transversal)

Solution:
s° = 53° (pair of corresponding angles)

Question 2.
Solve for x and give reasons. [Page No. 88]

11x + 2 = 75°
11x = 75 – 2 = 73
∴ x = \(\frac{73}{11}\)
(∴ pair of corresponding angles are equal).

Solution:
8x – 4 = 60°
8x = 60 + 4 = 64
∴ x = \(\frac{64}{8}\) = 8
(∴ alternate interior angles are equal)

Solution:
(14x- 1)°.= (12x + 17)°
14x – 12x = 17 + 1
2x = 18
x = \(\frac{18}{2}\) = 9
(∴ alternate exterior angles are equal)

Solution:
13x- 5 = 17x + 5
13x- 17x = 5 + 5
– 4x = 10
x = \(\frac{10}{-4}=\frac{-5}{2}\)
(∴ Pair of, corresponding angles are equal).

Activity

Question 1.
Take a scale and a ‘set sqaure’. Arrange the set sqare on the scale as shown in figure. Along the slant edge of set sqare draw a line with the pencil. Now slide your set square along its horizontal edge and again draw a line. We observe that the lines are parallel. Why are they parallel ? Think and discuss with your friends. [Page No. 88]
Solution:

A. Student Activity (For Reference .)
All slant lines are parallel making an angle of 60° with the horizontal line. In this figure horizontal line is transversal to the slant line and corresponding angles are equal.

Do This

Question
Draw a line \(\overline{\mathbf{A D}}\) and mark points B and C on it. At B and C, construct ∠ABQ and ∠BCS equal to each other as shown. Produce QB and SC on the other side of AD to form two lines PQ and RS.

Draw common perpendiculars EF and GH for the two lines PQ and RS. Measure the lengths of EF and GH. What do you observe ? What can you conclude from that ? Recall that if the perpendicular distance between two lines is the same, then they are parallel lines. [Page No. 89]


Solution:
As ∠ABQ =∠BCS and they lie on the same line AD we can say that BQ // CS. Now EF and GH are the perpendicular distances between two parallel lines PQ and R, we say EF = GH.

Try This

Question i)
Find the measure of the question marked angle in the given figure. [Page No. 90]

Solution:
? = 70°
[ ∵ from the figure, these two angles are exterior angles on the same side of the transversal]

ii) Find the angles which are equal to ∠P.
Solution:
∠P = ∠Q = ∠R = 110°
(corresponding angles)

Activity

Question
Draw and cut out a large triangle as shown in the figure.
Number the angles and tear them off.
Place the three angles adjacent to each other to form one angle as shown below. [Page No. 97]

1. Identify angle formed by the three adjacent angles ? What is its mea-sure ?
2. Write about the sum of the measures of the angles of a triangle. Now let us prove this statement
using the axioms; and theorems related to parallel lines.
Solution:
Student Activity.

Think, Discuss and Write

Question
If the sides of a triangle are produced in order, what will be the sum of exterion angles formed ? [Page No. 99]
Solution:
Let ΔABC and the sides of the triangle is formed by exterior angles.

∠3 = ∠B + ∠C
∠1 + ∠2 + Z∠3 = 2[∠A + ∠B + ∠C]
= 2 x 180° = 360°
∴ Sum of the exterior angles are 360°.

AP Board 9th Class Maths Solutions

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

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry InText Questions

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Can you give any two axioms from your daily life. [Page No. 63]
Solution:
Example -1 : If we flip coin then there are only two chances that is head or tail.
Example – 2 : Things which are equal to the same thing are also equal to one another.
Example – 3 : The whole is greater than the part.

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

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation InText Questions

Think, Discuss and Write

Question
Which of the following expressions are polynomials ? Which are not ? Give reasons. [Page No. 28]
Solution:
i) 4x² + 5x – 2 is a polynomial.
ii) y² – 8 is a polynomial.
iii) 5 is a constant polynomial.
iv) \(2 x^{2}+\frac{3}{x}-5\) is not a polynomial as x is in denominator.
v) √3x² + 5y is a polynomial.
vi) \(\frac{1}{x+1}\) is not a polynomial as the variable x is in denominator.
vii) √x is not a polynomial as its exponent is not an integer.
viii) 3xyz is a polynomial.

Do These

Question
Write two polynomials with variable ‘x’. [Page No. 29]
Solution:
5x² + 2x – 8 and 3x² – 2x + 6.

Question
Write three polynomials with variable ‘y’.
Solution:
y³ – y² + y ; 2y² + 7y – 9 + 3y³; y⁴ – y + 6 + 2y².

Question
Is the polynomial 2x² + 3xy + 5y² in one variable ?
Solution:
No. It is in two variables x and y.

