Category: Class 9

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas InText Questions

Activity

Question

Observe Figure I and II. Find the area of both. Are the areas equal?
Trace these figures on a sheet of paper, cut them. Cover fig. 1 with fig. II. Do they cover each other completely? Are they congruent?
Observe fig. Ill and IV. Find the areas of both. What do you notice?
Are they congruent?
Now, trace these figures on sheet of paper. Cut them let us cover fig. Ill by fig. IV by coinciding their bases (length of same side).
As shown in figure V are they covered completely?
We conclude that Figures I and II are congruent and equal in area. But figures III and IV are equal in area but they are not congruent

Think, Discuss and Write

Question 1.
If 1 cm represents 5 in, what would be an area of 6 cm² represents ?
[Page No. 247]
Solution:
1 cm² = 5 m
1 cm² = 1 cm × 1 cm = 5m × 5m = 25m²
∴ 6 cm² = 6 × 25 m² = 150 m²

Question 2.
Rajni says 1 sq. m = 100² sq. cm. Do you agree ? Explain.
Solution:
No
1 sq. m = 100 cm²

Think, Discuss and Write

Question
Which of the following figures lie on the same base and between the same parallels? In such cases, write the common base and two parallels. [Page No. 249]

Solution:
a) In figure (a) ΔPCD and □ ABCD lie on same base CD and between the same parallels AB//CD.
b) No,
c) ΔTRQ and □ PQRS lie on the same base QR and between the same par-allels PS//QR.
d) ΔAPD and □ ABCD lie on the same base AD and between the same par-allels AD//BC.
e) No.

Activity

Question
Take a graph sheet and draw two par-allelograms ABCD and PQCD on it as show in the Figure, [Page No. 250]

The parallelograms are on the same base DC and between the same parallels PB and DC. Clearly the part DCQA is common between the two parallelo-grams. So if we can show that ΔDAP and ΔCBQ have the same area then we can say ar(PQCD) = ar(ABCD)

Activity

Draw pairs of triangles one the same base or ( equal bases) and between the same parallels on the graph sheet as shown in the Figure.
Let AABC and ADBC be the two triangles lying on the same base BC and between parallels BC and FE.
Draw CE II AB and BF II CD. Parallelograms AECB and FDCB are on the same base BC and are between the same parallels BC and EF.

Thus ar (AECB) = ar (FDCB).
We can see ar (ΔABC) = \(\frac { 1 }{ 2 }\) ar (parallelogram AECB) …………….(i)
and ar (ΔDBC) = \(\frac { 1 }{ 2 }\) ar (parallelogram FDCB) ……………..(ii)
From (i) and (ii), we get ar (ΔABC) = ar (ΔDBC)
You can also find the areas of ΔABC and ΔDBC by the method of counting the squares in graph sheet as we have done in the earlier activity and check the areas are whether same.[Page No. 254]

Think, Discus and Write

Draw two triangles ABC and DBC on the same base and between the same parallels as shown in the figure with P as the point of intersection of AC and BD. Draw CE//BA and BF//CD such that E and F lie on line AD.
Can you show ar(ΔPAB) = ar(ΔPDQ) ?[Page No. 254]

[Hint: These triangles are not congruent but have equal areas.
Solution:
□ ABCE = 2 × ΔABC
[∵ ΔABC; □ABCE lie on the same base BC and between the same parallels BC // AE]
ΔABC = \(\frac { 1 }{ 2 }\) × □ ABCE ……………(1)
Also □ BCDF = 2 × ΔBCD…………..
[∵ΔBCD and □ BCDE lie on the same base BC and between , the same parallels BC//DE]
ΔBCD = \(\frac { 1 }{ 2 }\) × □ BCDF ……………… (2)
But □ABCE = □BCDF
[ ∵ □ABCE and □BCDF lie on the same base BC and between the same parallels BC//FE]
From (1) & (2); ΔABC = ΔBCD
ΔPAB + ΔPBC = ΔPBC + ΔPDC
⇒ ΔPAB = ΔPDC
Hence proved.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 10 Surface Areas and Volumes InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 10th Lesson Surface Areas and Volumes InText Questions

Try This

Take a cube of edge T cm and cut it as we did in the previous activity and find total surface area and lateral surface area of cube. [Page No. 216]
Solution:

If we cut and open a cube of edgewe obtain a figure as shown above.
In the figure, A, B, C, D, E, F are squares of side
The faces A, C, D, F forms the lateral surfaces of the cube.
∴ Lateral surface area of the cube = 4l²
And all six faces form the cube.
∴ Total surface area of the cube = 6l²

