Srinivas

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.1

Question
Solve the following Simple Equations:

(i) 6m = 12
(ii) 14p -42
(iii) -5y = 30
(iv) – 2x = – 12
(v) 34x = – 51
(vi) \(\frac{n}{7}\) = -3
(vn) \(\frac{2x}{3}\) = 8
(vui) 3x+1 = 16
(ix) 3p – 7 = 0
(x) 13 – 6n = 7
(xi) 200y – 51 = 49
(xii) 11n + 1 = 1
(xiii) 7x – 9 = 16
(xiv) 8x + \(\frac{5}{2}\) =13
(xv) 4x – \(\frac{5}{3}\) = 9
(xvi) x – \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
Solution:
i) 6m = 12 ⇒ m = \(\frac{12}{6}\) ⇒ m = 2

ii) 14p = – 42p ⇒ P = \(\frac{-42}{14}\)
∴ p = -3

iii) -5y = 30 ⇒ y = \(\frac{30}{-5}\) = -6
∴ y = -6

iv) -2x = -12
⇒ 2x = 12
x = \(\frac{30}{-5}\)
= 6
∴ x = 6

v) 34x = -51
⇒ \(\frac{-3}{2}\) = \(\frac{-3}{2}\)
∴ x = \(\frac{-3}{2}\)

vi) \(\frac{n}{7}\) = -3
⇒ n = -3 x 7 = -21
∴ n = -21

vii) \(\frac{2x}{3}\) = 18 ⇒ 18 x \(\frac{3}{2}\) = 27
∴ x = 27

viii) 3x + 1 = 16
3x = 16 – 1 = 15
3x = 15
x = \(\frac{15}{3}\)
∴ x = 5

ix) 3p – 7 = 0
⇒ 3p = 7
∴ p = \(\frac{7}{3}\)

x) 13 – 6n = 7 ⇒ -6n = 7 – 13
⇒ -6n = -6 ⇒ n= \(\frac{-6}{-6}\)
∴ n = 1

xi) 200y – 51 = 49
⇒ 200y = 49 + 51
⇒ 200y = 100
⇒ y = \(\frac{100}{200}\)
∴ y = \(\frac{1}{2}\)

xii) 11n + 1 = 1
⇒ 11n = 1 – 1
= 11n = 0
⇒ n = \(\frac{0}{11}\) = 0
∴ n = 0

xiii) 7x – 9 = 16
⇒ 7x = 16 + 9
⇒ 7x = 25
∴ x = \(\frac{25}{7}\)

xiv) 8x + \(\frac{5}{2}\) = 13

xv) 4x – \(\frac{5}{3}\)

xvi) x + \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
⇒ \(x+\frac{4}{3}=\frac{7}{2}\)
⇒\(\frac{7}{2}-\frac{4}{3}=\frac{21-8}{6}\)
∴ x = \(\frac{13}{6}\)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.3

Question 1.
Express each of the following decimal in the \(\frac{p}{q}\) form.
(i) 0.57 (ii) 0.176 (iii) 1.00001 (iv) 25.125
Solution:
(i) 0.57 = \(\frac{57}{100}\) (∵ two digits are there after the decimal poing)
(ii) 0.176 = \(\frac{176}{1000}\)
(iii) 1.00001 = \(\frac{100001}{100000}\)
(iv) 25.125 = \(\frac{25125}{1000}\)

Question 2.
Express each of the following decimals in the rational form \(\frac{p}{q}\)
(1) \(0 . \overline{9}\)
(ii) \(0 . \overline{57}\)
(iii) \(0 .7 \overline{29}\)
(iv) \(12.2 \overline{8}\)
Solution:
(i) \(0 . \overline{9}\)
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ………………. (1)
Here periodicity is 1. So, equation (1) should be multiplied both sides with
= 10 × x = 10 × 0.999
10 x = 9.999 ………….. (2)

\(0 . \overline{9}\) = 1

Second Method:
\(0 . \overline{9}=0+\overline{9}=0+\frac{9}{9}\)
= 0 + 1 = 1

(ii) \(0 . \overline{57}\)
x = \(0 . \overline{57}\) ⇒ x = 0.5757…………(1)
Here periodicity is 2. So, we should multiply with 100
⇒ 100 × x = 100 x 0.5757 …………..
100 × =57.57 ……………………. (2)