Question
Write the formulae of area and volume of different solid shapes. Find out the variables and constants in them. [Page No. 29]
Solution:

Question 1.
Write the degree of each of the following polynomials. [Page No. 30]
Solution:
i) 7x³ + 5x² + 2x – 6 – degree 3
ii) 7 – x + 3x² – degree 2
iii) 5p – √3 – degree 1
iv) 2 – degree 0
v) – 5 xy² – degree 3

Question 2.
Write the co-efficient of x² in each of the following. [Page No. 30]
Solution:
i) 15 – 3x + 2x² : co-efficient of x² is 2
ii) 1 -x² : co-efficient of x² is -1
iii) πx² – 3x + 5 : co-efficient of x² is π
iv) √2x² + 5x – 1 : co-efficient of x² is √2

Think, Discuss and Write

Question
How many terms a cubic (degree 3) polynomial with one variable can have? Give examples. [Page No. 31]
Solution:
A cubic polynomial can have atmost 4 terms.
E.g.: 5x³ + 3x² – 8x + 4; x³ – 8

Try These

Question 1.
Write a polynomial with 2 terms in variable x. [Page No. 31]
Solution:
2x + 3x²

Question 2.
How can you write a polynomial with 15 terms in variable ‘x’. [Page No. 31]
Solution:
a₁₄p¹⁴ + a₁₃p¹³ + a₁₂p¹²+ …………….+ a₁p + a₀

Do This

Question
Find the value of each of the follow ing polynomials for the indicated value of variables. [Page No. 33]
(i) p(x) = 4x² – 3x + 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
4(1)² – 3(1) + 7 = 8

ii) q(y) = 2y³ – 4y + √11 at y = 1.
Solution:
The value of q(y) at y = 1 is
2(1)³ – 4(1) + √11 = -2 + √11

iii) r(t) = 4t⁴ + 3t³ – t² + 6 at t = p, t ∈ R.
Solution:
The value of r(t) at t = p is
4p⁴ + 3p³ – p² + 6

iv) s(z) = z³ – 1 at z – 1.
Solution:
The value of s(z) at z = 1 is 1³ – 1 = 0

v) p(x) = 3x² + 5x – 7 at x = 1.
Solution:
The value of p(x) at x = 1 is
3(1)² + 5(1) — 7 = 1.

vi) q(z) = 5z³ – 4z + √2 at 7. = 2.
Solution:
The value of q(z) at z = 2 is
5(2)³ – 4(2) + √2 = 40 – 8 + √2
= 32 + √2

Try These

Question
Find zeroes of the following polyno¬mials. [Page No. 34]

1. 2x-3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)
∴ x = \(\frac{3}{2}\) is the zero of 2x – 3.

2. x² – 5x + 6
Solution:
x² – 5x + 6 = 0
⇒ x² – 3x – 2x + 6 = 0
⇒ x (x – 3) – 2 (x – 3) = 0
⇒ (x – 2) (x – 3) = 0
⇒ x – 2 = 0 or x – 3 = 0
⇒ x = 2 or x = 3
∴ x = 2 or 3 are the zeroes of x² – 5x + 6.

3. x + 5
Solution:
x + 5 = 0
x = – 5
∴ x = – 5

Do This

Fill in the bianks : [Page No. 35]

Linear polynomial	Zero of the polynomial
x + a	– a
x – a	a
ax + b	\( \frac{-b}{a} \)
ax – b	\( \frac{b}{a} \)

Solution:

Linear polynomial	Zero of the polynomial
x + a	– a
x – a	a
ax + b	\( \frac{-b}{a} \)
ax – b	\( \frac{b}{a} \)

Think, and Discuss

Question 1.
x² + 1 has no zeroes. Why ? [Page No. 36]
Solution:
x² + 1 = 0 ⇒ x² = -1
No real number exists such that whose root is – 1.
∴ x² + 1 has no zeroes.

Question 2.
Can you tell the number of zeroes of a polynomial of degree ‘n’ will have? [Page No. 36]
Solution:
A polynomial of degree n will have n- zeroes.

Do These

Question 1.
Divide 3y³ + 2y² + y by ‘y’ and write division fact. [Page No. 38]
Solution:
(3y³ + 2y² + y) ÷ y = \(\frac{3 y^{3}}{y}+\frac{2 y^{2}}{y}+\frac{y}{y}\)
= 3y² + 2y + 1
Division fact = (3y² + 2y + 1) y
= 3y³ + 2y² + y

Question 2.
Divide 4p² + 2p + 2 by ‘2p’ and write division fact.
Solution:
4p² + 2p ÷ 2 = \(\frac{4 p^{2}}{2 p}+\frac{2 p}{2 p}+\frac{2}{2 p}\)
= 2p + 1 + \(\frac{1}{\mathrm{P}}\)
Division fact:
(2p + 1 + \(\frac{1}{\mathrm{P}}\)).2p = 4p² + 2p + 2

Try These

Show that (x – 1) is a factor of xⁿ – 1. [Page No. 45]
Solution:
Let p(x) = xⁿ – 1
Then p(1) = 1ⁿ – 1 = 1 – 1 = 0
As p(1) = 0, (x – 1) is a factor of p(x).