Do This

Question 1.
Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]1. Find the total surface area and lateral surface area of the cube with side 4 cm. By using the formulae deduced in above. Try this. [Page No. 216]

Solution:
Total surface area of a cube = 6l²
Given l = 4crn
∴ T.S.A. = 6 × 4² = 6 × 16 = 96 cm²
∴ L.S.A. = 4l² = 4 × 4² = 64cm²

Question 2.
Each edge of a cube is increased by 50%. Find the percentage increase in the surface area. [Page No. 216]
Solution:
Let the edge of a cube x units
Its surface area = 6l² = 6x² sq. units
If its edge is increased by 50%, then
new edge = x + 50% of x


Where x = increase/decrease

Try These

Question
a) Find the volume of a cube whose edge is ‘a’ units. [Page No. 217]

Solution:
V = edge³ = a³ cubic units.

b) Find the edge of a cube whose volume is 1000 cm³. [Page No. 217]
Solution:
V = edge³ = 1000 = 10 × 10 × 10 = 10³
∴ Edge =10 cm

Do These

Question 1.
Find the volume of a cuboid if l = 12 cm, b = 10 cm, h = 8 cm. [Page No. 218]
Solution:
Volume V = lbh = 12 × 10 × 8 = 960cm³

Question 2.
Find the volume of cube, if its edge is 10 cm [Page No. 218]
Solution:
Volume V = l³ = 10 × 10 × 10 = 1000cm³

Question 3.
Find the volume of isosceles right angled triangular prism. [Page No. 218]

Solution:
Volume = Area of base × height
= Area of isosceles triangle × height

Activity

Question
Take the square pyramid and cube containers of same base and with equal heights. [Page No. 218]

Fill the pyramid with a liquid and pour into the cube (prism) completely. How many times it takes to fill the cube? From this, what inference can you make?

Thus volume of pyramid = \(\frac { 1 }{ 3 }\) × of the volume of right prism
= \(\frac { 1 }{ 3 }\) × Area of the base × height.

Note : A right prism has bases perpendicu¬lar to the lateral edges and all lateral faces are rectangles.

Do These

Question 1.
Find the volume of a pyramid whose square base is 10 cm and height is 8 cm. (Page No. 219)
Solution:
Volume of a pyramid
= \(\frac { 1 }{ 3 }\) × Area of the base × height
= \(\frac { 1 }{ 3 }\) × 10 x 10 × 8 = \(\frac { 800 }{ 3 }\) cm³

Question 2.
The volume of a cube is 1200 cubic cm. Find the volume of square pyra¬mid of the same height. (Page No. 219)
Solution:
Volume of the square pyramid
= \(\frac { 1 }{ 3 }\) × volume of the square prism
= \(\frac { 1 }{ 3 }\) × 1200 = 400 cm³

Activity

Cut out a rectangular sheet of paper. Paste a thick string along the line as shown in the figure. Hold the string with your hands on either sides of the rectangle and rotate the rectangle sheet about the string as fast as you can.

Do you recognize the shape that the rotating rectangle is forming ?
Does it remind you the shape of a cylinder ? [Page No. 220]

Do This

Question 1.
Find C.S.A. of each of the following cylinders. [Page No. 221]
i) r = x cm; h = y cm
Solution:
CSA = 2πrh = 2πxy cm²

ii) d = 7 cm; h = 10 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 10
= 220 cm²

iii) r = 3 cm; h = 14 cm
Solution:

CSA = 2πrh
= 2 × \(\frac{22}{7} \times \frac{7}{2}\) × 3 × 14
= 264 cm²

Question 2.
Find the total surface area of each of the following cylinder. [Page No. 222]
i)

Solution:
r = 7 cm; h = 10 cm
T.S.A. = 2πr (r + h)
=2 × \(\frac{22}{7}\) × 10 (7 + 10)
= 2 × \(\frac{22}{7}\) × 10 × 17
= 1068.5 cm²

ii)

h = 7cm; πr² = 250
πr² = 250 = 250
\(\frac{22}{7}\) × r² = 250 = 250
r² = 125 x \(\frac{7}{11}\)

Try These

Question 1.
If the radius of a cylinder is doubled keeping Its lateral surface area the same, then what is its height? [Page No. 225]
Solution:
Let the initial radius and height of the
cylinder be r and h.
Then L.S.A. = 2πrh
When r is doubled and the L.S.A. remains the same, then the height be hr By problem new L.S.A. = 2πrh
= 2π (2r) (h₁)
⇒ 2πrh = 4πrh₁
∴ \(\frac{2 \pi \mathrm{rh}}{4 \pi \mathrm{r}}=\frac{1}{2} \mathrm{~h}\)
Height becomes its half.