(iii) \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\) ⇒ x = 0.7979…………(1)
Here periodicity is 2. So, equation (1) should multiply with 100
⇒ 100 × x = 100 × 0.72929 …………..
100 × = 72.929 …………………… (2)

(iv) \(12.2 \overline{8}\)
x = (iv) \(12.2 \overline{8}\)
⇒ x = 12.288 ………..(1)
Here periodicity is 1. So, equation (1) should multiply with 10
⇒ 100 × x = 100 × 12.288 …………..
10 x = 122.888 …………………… (2)

Question 3.
Find(x + y) ÷ (x – y) if
(i) x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\)
(ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)
Solution:
If x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\) then

ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)

Question 4.
Divide the sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\) by the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)
Solution:
Sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\)

the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)

Question 5.
If \(\frac{2}{5}\) of a number exceeds \(\frac{1}{7}\) of the same number by 36. Find the number.
Solution:
Let the number be ‘x’ say.
\(\frac{2}{5}\) part of x = \(\frac{2}{5}\) × x = \(\frac{2x}{5}\)
\(\frac{1}{7}\) part of x = \(\frac{1}{7}\) × x = \(\frac{x}{7}\)
∴ According to the sum,

Question 6.
Two pieces of lengths 2\(\frac{2}{5}\) m and 3\(\frac{3}{10}\) mare cut off from a rope 11 m long. What is the length of the remaining rope?
Soltuion:
The length of the remaining rope

∴ The length of remaining rope
= 5\(\frac{1}{10}\) mts.

Question 7.
The cost of 7\(\frac{2}{3}\) meters of cloth is ₹12\(\frac{3}{4}\) . Find the cost per metre.
Solution:
The cost of 7\(\frac{2}{3}\) mts (\(\frac{23}{3}\) mts ) of cloth
= ₹ \(12 \frac{3}{4}\) = ₹ \(\frac{51}{4}\)
∴ The cost of 1m cloth
= \(\frac{51}{4} \div \frac{23}{3}=\frac{51}{4} \times \frac{3}{23}=\frac{153}{92}\) = ₹ 1.66

Question 8.
Find the area of a rectangular park which is 18\(\frac{3}{5}\)m long and 8\(\frac{2}{3}\) in broad.
Solution:
The length of the rectangular park
= 18\(\frac{3}{5}\)m = \(\frac{93}{5}\)
Its width / breath = 8\(\frac{2}{3}\) m = \(\frac{26}{3}\) m
∴ Area of the rectangular park
(A) = l × b

Question 9.
What number should \(-\frac{33}{16}\) be divided by to get \(-\frac{11}{4}\)
Solution:
Let the dividing number be ‘x’ say.

Question 10.
If 36 trousers of equal sizes can be stitched with 64 meters of cloth. What is the length of the cloth required for each trouser?
Solution:
36 trousers of equal sizes can he stitched with 64 mts of cloth, then the length of the cloth ¡s required for each trouser
= 64 ÷ 36
= \(\frac{64}{36}=\frac{16}{9}\) = 1 \(\frac{7}{9}\)

Question 11.
When the repeating decimal 0.363636 …. is written in simplest fractional form\(\frac{p}{q}\) , find the sum p+ q.
Solution:
x = 0.363636………………………….. (1)
Here periodicity is ‘2’. So, equation (1) should be multiplied both sides with 100.
⇒ 100 × x = 100 × 0.363636 …………..
100 x = 36.3636 ………..(2)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.2

Question 1.
Represent these numbers on the number line.
(i) \(\frac{9}{7}\)
(ii) \(-\frac{7}{5}\)
Solution:
(i) \(\frac{9}{7}\)

(ii) \(-\frac{7}{5}\)

Question 2.
Represent \(-\frac{2}{13}, \frac{5}{13}, \frac{-9}{13}\) on the number line.
Solution:

Question 3.
Write five rational numbers which are smaller than \(\frac{5}{6}\)
Solution:
The rational number which are less than
\(\frac{5}{6}=\left\{\frac{4}{6}, \frac{3}{6}, \frac{2}{6}, \frac{1}{6}, \frac{0}{6}, \frac{-1}{6}, \frac{-2}{6} \ldots \ldots .\right\}\)