Do These

Question
Factorise the following. [Page No. 46]

1. 6x² + 19x + 15
Solution:
6x² + 19x + 15 = 6x² + 10x + 9x + 15
= 2x (3x + 5) + 3 (3x + 5)
= (3x + 5) (2x + 3)

2. 10m² – 31m – 132
Solution:
10m² – 31m – 132
= 10m² – 55m + 24m – 132
= 5m (2m- 11) + 12 (2m- 11)
= (2m – 11) (5m + 12)

3. 12x² + 11x + 2
Solution:
12x² + 11x + 2
= 12x² + 8x + 3x + 2
= 4x (3x + 2) + 1 (3x + 2)
= (3x + 2) (4x + 1)

Try This

Question
Try to draw the geometrical figures for other identities. [Page No. 49]
i) (x + y)² ≡ x² + 2xy + y²

Step – 1 : Area of fig. I = x x = x²
Step – 2 : Area of fig. II = x y = xy
Step – 3 : Area of fig. III = x y = xy
Step – 4 : Area of fig. IV = y y = y²

Area of big square = sum of the areas of figures I, II, III and IV
∴ (x + y) (x + y) = x² + xy + xy + y²
(x + y)² = x² + 2xy + y²

ii) (x + y) (x – y) ≡ x² – y²

Step -1: Area of fig! I = x (x – y) = x² – xy
Step – 2: Area of fig. II = (x – y) y = xy – y²
Area of big rectangle = sum of areas of figures I & II
(x + y) (x – y) = x² – xy + xy – y²
= x² – y²
∴ (x + y) (x-y) = x2-y2

iii) (x + a) (x + b) ≡ x² + (a + b) x + ab

Step – 1 : Area of fig. I = x²
Step – 2 : Area of fig. II = ax
Step – 3 : Area of fig. Ill = bx
Step – 4 : Area of fig. IV = ab
∴ Area of big rectangle = Sum of areas of four small figures.
∴ (x + a) (x + b) = x² + ax + bx + ab
(x + a) (x + b) = x² + (a + b) x + ab

Do These

Question
Find the following product using appropriate identities. [Page No. 49]
i) (x + 5) (x + 5)
Solution:
(x + 5) (X + 5) = (x + 5)²
= x² + 2(x) (5) + 5²
= x² + 10x + 25

ii) (p – 3) (p + 3)
Solution:
(p – 3) (p + 3)
= p² – 3²
= p² – 9

iii) (y – 1) (y – 1)
Solution:
(y – 1) (y – 1)
= (y – 1)²
= y² – 2y + 1

iv) (t + 2) (t + 4)
Solution:
(t + 2) (t + 4)
= t² + t(2 + 4) + 2 x 4
= t² + 6t + 8

v) 102 x 98
Solution:
102 x 98 = (100 + 2) (100 -2)
= 100² – 2²
= 10000 – 4
= 9996

Do These

Question
Factorise the following using appro-priate identities. [Page No. 50]
i) 49a² + 70ab + 25b²
Solution:
49a² + 70ab + 25b²
= (7a)² + 2 (7a) (5b) + (5b)²
= (7a + 5b)²
= (7a + 5b)(7a + 5b)

ii) \(\frac{9}{16} x^{2}-\frac{y^{2}}{9}\)
Solution:

iii) t² – 2t + 1
= (t)² – 2(t) (1) + (1)²
= (t – 1)² = (t – 1) (t – 1)

iv) x² + 3x + 2
Solution:
x² + 3x + 2 = x2 + (2 + 1) x + (2 x 1)
(x + 2) (x + 1)

Do These

Question i)
Write (p + 2q + r)² in expanded form. [Page No. 52]
Solution:
(p + 2q + r)² = (p)² + (2q)² + (r)²
+ 2 (P) (2q) + 2 (2q) (r) + 2(r) (p)
= p² + 4q² + r² + 4pq + 4qr + 2rp

Question ii)
Expand (4x – 2y – 3z)² using identity. [Page No. 52]
Solution:
(4x – 2y – 3z)² = (4x)² + (- 2y)² + (- 3z)² + 2 (4x) (- 2y) + 2 (- 2y) (- 3z) + 2 (- 3z) (4x)
= 16x² + 4y² + 9z² – 16xy + 12yz – 24zx.

Question iii)
Factorise 4a² + b² + c² – 4ab + 2bc – 4ca
using identity. [Page No. 52]
Solution:
4a² + b² + c² – 4ab + 2bc – 4ca
= (2a)² + (- b)² + (- c)² + 2(2a) (- b) + 2 (- b) (- c) + 2(- c) (2a)
= (2a – b – c)² = (2a – b – c) (2a – b – c)

Try These

Question
How can you find (x – y)³ without actual multiplication ? Verify with actual multiplication. [Page No. 52]
Solution:
(x – y)³ = x³ – 3x²y + 3xy² – 3y³ from identity.
By actual multiplication
(x – y)³ = (x – y)² (x – y)
= (x² – 2xy + y²) (x – y)
= x³ – 2x²y + xy² – x²y + 2xy² – y³
= x³ – 3x² y + 3xy² – y³
Both are equal.

Do These

Question 1.
Expand (x + 1)³ using an identity. [Page No. 54]
Solution:
(x + 1)³ = (x)³ + (1)³ + 3 (x) (1) (x + 1)
= x³ + 1 + 3x (x + 1)
= x³ + 1 + 3x² + 3x = x³ + 3x² + 3x + 1

Question 2.
Compute (3m – 2n)³. [Page No. 54]
Solution:
(3m-2n)³
=(3m)³ – 3 (3m)² (2n) + 3 (3m) (2n)² – (2n)³
= 27m³ – 54m²n + 36mn² – 8n³

Question 3.
Factorise a³ – 3a²b + 3ab² – b³. [Page No. 54]
Solution:
a³ – 3a²b + 3ab² – b³
= (a)³ – 3 (a)² (b) + 3 (a) (b)² – (b)³
= (a – b)³
= (a – b) (a – b) (a – b)

Do These

Question 1.
Find the product (a – b – c) (a² + b² + c² – ab + be – ca) without actual multi-plication. [Page No. 55]
Solution:
The problem is incorrect.