Question 2.
A hot water system (Geyser) consists of a cylindrical pipe of length 14 m and diameter 5 cm. Find the total radiating surface of hot water system. [Page No. 225]
Solution:
Radius (r) = \(\frac{\text { diameter }}{2}=\frac{5}{2}\) = 2.5 cm
Length of the pipe = height = 14 m Radiating surface = 2πrh
= 2 × \(\frac{22}{7}\) × 2.5 × 1400
= 22000 cm³

Activity

Question
Making a cone from a sector. [Page No. 227] Follow the instructions and do as shown in the figure.

i) Draw a circle on a thick paper Fig (a).
ii) Cut a sector AOB from it Fig (b).
iii) Fold the ends A, B nearer to each other slowly and join AB. Remember A, B should not overlap on each other. After joining A, B attach them with cello tape Fig (c).
iv) What kind of shape you have obtained? Is it a right cone?
While making a cone observe what hap-pened to the edges ‘OA’ and ‘OB’ and length of arc AB of the sector?

Try This

A sector with radius r and length of its arc / is cut from a circular sheet of paper. Fold it as a cone. How can you derive the formula of its curved surface area A = πrl. [Page No. 228]

C.S.A = πrl
Solution:
When a sector of radius ‘r’ and whose length of arc l is folded to form a cone. Radius ‘r’ becomes slant height ‘l’ and arc ‘l’ becomes perimeter of the base 2πr.

∴ Area of the sector = \(\frac{l r}{2}\) = Area of the cone
\(\frac{2 \pi \mathrm{r} l}{2}\) = Surface area of the cone
C.S.A = πrl

Do This

Question 1.
Cut a right angled triangle. Stick a string along its perpendicular side, as shown in fig. (T) hold both the sides of a string with your hands and rotate it with constant speed. What do you observe ? [Page No. 229]

Solution:
A right circular cone is observed.

Question 2.
Find the curved surface area and total surface area of the each following right circular cones.[Page No. 229]
Solution:

OP = 2 cm; OB = 3.5 cm
OP = h = 2 cm

OP = 3.5 cm; AB = 10 cm
r = \(\frac{\mathrm{AB}}{2}\) = 5cm; h = 3.5cm
r = OB = 3.5 cm
C.S.A. = πrl

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 3.5(3.5 + 4.03)
= \(\frac{22}{7}\) × 3.5 × 7.53 = 82.83cm²

C.S.A. = πrl
\(\frac{22}{7}\) × 5 × 6.10
= 95.90cm²

T.S.A. = πr (r + l)
= \(\frac{22}{7}\) × 5 × (5 + 6.10)
= 174.42 cm²

Activity

Draw a circle on a thick paper and cut it neatly. Stick a string along its diam¬eter. Hold the both the ends of the string with hands and rotate with con¬stant speed and observe the figure so formed. [Paper 235]

Try This

Question 1.
Can you find the surface area of sphere in any other way ? (Page No. 235)
Solution:

Height of the pyramid is equal to r.
To derive the formula of the surface area of a sphere, we imagine a sphere with many pyramids inside of it until the base of all the pyramids cover the entire surface area of the sphere. In the figure below, only one of such pyra¬mid is shown.
Then, do a ratio of the area of the pyramid to the volume of the pyramid.
The area of the.pyramid is A.
The volume of the pyramid is V = (1/3) × A × r = (A × r)/3
So, the ratio of area to volume is A/V = A + (A × r) / 3 = (3 × A) / (A × r) = 3 / r

For a large number of pyramids, let say that n is such large number, the ratio of the surface area of the sphere to the volume of the sphere is the same as 3/r.
For n pyramids, the total area is n × A- Also for n pyramids, the total volume is n × V.
Therefore, ratio of total area to total volume is n × A/n × V = A/ V and we already saw before that A / V = 3 / r

Further more, n × A_pyramid = A_sphere (The total area of the bases of all pyramids or n pyramids is approximately equal to the surface area of the sphere).
n × V_pyramid = V_sphere (The total Volume of all pyramids or n pyramids is approximately equal to the volume of the sphere.
Putting observation # 1 and # 2 together, we get