Question 4.
Find 12 rational numbers between -1 and 2.
Solution:

Question 5.
Find a rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\)
[Hint : First write the rational numbers with equal denominators.]
Solution:
The given rational numbers are \(\frac{2}{3}\) and \(\frac{3}{4}\)
\(\frac{2}{3} \times \frac{4}{4}=\frac{8}{12}, \frac{3}{4} \times \frac{3}{3}=\frac{9}{12}\)
The rational numbers between \(\frac{8}{12}, \frac{9}{12}\) is
\(\frac{\left(\frac{8}{12}+\frac{9}{12}\right)}{2}=\frac{\frac{17}{12}}{2}=\frac{17}{24}\)
(∵ the rational number between a, b is \(\frac{a+b}{2}\) )
∴ the rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\) is \(\frac{17}{24}\)

Question 6.
Find ten rational numbers between \(-\frac{3}{4}\) and \(\frac{5}{6}\)
Solution:

The 10 rational numbers between \(-\frac{9}{12}\) and \(\frac{10}{12}\) are

∴ We can select any 10 rational numbers from the above number line.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.1

Question 1.
Name the properly Involved in the following examples.

vii) 7a + (-7) = 0
viii) x + \(\frac{1}{x}\) = 1(x ≠ 0)
ix) (2 x x) + (2 x 6) = 2 x (x + 6)
Solution:
i) Additive identity
ii) Distributive law
iii) Multiplicative identity
iv) Multiplicative identity
v) Commutative law of addition
vi) Closure law in multiplication
vii) Additive inverse
viii) Multiplicative inverse
ix) Distributive

Question 2.
Write the additive and the multiplicative inverses of the following.
i) \(\frac{-3}{5}\)
ii) 1
iii) 0
iv) \(\frac{7}{9}\)
v) -1
Solution:

Question 3.
Fill in the blanks



Solution:
i) \(\left(\frac{-12}{5}\right)\)
ii) \(\left(\frac{4}{3}\right)\)
iii) \(\left(\frac{9}{11}\right)\)
iv) \(\left(\frac{6}{7}\right)\)
v) \(\left(\frac{3}{4}, \frac{1}{3}\right)\)
vi) 0

Question 4.
Multiply \(\frac{2}{11}\) by the reciprocal of \(\frac{-5}{14}\)
Solution:
The reciprocal of \(\frac{-5}{14}\) is \(\frac{-14}{5}\)
( ∵ \(\left(\frac{-5}{14}\right) \times\left(\frac{-14}{5}\right)=1\) )
∴ The product of \(\frac{2}{11}\) and \(\frac{-14}{5}\) is
\(\frac{2}{11} \times\left(\frac{-14}{5}\right)=\frac{-28}{55}\)

Question 5.
Which properties can be used computing \(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
Solution:
The following properties are involved in the product of
\(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
i) Multiplicative associative property.
ii) Multiplicative inverse.
iii) Multiplicative identity.
iv) Closure with addition

Question 6.
Verify the following
\(\left(\frac{5}{4}+\frac{-1}{2}\right)+\frac{-3}{2}=\frac{5}{4}+\left(\frac{-1}{2}+\frac{-3}{2}\right)\)
Solution:

Question 7.
Evaluate \(\frac{3}{5}+\frac{7}{3}+\left(\frac{-2}{5}\right)+\left(\frac{-2}{3}\right)\) after rearrangement.
Solution:

Let x = \(1.2 \overline{4}\)
⇒ x = 1.244……. …………………(1)
Here periodicity of equation (1) is 1. So
it should be multiplied by 10 on both
sides.
⇒ 10 x x = 10 x 1.244
10x = 12.44 …………..(2)

Question 8.
Subtract
(i) \(\frac{3}{4}\) from \(\frac{1}{3}\)
(ii) \(\frac{-32}{13}\) from 2
(iii) -7 from \(\frac{-4}{7}\)
Solution:

Question 9.
What numbers should be added to \(\frac{-5}{8}\) so as to get \(\frac{-3}{2}\) ?
Solution:
Let the number to be add ‘x’ say

∴ \(\frac{-7}{8}\) should be added to \(\frac{-5}{8}\) then we will get \(\frac{-3}{2}\)

Question 10.
The sum of two rational numbers is 8 If one of the numbers is \(\frac{-5}{6}\) find the other.
Let the second number be ‘x’ say
⇒ \(x+\left(\frac{-5}{6}\right)=8\)
⇒\(8+\frac{5}{6}=\frac{48+5}{6}=\frac{53}{6}\)
∴ The other number (x) = \(\frac{53}{6}\)

Question 11.
Is subtraction associative in rational numbers? Explain with an example.
Solution:
Let \(\frac{1}{2}, \frac{3}{4}, \frac{-5}{4}\) are any 3 rational numbers.
Associative property under subtraction
a – (b – c) = (a – b) – c

∴ L.H.S. ≠ R.H.S.
∴ a – (b – c) ≠ (a – b) – c
∴ Subtraction is not an associative in rational numbers.

Question 12.
Verify that – (-x) = x for
(i) x = \(\frac{2}{15}\)
(ii) x = \(\frac{-13}{15}\)
Solution:

Question 13.
Write-
(i) The set of numbers which do not have any additive identity
(ii) The rational number that does not have any reciprocal
(iii) The reciprocal of a negative rational number.
Solution:
i) Set of natural numbers ’N’ doesn’t possesses the number ‘0’.
ii) The rational number ‘0’ has no multiplicative inverse.
[ ∵ 1/0 is not defined]
iii) The reciprocal of a negative rational number is a negative rational number.
Ex : Reciprocal of \(\frac{-2}{5}=\frac{-5}{2}\)

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

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics InText Questions

Do This

Question
Make 5 more sentences and check whether they are statements or not ? * Give reasons. [Page No. 311]
Solution:
1) 9 is a prime number – False
This is a statement because we can judge the truthness of this sentence. Clearly it is a false statement as 9 has factors other than 1 and 9, hence it is a composite number.
2) x is less than 5 – can’t say True or False
This is not a statement. The truthness can t be verified unless the value of x is known. Hence it is a sentence only.
3) 3 + 5 = 8 – True
The above sentence is a statement. It is a true statement as 5 + 3 = 8.
4) Sum of two odd numbers is even – True
The above sentence can be verified as a true sentence by taking ex¬amples like 3 + 5 = 8, 5 + 7 = 12 etc. Hence it is a true statement.
5) \(\frac{\mathrm{X}}{2}\) +3 = 9- can’t say True or False.
The above sentence is not a state¬ment. Its truthness can’t be
verified without the value of x.

Try This

1. 3 is a prime number.
2. Product of two odd integers is even.
3. For any real number x; 4x T x = 5x
4. The earth has one moon.
5. Ramu is a good driver.
6. Bhaskara has written a book “Leelavathi ”.
7. All even numbers are composite.
8. A rhombus is a square.
9. x > 7.
10. 4 and 5 are relative primes.
11. Silver fish is made of silver.
12. Humans are meant to rule the earth.
13. For any real number .v. 2x > x.
14. Havana is the capital of Cuba.

Question
Which of the above statements can be tested by giving counter example ?
[Page No. 312]
Solution:
Statements 2, 7, 8, 13 can be tested by giving counter examples 2) Product of two odd integers is even. Counter example.
2) Product of two odd integers 3 and 5 is 3×5 = 15 is not an even number.
7) All even numbers are composite. Counter example : 2 is an even prime.
8) A rhombus is a square.