Question 2.
Factorise 27a³ + b³ + 8c³ – 18abc using identity. [Page No. 55]
Solution:
27a³ + b³ + 8c³ – 18abc
= (3a)³ + (b)³ + (2c)³ – 3(3a) (b) (2c)
= (3a + b + 2c) (9a² + b² + 4c² – 3ab – 2be – 6ca)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers InText Questions

Do This

Question 1.
Represent \(\frac{-3}{4}\) on a number line. (Page No.3)
Solution:

Step – 1: Divide each unit into four equal parts to the right and left side of zero on the number line.
Step – 2 : Take 3 parts after zero on its left side.
Step -3 : It represents \(\frac{-3}{4}\)

Question 2.
Write 0, 7, 10, -4 in p/q form (Page No.2)
Solution:

Question 3.
Guess my number : Your friend chooses an integer between 0 and 100. You have to find out that number by asking questions, but your friend can only answer ‘Yes’ or ‘No’. What strategy would you use ? (Page No.3)
Solution:
Let my friend choosen 73; then my questions may be like this.

Q : Does it lie in the first 50 numbers ?
A: No .

Q : Does it lie between 50 and 60 ?
A: No

Q : Does it lie between 60 and 70 ?
A: No

Q : Does it lie between 70 and 80 ?
A: Yes
[then my guess would be “it is a number from 70 to 80]

Q : Is it an even number ?
A: No
[my guess : it should be one of the numbers 71, 73, 75, 77 and 79]

Q : Is it a prime number ?
A: Yes
[my guess : it may be 71, 73 or 79]

Q : Is it less than 75 ?
A: Yes
[my guess : it may be either 71 or 73]

Q : Is it less than 72 ?
A: No

Then the number is 73.
Therefore the strategy is
→ Use number properties such as even, odd, composite or prime to determine the number.

Do This

Question (i)
Find five rational numbers between 2 and 3 by mean method. (Page No.4)
Solution:
We know that \(\frac{a+b}{2}\) is a rational
between any two numbers a and b.
Let a = 2 and b = 3

Question (ii)
Find 10 rational numbers between \(\frac{-3}{11}\) and \(\frac{8}{11}\) (Page No.4)
Solution:
\(\frac{-3}{11}<\left[\frac{-2}{11}, \frac{-1}{11}, \frac{0}{11}, \frac{1}{11}, \frac{2}{11}, \frac{3}{11}, \frac{4}{11}, \frac{5}{11}, \frac{6}{11}, \frac{7}{11}\right]<\frac{8}{11}\)

Do This

Question
Find the decimal form of
(i) \(\frac{1}{17}\) (Page No.5)
Solution:

\(\frac{1}{19}\) = 0.052631

(ii) \(\frac{1}{19}\)(Page No.5)
Solution:

Try These

Question 1.
Find the decimal values of the following. (Page No.6)
Solution:
i) \(\frac{1}{2}\) = 0.5
ii) \(\frac{1}{2^{2}}=\frac{1}{4}\) = 0.25
iii) \(\frac{1}{5}\) = 0.2
iv) \(\frac{1}{5 \times 2}=\frac{1}{10}\) = 0.1
v) \(\frac{3}{10}\) = 0.3
vi) \(\frac{27}{25}\)

vii) \(\frac{1}{3}\)
Solution:

viii) \(\frac{7}{6}\)
Solution:

ix) \(\frac{5}{12}\)
Solution:

x) \(\frac{1}{7}\)
Solution:

Think, Discuss and Write

Question 1.
Kruthi said √2 can be written as \(\frac{\sqrt{2}}{1}\) which is in form. So √2 is a rational number. Do you agree with her argument ? (Page No.10)
Solution:
No.
Writing √2 = \(\frac{\sqrt{2}}{1}\) is not in the \(\frac { p }{ q }\) form.
Since p and q are integers and √2 is not an integer.

Try These

Question 1.
Find the value of √3 upto six decimals.(Page No.10)
Solution:

Step 1 :
Write 3 as 3.00 00 00 00 00 00 00
Step – 2 :
Group the zeros in pairs (i.e.) make periods.
Step – 3 :
Find the square root using long division method.
∴ √3 = 1.732050

Try These

Question 1.
Locate √5 and -√5 on the number line [Hint 5 = 2² + 1²](Page No.12)
Solution:
Step – 1 :
At zero draw a rectangle of length 2 units and breadth 1 unit.

Step – 2 :

Step – 3 :
Using compass draw an arc of radius OB with centre ‘O’ which cuts the num-ber line at D and D¹.

Step – 4 :
D represents √5 and D¹ represents – √5 on the number line.