Therefore, the total surface area of a sphere, call it S.A. is 4πr ²

Do These

Question 1.
A right circular cylinder just encloses a sphere of radius r (see figure). Find i) surface area of the sphere
ii) curved surface area of the cylinder
iii) ratio of the areas obtained in (i) and (ii) [Page No. 236]

[Page No. 236]
Solution:
i) Radius of the sphere = radius of the cylinder = r
∴Surface area of the sphere = 4πr²
ii) C.S.A. of cylinder = 2πr (2r) [∵ h = 2r]
= 4πr²
iii) Ratio of (i) and (ii) = 4πr² : 4πr² = 1:1

Question 2.
Find the surface area of each of the following figure.
i)

Surface area = 4πr²
C.S.A = 4 × \(\frac{22}{7}\) × 7 × 7
= 616cm²

ii)

C.S.A = 2πr² = 2 × \(\frac{22}{7}\) × 7 × 7 = 308cm²
Total Surface area = 3πr²
= 3 × \(\frac{22}{7}\) × 7 × 7 = = 462cm²

Do This

Question 1.
Find the volume of the sphere given below[Page No. 238]

Solution:
r = 3 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 3 × 3 × 3
= 113.14cm³

d = 5.4 cm
r = \(\frac{\mathrm{d}}{2}=\frac{5.4}{2}\) = 2.7 cm
V = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) × 2.7 × 2.7 × 2.7 = 82.48cm³

Question 2.
Find the volume of sphere of radius 6.3 cm. [Page No. 238]
Solution:
r = 6.3 cm
V = \(\frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7}\) × 6.3 × 6.3 × 6.3 = 1047.81cm³

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

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics InText Questions

Activity

Divide the students of your class into four groups. Allot each group the work of collecting one of the following kinds of data: [Page No. 195]
Solution:
i) Weights of all the students in your class
ii) Number of siblings that each student have.
iii) Day wise number of absentees in your class during last month.
iv) The distance between the school and home of every student of your class.

Do This

Which of the following are primary and secondary data ? (Page No. 195)
i) Collection of the data about enroll¬ment of students in your school for a period from 2001 to 2010.
Solution:
Secondary data.

ii) Height of students in your class re¬corded by physical education teacher.
Solution:
Primary data.

Activity

Make frequency distribution table of the initial letters of that denotes surnames of your classmates and answer the following questions.
i) Which initial letter occured mostly among your classmates?
ii) How many students names start with’ the alphabat T?
iii) Which letter occured least as an initial among your classmates? [Page No. 197]

Think, Discuss anil Write

Question 1.
Give 3 situations, where mean, median and mode are separately appropriate and counted. [Page No. 202]
Solution:
Mean :
a) Rice required for a certain number of students for a given period.
b) To compare the marks of students of a class.
c) To calculate the daily income of a shop during a month.

Median :
a) To study the salaries of staff of an institution.
b) To compare heights of boys and girls.

Mode :
a) To find the size of the shoe with heighest sale.

Try These

Question 1.
Find the median of the scores 75, 21, 56, 36, 81, 05, 42. [Page No. 207]
Solution:
Arranging the data in ascending order
05, 21, 36, 42, 56, 75, 81
Number of terms in the data = 7 – odd
∴ \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\frac{7+1}{2}=\frac{8}{2}\) = 4th term is the median = 42.

Question 2.
Median of a data arranged in ascending order 7, 10, 15, x, y, 27, 30 is 17 and when one more observation 50 is added to the data the median has become 18. Find x and y. [Page No. 207]
Solution:
The given data in ascending order is 7, 10, 15, x, y, 27, 30
Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{7+1}{2}\right)\) = 4th term = x
∴ x = 17 by problem
When 50 is added, the data becomes 7, 10, 15, 17, y, 27, 30, 50
Median = \(\frac{\left(\left(\frac{n}{2}\right)+\left(\frac{n}{2}+1\right)\right)}{2}\) terms
18 = \(\frac{17+y}{2}\) (by problem)
17 + y = 36
y = 36 – 17 = 19

Question 3.
Find the median marks in the data.[Page No. 208]

Solution:

Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term as N = 29 is odd
∴ \(\frac{29+1}{2}\) = 15th term.
15th observation is 15. (from the table)

Question 4.
In finding the median, the given data must be in order ? Why ? [Page No. 208]
Solution:
In finding the median, the observations must be in order. The data is to be arranged either in ascending/descending order to divide the data into two equal groups.