Counter example: (40°, 140°, 40°. 140°) is a rhombus.
13) For anyx; 2x > x
Counter example : for x = -3:
2x = 2(-3) = – 6
here -6 < – 3

b

Try This

Envied by the popularity of Pythagoras, his younger brother claimed a different relation between the sides of a right angled triangles. [Page No. 319]

Liethagoras Theorem: In any right angled triangle the square of the smallest side equals the sum of the other sides. Check this conjecture, whether It is right or wrong.
Solution:
This conjecture is true for the above
triangles.
i) 3² 5 + 4 ⇒ 9 = 5 + 4
ii) 52 = 25 = 12 + 13
iii) 7² = 49 = 24 + 25
But, when the smallest side happens to be an even integer the conectiire may not hods good.
Eg: 1) 6² = 36 ≠ 10 + 8

ii) 12² = 144 ≠ 20 + 16

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

AP State Syllabus 9th Class Maths Solutions 14th Lesson Probability InText Questions

Do This

Question 1.
Observe the table given in the previous page (Textbook Page No. 293) and give some other example for each term. [Page No. 294]
Solution:
Certain : Independence day on 15th Aug.
More likely : When a die is thrown, the chance of getting a number less than or equal to 5. Equally likely : When a coin is tossed, getting a head.
Less likely : When a die is thrown, the chance of getting neither prime nor composite. Impossible : When a die is thrown, getting a negative number.

Question 2.
Classify the following statements into the categories less likely, equally . likely, more likely. [Page No. 294]

a) Rolling a die and getting a number 5 on the top face.
b) Getting a cold wave in your village in the month of November.
c) India winning the next soccer (foot ball) world cup.
d) Getting a tail or head when a coin is tossed.
e) You buy a lottery ticket and win the jackpot.
Solution:
a) less likely
b) more likely
c) less likely
d) equally likely
e) more likely

Try These

Question 1.
If you try to start a scooter, what are the possible outcomes ? [Page No. 295]
Solution:
[Starts], [Doesn’t start]

Question 2.
When you roll a die, what are the six possible outcomes ? [Page No. 295]
Solution:
1, 2, 3, 4, 5 and 6.

Question 3.
When you spin the wheel shown, what are the possible outcomes ? (Outcomes here means the possible sector where the pointer stops) [Page No.295]

Solution:
A, B and C.

Question 4.
You have ajar with five identical balls of different colours. [White, Red, Blue, Grey and Yellow] and you have to pick up (draw) a ball without looking at it. List the possible outcomes you get. [Page No. 295]

Solution:
White ball, Red ball, Blue ball. Grey ball and Yellow ball.

Think, Discuss and Write

In rolling a die [Page No. 295]

Question
Does the first player have a greater chance of getting a six on the top face ?
Solution:
No. The chance of getting 6 on the top face is independent of the turn of the player.

Question
Would the player who played after him have a lesser chance of getting a six on the top face?
Solution:
No.

Question
Suppose the second player got a six on the top face. Does it mean that the third player would not have a chance of getting a six on the top face ?
Solution:
No. The third player may or may not get six on the top face. It is indepen¬dent of 2nd player’s outcome.

Do This

Question
Toss a coin for number of times as shown in the table. And record your findings in the table [Page No. 296]
Solution:

Question
What hapens if you increase the num- her of tosses more and more.
Solution:
If you increase the number of tosses more and more they are equally likely chances to get a head or a tail.
Note : This could also be done by the students with a die, roll it for large number of times and observe!

Do This

If three coins are tossed simulta¬neously : [Page No. 299]

a) Write all possible outcomes.
Solution:
HHH, HHT, HTH, THH, HTT, THT, TTH, TTT, total 8 outcomes.

b) Number of possible outcomes.
Solution:
8

c) Find the probability of getting at least one head, (getting one or more than one head)
Solution:
P = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}=\frac{7}{8}\)

d) Find the probability of getting at most two heads, (getting two or less than two heads)
Solution:
P = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}=\frac{7}{8}\)

e) Find the probability of getting no tails.
Solution:
P = \(\frac{\text { favourable outcomes }}{\text { total outcomes }}=\frac{1}{8}\)

Try This

Find the probability of each event when a die is rolled once. [Page No. 300]

Try These

From the figure given below [Page No. 306]

Question 1.
Find the probability of the dart hit¬ting the board in the circular region B. (i.e., ring B)
Solution:
Area of innermost ‘C’ circle = πr²
= π x 1² = π sq. units.
Area of the middle ‘B’ circle
= π (2² – 1²) = π (4 – 1) = 3π sq.units.
Area of the outermost ’A’ circle
= π (3² – 2²) = π (9 – 4) = 5π sq.units.
Probability of hitting the circle B