Activity

Question
Constructing the ‘Square root spiral’. (Page No. 15)
Solution:
Take a large sheet of paper and construct the ‘Square root spiral’ in the following manner.
Step – 1 : Start with point ‘O’ and draw a line segment \(\overline{\mathrm{OP}}\) of 1 unit length.
Step – 2 : Draw a line segment \(\overline{\mathrm{PQ}}\) perpendicular to \(\overline{\mathrm{OP}}\) of unit length (where OP = PQ = 1) (see Fig.)
Step – 3 : Join 0, Q. (OQ = √2 )
Step – 4 : Draw a line segment OR of unit length perpendicular to \(\overline{\mathrm{OQ}}\)
Step – 5 : Join O, R. (OR = √3 )
Step – 6 : Draw a line segment RS of unit length perpendicular to \(\overline{\mathrm{OR}}\).
Step – 7 : Continue in this manner for some more number of steps, you will create a
beautiful spiral made of line segment \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}, \overline{\mathrm{RS}}, \overline{\mathrm{ST}}, \overline{\mathrm{TU}}\) …………etc. Note that the line segments \(\overline{\mathrm{OQ}}, \overline{\mathrm{OR}}, \overrightarrow{\mathrm{OS}}, \overline{\mathrm{OT}}, \overline{\mathrm{OU}}\) …… etc., denote the lengths √2,√3, √4, √5, √6 respectively.

Do This

Question 1.
Find rationalising factors of the denominators of (Page No. 20)
i) \(\frac{1}{2 \sqrt{3}}\)
Solution:
\(\frac{1}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{2 \times 3}=\frac{\sqrt{3}}{6}\)
∴ The R.F is √3

ii) \(\frac{1}{\sqrt{5}}\)
Solution:
\(\frac{3}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}\)
∴ The R.F is √5

iii) \(\frac{1}{\sqrt{8}}\)
Solution:

Do This;

Question
Simplify : (Page No. 23)

i) (16) ^1/2
Solution:
(16)^1/2 = (4 x 4)^1/2 = (42)^1/2 = 4^2/2 = 4

ii) (128)^1/7
Solution:
(128)^1/7 =(2 X 2 X 2 X 2 X 2 X 2 X 2)^1/7 = (2⁷)^1/7 = 2

iii) (343)^1/5
Solution:
(343)^1/5= (3 x 3 x 3 x 3 x 3)^1/5 = (3⁵)^1/5= 3

Question 1.
Write the following surds in exponen-tial form (Page No. 24)
i) √2
Solution:
√2 = 2^1/2

ii) 3√9
Solution:
\(\sqrt[3]{9}=\sqrt[3]{3 \times 3}=\sqrt[3]{3^{2}}=3^{\frac{2}{3}}\)

iii) \(\sqrt[5]{20}\)
Solution:
\(\begin{aligned}
\sqrt[5]{20}=\sqrt[5]{2 \times 2 \times 5} &=\sqrt[5]{2^{2} \times 5} \\
&=2^{\frac{2}{5}} \times 5^{\frac{1}{5}}
\end{aligned}\)

iv) 17√19
Solution:
\(\sqrt[17]{19}=19^{\frac{1}{17}}\)

Question 2.
Write the surds in radical form. (Page No. 24)
i) 5^1/7 = \(\sqrt[7]{5}\)
ii) 17^1/6 = \(\sqrt[6]{17}\)
iii) 5^2/5 = \(\sqrt[5]{5^{2}}=\sqrt[5]{5 \times 5}=\sqrt[5]{25}\)
iv) 142^1/2 = \(\sqrt{142}\)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.4

Question 1.
State which of the following are mathematical statements and which are not ? Give reason.
i) She has blue eyes.
Solution:
This is not a mathematical statement. no mathematics is involved in it.

ii) x + 7 = 18
Solution:
This is not a statement, as its truthness cant be determined.

iii) Today is not Sunday.
Solution:
This is not a statement. This is an am biguous open sentence.

iv) For each counting number x, x + 0 = x.
Solution:
This is a mathematical statement.

v) What tune is it?
Solution:
This is not a riathematical statement.

Question 2.
Find counter examples to disprove the following statements.
i) Every rectangle is a square.
Solution:

A rectangle and square are equiangular i.e., all the four angles are right angles. This doesn’t mean that they have equal sides.

ii) For any integers x and y,
\(\sqrt{x^{2}+y^{2}}\) = x + y
Solution:
Let x = 3; y = 8
\(\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+8^{2}}\)
= \(\sqrt{9+64}=\sqrt{73}\)

x + y = 3 + 8 = 11
Here, √73 ≠ 11
i.e., \(\sqrt{x^{2}+y^{2}}\) ≠ x + y

iii) If n is a whole number then 2n² +11 is a prime.
Solution:
If n = 11 then 2n²+ 11 = 2 (11)² + 11
= 11 (2 × 11 + 1) = 11 × (22 + 1)
= 11 × 23 is not a prime.

iv) Two triangles are congruent if all their corresponding angles are equal.
Solution:

If the corresponding angles are equal then the triangles are only similar.

v) A quadrilateral with all sides are equal is a square.
Solution:
A rhombus is not a square, but all its sides are equal.

Question 3.
Prove that the sum of two odd numbers is even.
Solution:

Steps	Reasons
1) (2m + 1); (2n + 1) be the two odd numbers	General form of an odd number.
2) (2m + 1) + (2n + 1) = (2m + 2n + 2) = 2 (m + n + 1) = 2K Hence proved.	Adding the two numbers General form of an even number.