Think and Discuss

Question 1.
Classify your classmates according to their heights and find the mode of it.
(Page No. 208)
Solution:
Student to find the mode of the classmates according their heights.

Question 2.
If shopkeeper has to place an order for shoes, which number shoes should he order more ? [Page No. 208]
Solution:
Number 7 as it has highest sales.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals InText Questions

Try This

Question
Extend AB to E. Find ∠CBE. What do you notice? What kind of angles are ∠ABC and ∠CBE?
Solution:
Given that □ABCD is a parallelogram and∠A = 40°
∴ ∠ABC = 180°-40° = 140°
∠CBE = 40° ( ∵ ∠A and, ∠CBE are corresponding angles) And ∠CBE and ∠ABC are linear pair of angles.

Do This

Question
Cut out a parallelogram from a sheet of paper again and cut along one of its diagonal. What kind of shapes you obtain ? What can you say about these triangles 7 [Page No. 179]
Solution:
We get two triangles.

The two triangles are congruent to each other.

Think, Discuss and Write

Question 1.
Show that the diagonals of a square are equal and right bisectors of each other. (Page No. 185)
Solution:

Let □ABCD be a square.
Thus AB = BC = CD = DA
In ΔABC and ΔBAD
AB = AB (common base)
∠B =∠A (each 90°)
BC = AD (equal sides)
∴ ΔABC = ΔBAD (SAS congruence)
⇒ AC = BD (CPCT)
Also in ΔAOB and ΔCOD
∠OAB = ∠OCD [∵ alt. int. angles]
∠OBA = ∠ODC [∵ alt. int. angles]
AB = DC (sides of a square)
∴ ΔAOB = ΔCOD (ASA congruence)
Thus AO = OC (CPCT) ⇒ O is midpoint of AC
Also BO = OD (CPCT) ⇒ O is midpoint of BD
∴ O is midpoint of both AC and BD.
∴ Diagonals bisect each other.
In ΔAOB and ΔCOB
AB = BC (given)
OB = OB (common)
AO = OC (proved)
∴ ΔAOB ≅ ΔCOB (SSS congruence)
∠AOB = ∠COB (CPCT)
But ∠AOB + ∠COB = 180° (∵ linear pair of angles)
∴ ∠AOB = ∠COB = 180°/2 = 90°
Also ∠AOB = ∠COD (∵ vertically opposite angles)
∠BOC = ∠AOD (∵ vertically opposite angles)
∴ AC ⊥ BD
(i.e.,) In a square diagonals bisect at right angles.

Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles. (Page No. 185)
Solution:

□ABCD is a rhombus
Let AC and BD meet at ‘O’
In ΔAOB and ΔCOD
∠OAB = ∠ODC (alt.int.angles)
AB = CD (def. of rhombus)
∠OBA = ∠ODC (alt. mt, angles)
∴ ΔAOB ≅ΔCOD …………(1)
(ASA congruence)
Thus AO = OC (CPCT
Also ΔAOD ≅ ΔCOD …………..(2)
[ ∵AO = OC; AD = CD; OD = OD SSS congruence]
Similarly we can prove
ΔAOD ≅ ΔCOB …………..(3)
From (1), (2) and (3) we have
ΔAOB = ΔBOC = ΔCOD = ΔAOD
∴ Diagonals of a rhombus divide it into four congruent triangles.

Try This

Question
Draw a triangle ABC and mark the mid points E and F of
two sides of triangle. \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{A C}}\) respectively. Join the point E and F as shown in the figure. Measure EF and the third side BC of triangle. Also measure ∠AEF and ∠ABC.
We find ∠AEF = ∠ABC and \(\overline{\mathrm{EF}}=\frac{1}{2} \overline{\mathrm{BC}}\)

As these are corresponding angles made by the transversal AB with lines EF and BC, we say EF//BC.
Repeat this activity with some more triangles. (Page No. 188)

Solution:

P, Q are mid points of XY and XZ
PQ // YZ
PQ = \(\frac{1}{2}\)YZ

X, Y are mid points of PQ and PR
XY // QR
XY = \(\frac{1}{2}\)QR

A, B are midpoints of DE and DF
AB // EF
AB = \(\frac{1}{2}\)EF
( ∵ In all cases the pairs of respective corresponding angles are equal.)

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

Do This

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

Try This

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

Do This

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

Try This

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.