Question 2.
Without calculating, write the percent¬age of probability of the dart hitting the board in circular region ‘C’ (Le., ring C).
Solution:
\(\frac{1}{9}\) x 100% = 11\(\frac{1}{9}\)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions InText Questions

Try These

Question
Observe the sides, angles and diagonals of quadrilateral BEFD. Name the figures given below and write properties of figures. [Page No. 283]

Solution:
In fig. (1)
\(\overrightarrow{\mathrm{BF}}\) is the bisector of ∠B and ∠F.
In quad BEFD
BE = BD = DF = EF
It is a rhombus

In fig. (2)
BD = BE
FD = FE
∴ BEFD is a kite.
BF is bisector of ∠B and ∠F.

Try This

Question
Draw a circle, identify a point on it. Cut arcs on the circle with the length of the radius in succession. How many parts can the circle be divided into ? Give reasons. [Page No. 284]
Solution:

Let P be the centre of the circle.
A is any point on its circumference.
It can be divided into 2π parts
∴ \(\frac{\text { Circumference }}{\text { Radius }}=\frac{2 \pi r}{r}=2 \pi\)

Think, Discuss and Write

Question
Can you construct a triangle ABC with BC = 6 cm, ∠B = 60° and AB + AC = 5 cm? If , not, give reasons. (Page No. 286)
Solution:
We can’t construct a triangle with measures ∠B = 60°; BC = 6 cm and AB + AC = 5 cm.
∵ AB + AC < BC
Sum of any two sides of a triangle must be greater than the third side.

Think, Discuss and Write

Question
Can you construct the triangle ABC with the same measures by changing the base angle ∠C instead of ∠B ? Draw a rough sketch and construct it.
BC = 4.2cm. ∠C = 30°, AB – AC = 1.6 cm (Page No. 287)

Solution:

Steps of construction:

1. Construct ΔBCD where BC = 4.2 cm and ∠C = 30° and AC – AB = 1.6 cm.
2. Draw perpendicular bisector of BD which meets \(\overline{\mathrm{CD}}\) produced at A.
3. Join B, D.
4. ΔABC is the required triangle.

Try These

Can you draw the triangle with the same measurements as shown in the figure in alternate way ? (Page No. 289)
[Measurements : ∠B = 6Q°, ∠C = 45° and AB + BC + CA =11 cm]
[Hint: Take ∠YXL = 60°/2 = 30° and ∠XYM = 45°/2 = 22 \(\frac { 1 }{ 2 }\) ]

Solution:
→ Draw XY = 11 cm [AB + BC + CA = 11 cm]
Construct ∠YXP = 30° at X \(\left[\frac{B}{2}=\frac{60^{\circ}}{2}=30^{\circ}\right]\)
Construct ∠XYQ = 22\(\frac { 1 }{ 2 }\) at Y \(\left[\frac{\mathrm{C}}{2}=\frac{45^{\circ}}{2}=22 \frac{1}{2}^{\circ}\right]\)
→ \(\overrightarrow{\mathrm{XP}}\) and \(\overrightarrow{\mathrm{YQ}}\) meet at A.
→ At A, draw \(\overrightarrow{\mathrm{AB}}\) such that ∠XAB = 30° where B is a point on XY.
→ Also draw \(\overrightarrow{\mathrm{AC}}\) such that ∠YAC = 22\(\frac{1}{2}^{\circ}\) where C is a point on XY.
→ Δ ABC is the required triangle.

Try These

Question
What happens if the angle in the. circle segment is right angle ? What kind of segment do you obtain ? Draw the figure and give reason. [Page No. 290]
Solution:
If the angle in the circle segment is right angle i.e., 90°, then the angle subtended by it at the centre is 2 x 90° = 180°
Thus the line segment becomes the diameter and the circle segment becomes the semi-circle.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles InText Questions

Activity

Question
Let us now do the following activity. Mark a point on a sheet of paper. Taking this point as centre draw a circle with any radius. Now increase or decrease the radius and.again draw some more circles with the same centre. What do you call the circles obtained in this activity ?[Page No. 261]

Circles having a common centre are called concentric circles.

Do This

Question 1.
In the figure which circles are congruent to the circle A ?
[Page No. 262]

Solution:
Circle ‘E’ is congruent to circle A.