Question 4.
Prove that the product of two even numbers is an even number.
Solution:

Steps	Reasons
1) Let 2m and 2n be two even numbers.	General form of an even number.
2) 2m.2n = 4mn = 2(2mn) = 2K	Taking the product Rearranging the numbers.
3) 2K where K = 2mn	K=2mn
4) Even number Hence proved.	General form of an even number.

Question 5.
Prove that if x is odd, then x² is also odd.
Solution:
Let x be an odd number.
Then x = 2m + 1
(general form of ah odd number) Squaring on both sides,
x² = (2m + 1)²
= 4m² + 4m +1
= 2 (2m² + 2m) + 1
= 2K + 1 where K = 2m²+ 2
Hence x² is also odd.

Question 6.
Examine why they work ?
Choose a number. Double it. Add nine. Add your original number. Divide by three. Add four. Subtract your original number. Your result is seven.
Solution:
Choose a number = x say
Double it = 2x
Add nine = 2x + 9
Add your original number
= 2x + 9 + x = 3x + 9
Divide by 3 = (3x + 9) ÷ 3
= \(\frac{3 x}{3}+\frac{9}{3}\) = x + 3
Add 4 ⇒ x + 3 + 4 = x + 7
Subtract your original number =
x + 7 – x = 7
Your result is 7 – True.

ii) Write down any three digit number (for example, 425). Make a six digit number by repeating these digits in the same order 425425. Your new number is divisible by 7, 11 and 13.
Solution:
Let a three digit number be xyz.
Repeat the digit = xyzxyz
= xyz × (1001)
= xyz × (7 × 11 × 13)
Hence the given conjecture is true.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.3

Question 1.
Take any three consecutive odd numbers and find their product, for example 1 × 3 × 5 = 15;
3 × 5 × 7 = 105: 5 × 7 × 9 = ……………
ii) Take any three consecutive even numbers and add them, say,
2 + 4 + 6 = 12; 4 + 6 + 8 = 18:
6 + 8 + 10 = 24; 8 + 10 + 12 = 30 ….
so on. Is there any pattern you can guess in these sums ? What can von conjecture about them ?
Solution:
i) 1 × 3 × 5 = 15
3 × 5× 7 = 105
5 × 7 × 9 = 315
7 × 9 × 11 = 693

1. The product of any three consecutive odd numbers is odd.
2. The product of any three consecutive odd numbers is divisible by ’3′.
3. 2 + 4 + 6 = 12; 4 + 6 + 8= 18;
   6 + 8 + 10 = 24; 8 + 10 + 12 = 30
4. The sum of any three consecutive even numbers is even.
5. The sum of any three consecutive even numbers is divisible by 6
6. The sum of any three consecutive even numbers is a multiple of 6.

Question 2.
Go back to Pascal’s triangle.
Line-1: 1=11°
Line-2: 11 = 11¹
Line-3 : 121 = 11²
Make a conjecture about line – 4 and line – 5.

Does your conjecture hold ? Does your conjecture hold for line – 6 too ?
Solution:
Line-4 : 1331 = 11³
Line-5 : 14641 = 11⁴
Line – 6 : 11⁵
∴ Line – n = 11^n-1
Yes the conjecture holds good for line – 6 too.

Question 3.
Look at the following pattern.
i) 28 = 2² × 7¹;
Total number of factors
(2+ 1)(1 + 1) = 3 × 2 = 6
28 is divisible by 6 factors i.e.,
1, 2, 4, 7, 14, 28.
ii) 30 = 2¹ × 3¹ × 5¹, Total number of .
factors (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 = 8
30 Is divisible by 8 factors i.e., 1, 2, 3, 5, 6, 10, 15 and 30
Find the pattern.
[Hint : Product of every prime base exponent +1)
Solution:
24 = 2³ × 3¹
24 has (3+1) (1 + 1) = 4 × 2 = 8 factors
[1, 2, 3, 4, 6, 8, 12 and 24]
36 = 2² × 3²
Number of factors = (2 + 1) (2 + 1)
3 × 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36]
If N = a^p. b^q . c^r…….. where N is a natural number.
a. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1)(q+ 1)(r + 1)

Question 4.
Look at the following pattern :
1² = 1
11² = 121
111² = 12321;
1111² = 1234321
11111² = 123454321
Make a conjecture about each of the following
111111² =
1111111² =
Check if your conjecture is true.
Solution:
111111² = 12345654321
1111111² = 1234567654321
(111………. n-times)² = (123 … (n- 1) n (n – 1) (n – 2) 1)
The conjecture is true.

Question 5.
List five axioms (postulates) used in text book.
Solution:

1. Things which are equal to the same things are equal to one another.
2. If equals are added to equals, the sums are equal.
3. If equals are subtracted from equals, the differences are equal.
4. When a pair of parallel lines are in-tersected by a transversal, the pairs of corresponding angles are equal.
5. There passes infinitely many lines through a given point.

Question 6.
In a polynomial p(x) = x² + x + 41 put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is ‘x’ an element of N ? Put x = 41 in p(x). Now what do you find ?
Solution:
p(x) = x² + x + 41
p(0) = 0² + 0 + 41 = 41 – is a prime
p(1) = 1² + 1 + 41 = 43 – is a prime
p(2) = 2² + 2 + 41 = 47 – is a prime
p(3) = 3² + 3 + 41 = 53 – is a prime
p(41) = 41² + 41 + 41
= 41(41 + 1 + 1)
= 41 x 43 is not a prime.
∴ p(x) = x²+ x + 41 is not a prime for all x.
∴ The conjecture “p(x) = x² + x + 41 is a prime” is false.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.2

Question 1.
Use deductive reasoning to answer the following.

i) Human beings are mortal. Jeevan is a human being. Based on these two statements, what can’ you conclude about Jeevan?
Solution:
From the above statements we can deduce that Jeevan is mortal as it is given that all humans are mortal and Jeevan is a human.

ii) All Telugu people are Indians. X is an Indian. Can you conclude that X belongs to Telugu people?
Solution:
No. X may belong to any other language like Tamil, Kannada, Malayali…. etc.

iii) Martians have red tongues. Gulag is a Martian. Based on these two statements, what can you conclude about Gulag?
Solution:
Gulag had red, tongue.

iv) What is the fallacy in the Raju’s reasoning in the cartoon below

Solution:
All smarts need not be a President. There could be some other persons who are smart too.

Question 2.
Once again you are given four cards. Each card has a number printed on one side and a letter on the other side. Which are the only two cards you need to turn .over to check whether the following rule holds ?
“If a card has a consonant on one side, then it has an odd number on the other side”.

Solution:
You need to turn over B and 8 only. If B has an even number then the rule has broken. Similarly if 8 has a consonant on the other side then also the rule has been broken.

Question 3.
Think of this puzzle. What do you need to find a chosen number from this square ? Four of the clues below are true but do nothing to help in finding the number. Four of the clues are necessary for finding it.
Here are eight clues to use:
a) The number is greater than 9.
b) The number is not a multiple of 10.
c) The number is a multiple of 7.
d) The number is odd.
e) The number is not a multiple of 11.
f) The number is less than 200.
g) Its ones digit is larger than its tens digit.
h) Its tens digit is odd.
What is the number ?

Solution:

Hint	Conclusion
a	The number may be from 10 to 99 (2nd clue implies, first clue is of no use) No use
b	The number is not any of (10, 20, 30, …………90)
c	The number may be any of (7, 14, 21,28, 35,42, …98)
d	The number may be any of (7,21,35,49, 63, 77,91)
e	The number may be any of (7, 21, 35, 49. 63. 91)
f	No use
g	The numbr may be 35, 49
h	35

 

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.1

Question 1.
State whether the following sentences are always true, always false or ambiguous. Justify your answer.
i) There are 27 days in a month.
Solution:
Always false. Generally 30 days or 31 days make a month except February.

ii) Makarasankranthi fells on a Friday.
Solution:
Ambiguous. Makarasankranthi may fall on any day of the week.

iii) The temperature in Hyderabad is 2°C.
Solution:
Ambiguous. Sometimes the temperature may go down to 2°C in winter.

iv) The earth is the only planet where life exist.
Solution:
We can’t say always true. To the known fact, so far we can say this.

v) Dogs can fly.
Solution:
Always false,’as dogs can never fly.

vi) February has only 28 days.
Solution:
Ambiguous. A leap year has 29 days for February.

Question 2.
State whether the following statements are true or false. Give reasons for your answers.

i) The sum of the interior angles of a quadrilateral is 350°.
Solution:
False. Sum of the interior angles of a quadrilateral is 360°.

ii) For any real number x, x² > 0
Solution:
True. This is true for all real numbers,

iii) A rhombus is a parallelogram.
Solution:
True. In a rhombus, both pairs of opposite sides are parallel and hence every rhombus is a parallelogram.

iv) The spm of two even numbers is even.
Solution:
True. This is true for any two even numbers.

v) Square numbers can be written as the sum of two odd numbers.
Solution:
Ambiguous. Since square of an odd number can’t be written as sum of two odd numbers.

Question 3.
Restate the following statements with appropriate conditions, so that they become true statements.

i) All numbers can be represented in prime factorization.
Solution:
Any natural number greater than 1 can be represented in prime factorization.

ii) Two times a real number is always even.
Solution:
Two times a natural number is always even.

iii) For any x, 3x + 1 > 4.
Solution:
For any x > 1; 3x + 1 > 4.

iv) For any x, x³ ≥ 0.
Solution:
For any x > 0; x³ ≥ 0.

v) In every triangle, a median is also an angle bisector.
Solution:
In an equilateral triangle, a median is also an angle bisector.

Question 4.
Disprove, by finding a suitable counter example, the statement
x² > y² for all x > y.
Solution:
If x = – 8 and y = – 10
Here x > y
x² = (- 8)² = 64 and y² = (- 10)² = 100
But x² > y² is false here. [ ∵ 64 < 100]
(This can be proved for any set of nega-tive numbers or a negative number and a positive number)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 14 Probability Ex 14.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 14th Lesson Probability Exercise 14.1

Question 1.
A die has six faces numbered from 1 to 6. It is rolled and number on the top face is noted. When this is treated as a random trial.
(Random trial: All possible outcomes are known before hand and the exact outcome can’t be predicted, then only the experiment is treated as a random experiment or a trial.)

a) What are the possible outcomes ?
Solution:
The possible outcomes are 1, 2, 3, 4, 5 and 6.

b) Are they equally likely ? Why ?
Solution:
Yes. All the outcomes are equally likely since every event has equal chance of occurrence or no event has priority to occur.