Question 2.
What measure of the circles make them congruent ? [Page No. 262]
Solution:
Radius of circles determines their congruency.

Activity

Take a thin circular-sheet and fold it to half and open. Again fold it along any other half and open. Repeat this activity for several times. Finally when you open it, what do you observe?
[Page No. 262]
Solution:
Student Activity.

Activity

Take a circular paper. Fold it along any diameter such that the two edges coincide with each other. Now open it and again fold it into half along another diameter. On open¬ing, we find two diameters meet at the centre ‘O’. There forms two pairs of vertically opposite angles which are equal. Name the end points of the diameter as A, B, C and D.

Draw the chords \(\overline{\mathrm{AC}}, \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) and \(\overline{\mathrm{AD}}\).
Now take cut-out of the four segments namely 1, 2, 3 and 4.
If you place these segments pair wise one above the other the edges of the pairs (1,3) and (2.4) coincide with each other.
Is \(\overline{\mathrm{AD}}=\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}=\overline{\mathrm{BD}}\) ?
Though you have seen it in this par¬ticular case, try it out for other equal angles too. The chords will all turn out to be equal because of the following theorem. [Page No. 265]

Activity

Take a circle shaped paper and mark centre ‘O’: Fold it into two unequal parts and open it. Let the crease represent a chord AB, and then make a fold such that ‘A’ coincides with B. Mark the point of intersection of the two folds as D. Is AD = DB?

∠ODA = ?; ∠ODB = ?. Measure ‘the angles between the creases. They are right- angles. So. we can make a hypothesis “the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord”.
[Page No. 267]

Try This

In a circle with centre ‘O’, \(\overline{\mathbf{A B}}\) is a chord and M is its midpoint. Now prove that \(\overline{\mathbf{O M}}\) is perpendicular to AB.
(Hint : Join OA and OB consider tri-angles OAM and OBM)
[Page No. 267]
Solution:

‘O’ is the centre of the circle.
AB is a chord, M is its midpoint.
Join A, B to’O’.
Now in ΔOMA and ΔOMB
OA = OB (radii)
OM = OM (common)
MA = MB (given)
∴ ΔOMA s ΔOMB (SSS congruence)
∴ ∠OMA = ∠OMB (C.P.C.T)
But ∠OMA and ∠OMB are linear pair
∴∠OMA = ∠OMB = 90°
i.e., OM ⊥ AB.

Question
If three points are eollinear, how many circles can be drawn through these points ? Now try to draw a circle passing through these, three points.
[Page No. 268]
Solution:
If three points are eollinear, we can’t draw a circle passing’ through these points.

Activity

Draw a big circle on a paper and take a cut-out of it. Mark its centre as ‘O’. Fold it in half. Now make another fold near semi-circular edge. Now unfold it. You will get two congruent folds of chords. Name them as AB and CD. Now make perpendicular folds pass¬ing through centre ‘O’ for them. Using di¬vider compare the perpendicular distances of these chords from the centre.

Repeat the above activity by folding congruent chords. State your observations as a hypothesis.
“The congruent chords in a circle are at equal distance from the centre of the circle”. [Page No. 269]

Try This

In the figure, O is the centre of the circle and AB = CD. OM is perpen-dicular on \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{O N}}\) is perpen-dicular on \(\overline{\mathbf{C D}}\). Then prove that OM = ON. [Page No. 269]

Solution:
O’ is the centre of the circle.
Chords AB = CD
OM ⊥ AB; ON ⊥ CD
In ΔOMB and ΔONC
OB = OC [∵ radii]
BM = CN \(\left[\because \frac{1}{2} \mathrm{AB}=\frac{1}{2} \mathrm{CD}\right]\)
∠OMB = ∠ONC [ ∵90° each]
∴ ΔOMB ≅ ΔONC [R.H.S congruence]
∴ OM = ON (CPCT)

Activity

Take a circle paper. Mark four points A, B, C and D on the circle paper. Draw cyclic quadrilateral ABCD and measure its angles and record it in the table. Repeat this activity for three more times

What do you infer from the table ?
Solution:
Student Activity

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.)