c) Find the probability of a composite number turning up on the top face.
Solution:
Even : Turning up of a composite number
Possible outcomes = 4, 6
No. of possible outcomes = 2
Total outcomes = 1, 2, 3, 4, 5 and 6
Number of total outcomes = 6
Probability = \( \frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }} \\\)
= \(\frac{2}{6}=\frac{1}{3}\)

Question 2.
A coin is tossed 100 times and the following outcomes are recorded. Head : 45 times; Tails : 55 times from the experiment.
a) Compute the probability of each outcome.
Solution:
Head = 45 times; Tails = 55 times
Total = 100
P(H),
Probability of getting Head = \(\frac{45}{100}\)
P(T),
Probability of getting Tail = \(\frac{55}{100}\)
\(\left[P=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\right]\)

b) Find the sum of the probabilities of all outcomes.
Solution:
P(H) + P(T) = \(\frac{45}{100}+\frac{55}{100}=\frac{100}{100}\) = 1

Question 3.
A spinner has four colours as shown in the figure. When we spin it once, find

a) At which colour, is the pointer more likely to stop ?
Solution:
Red = 5 sectors; Blue = 3 sectors;
Green = 3 sectors; Yellow = 1 sector
Total = 5 + 3 + 3+1 = 12 sectors
∴ Pointer is more likely to stop at Red.

b) At which colour, is the pointer less likely to stop ?
Solution:
Yellow; as only one sector is shaded in Yellow.

c) At which colours, is the pointer equally likely to stop ?
Solution:
Blue and Green have equal chances; as they are shaded in equal number of sectors.

d) What is the chance the pointer will stop on white ?
Solution:
No chance. Since no sector is shaded in white.

e) Is there any colour at which the pointer certainly stops ?
Solution:
No; as the experiment is a random ex-periment.

Question 4.
A bag contains five green marbles, three blue marbles, two red marbles and two yellow marbles. One marble is drawn out randomly. M)
a) Are the four different colour outcomes equally likely ? Explain.
Solution:
No. As they are not in equal number, they have different chances of occurrence.

b) Find the probability of drawing each colour marble, i.e., P (green), P (blue), P(red) and P (yellow).
Solution:

c) Find the sum of their probabilities.
Solution:
P(Green) + P(Blue) + P(Red) +P(Yellow)
= \(\frac{5}{12}+\frac{3}{12}+\frac{2}{12}+\frac{2}{12}\)
= \(\frac{5+3+2+2}{12}=\frac{12}{12}\) = 1

Question 5.
A letter is chosen from English alphabet. Find the probability of the letters being
a) A vowel
b) A letter comes after P
c) A vowel or a consonant
d) Not a vowel
Solution:
Total letters ;= 26 [A, B, C ….. Z]
Probability = \(\frac{\text { No. of favourables }}{\text { Total no. of outcomes }}\)
a) Vowels = a, e, i, o, u [5]
∴ (vowels) = \(\frac{5}{26}\)

b) Letter after P = 10
[Q, R, S, T, U, V, W, X, Y, Z]
Probability of a letter that comes after
’P’ = \(\frac{10}{26}\) = \(\frac{5}{13}\)

c) A vowel or a consonant
Vowel or consonants = 26
[all letters from A to Z]
P(vowel or consonant) = \(\frac{26}{26}\) = 1

d) Not a vowel = 21
[other than A, E, I, O, U]
∴ P(not a vowel) = \(\frac{21}{26}\)

Question 6.
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98,5.04,5.07,5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution:
Total number of bags = 11
No. of bags with a weight more than 5 kg = 7
[5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07]
∴ Probability = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}\)
P(E) = \(\frac{7}{11}\)

Question 7.
An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained is given in the following table.

Find the probabilities of the following events for a driver chosen at random from the city:

i) The driver being in the age group 18-29 years and having exactly 3 accidents in one year.
Solution:
Total number of accidents = (440 + 160 + 110 + 61 + 35 + 505 + 125 + 60 + 22 + 18 +
360 + 45 + 35 + 15 + 9) = 2000
Event: The driver being in the age group (18 – 29) years and having exactly 3 accidents
= 61
P(E) = \(\frac{\text { No. of favourables }}{\text { Total outcomes }}=\frac{61}{2000}\)

ii) The driver being in the age group of 30 – 50 years and having one or more accidents in a year.
Solution:
Favourable outcomes = 125 + 60 + 22 + 18 = 225
Total outcomes = 2000
P(E) = \(\frac{225}{2000}=\frac{9}{80}\)

iii) Having no accidents in the year.
Solution:
Favourable outcomes = 440 + 505 + 360 = 1305
Total outcomes = 2000
P(E) = \(\frac{1305}{2000}=\frac{261}{400}\)

Question 8.
What is the probability that a randomly thrown dart hits the square board in shaded region ( Take π = \(\frac{22}{7}\) and express in percentage )
Solution:

Radius of the circle r = 2 cm
Area of the circle A = πr² = \(\frac{22}{7}\) × 2 × 2 = \(\frac{88}{7}\) cm²
Side of the square = 2 × radius
= 2 × 2 = 4cm
Area of the square = S² = 4 × 4 = 16 cm²
Area of the shaded region = Area of the square – Area of the circle