Category: Class 9

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
Use deductive reasoning to answer the following.

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

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

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

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

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

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

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

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

Solution:

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

 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.2

Question 1.
Construct AABC in which BC = 7 cm,∠B = 75° and AB + AC =12 cm.
Solution:
A.

Steps:

• Draw a line segment BC = 7 cm.
• Erect ∠B = 75°
• MarkapointDon \(\overrightarrow{\mathrm{BX}}\) suchthat BD = AB + AC.
• Join D, C and draw the perpendicular bisector of \(\overline{\mathrm{CD}}\) meeting BD at A.
• Join A to C to form the ΔABC.

Question 2.
Construct ΔPQR in which QR = 8 cm, ∠B = 60° and AB – AC = 3.5 cm.
Read ∠Q = 60°and PQ – PR = 3.5 cm
Solution:
A.

Steps: I

• Draw QR = 8 cm.
• Construct ∠RQX = 300 at Q.
• Mark a point S on \(\overrightarrow{\mathrm{QX}}\) such that QS = PQ  –  PR = 3.5 cm.
• Join S, R and draw the perpendicular bisector to \(\overline{\mathrm{QR}}\) meeting \(\overrightarrow{\mathrm{QX}}\) at P.
• Join P, R to form the ΔPQR.

Question 3.
Construct ΔXYZ in which ∠Y = 30 °; ∠Z = 60 ° and XY + YZ + ZX = 10 cm.
Solution:
A.

Steps:

• Draw a line segment AB = XY + YZ + ZX = 10 cm.
• Construct ∠BAP = \(\frac { 1 }{ 2 }\) ∠Y at A and ∠ABQ = \(\frac { 1 }{ 2 }\) ∠Z at B meeting at X.
• Draw the perpendicular bisectors to XA and XB meeting \(\overline{\mathrm{AB}}\) at Y and Z respectively.
• Join X to Y and Z to form the ΔXYZ.

Question 4.
Construct a right triangle whose base is 7.5 cm and sum of its hypotenuse and otherside is 15 cm.
Solution:
A.

Steps:

• Draw BC = 7.5 cm.
• Construct ∠CBX = 90°
• Mark a point D on \(\overrightarrow{\mathrm{BX}}\) such that BD = 15 cm.
• Join C, D. ‘
• Draw the perpendicular bisectors of \(\overline{\mathrm{CD}}\) meeting BD at A.
• Join A, C to form the ΔABC.

Question 5.
Construct a segment of a circle on a chord of length 5 cm containing the following angles i) 90° ii) 45° iii) 120°
Solution:
i) 90°
A.

Steps:

• Draw a rough sketch of ∠BAC = 90° and ∠BOC = 180°.
• Draw a line segment BC = 5 cm.
• Draw the perpendicular bisector of BC meeting \(\overline{\mathrm{BC}}\) at O
• Draw an arc of radius OB or OC with centre O.
• Mark any point A on the arc and join it with B and C.
• ∠BAC = 90°

ii) 45°

Steps:

• Draw a line segment BC = 5 cm.
• Construct ABOC such that BC = 5 cm, ∠B = 45° = ∠C.
• Draw a circle segment of radius OB or OC with centre ’O’.
• Mark any point A on the segment and join it with B and C.
• ∠BAC = 45°

iii) 120°

Steps:

• Draw a line segment AB = 5 cm. ,
• Construct ΔAOB in which ∠A = 30°; ∠B = 30°; AB = 5 cm.
• With ‘O’ as centre draw a circle segment.
• On the opposite side make any point C and join it with B and C.
• ∠ACB = 120°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 13 Geometrical Constructions Ex 13.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 13th Lesson Geometrical Constructions Exercise 13.1

Question 1.
Construct the following angles at the initial point of a given ray and justify the construction.
a) 90°
Solution:

• Let AB be the given ray.
• Produce BA to D.
• Taking A as centre draw a semi circle with some radius.
• With X and Y as Center draw two intersecting arcs of same radius.

Or

|

• Let \(\overrightarrow{\mathrm{AB}}\) be the given ray.
• With A as centre draw an arc of any radius.
• Mark off two equal arcs from X as shown in the figure with the same radius taken as before.
• Bisect the second segment.
• Join the point of intersection of above arcs, with A.
• ∠BAC is the required right angle.
• Join the point of intersection ‘C’ and ‘A’.
• ∠BAC = 90°

In ΔAXY; ∠YAX = 60° and
in ΔAYC ∠YAC = 30° ∠BAC = 90°

b) 45°
Solution:
Steps:

• Construct 90° with the given ray AB.
• Bisect it from ∠BAD = 45°

Or

Steps:

• Construct ∠BAC = 60°
• Bisect ∠BAC = ∠DAC = 30°
• Bisect ∠DAC such that ∠DAE = ∠FAC = 15°
• ∠BAE=45°

ΔAXZ is equilateral
and ∠YAZ = 15°
∴∠XAY = 45°

Question 2.
Construct the following angles using ruler and compass and verify by measuring them by a protractor.
a) 30°
Solution:

• Construct ∠ABY = 60°
• Bisect ∠ABY = 60°
• Such that ∠ABC = ∠CBY = 30°

b) 22\(\frac{1}{2}^{\circ}\)
Solution:

• Construct ∠ABD = 90°.
• Bisect ∠ABD such that ∠ABC = ∠CBD = 45°
• Bisect ∠ABC such that
  ∠ABE = ∠EBC = 22\(\frac{1}{2}^{\circ}\)

c) 15°
Solution:

Steps of construction : ,

• Construct ∠BAE = 60°
• Bisect ∠BAE such that ∠BAC = ∠CAE = 30°
• Bisect ∠BAC such that ∠BAF = ∠FAC = 15°

d) 75°
Solution:

Steps of construction :

• Construct ∠BAC = 60°
• Construct ∠CAD = 60°
• Bisect ∠CAD such that ∠BAE = 90°
• Bisect ∠CAE such that ∠BAF = 75°

e) 105°
Solution:

Steps of construction:

• Construct ∠ABC = 90°
• Construct ∠CBE = 30°
• Bisect ∠CBE such that the angle formed ∠ABD = 105°

f) 135°
Solution:

Steps of construction:

• Construct ∠ABC = 120°
• Construct ∠CBD = 30°
• Bisect ∠CBD such that the angle formed ∠ABE = 135°

Question 3.
Construct an equilateral triangle, given its side of length of 4.5 cm and justify the constraction.
Solution:
A.

• Draw a line segment AB = 4.5 cm.
• With B and A as centres draw two arcs of radius 4.5 cm meeting at C.
• Join C to A and B.
• ΔABC is the required triangle.

Justification:
In ΔABC
AB = ∠C ⇒ ∠C = ∠B
Also AB = BC ⇒ ∠C = ∠A
Hence ∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
∴ ∠A = ∠B = ∠C = \(\frac{180^{\circ}}{3}\) = 60°

Question 4.
Construct an isosceles triangle, given its base and base angle and justify the construction. [Hint: You can take any measure of side and angle]
Solution:
A.

Steps:

• Draw a line segment AB of any given length.
• Construct ∠BAX and ∠ABY at A and B such that ∠A = ∠B.
• \(\overrightarrow{\mathrm{AX}}\) and \(\overrightarrow{\mathrm{BY}}\) will intersect at C.
• ΔABC is the required triangle.

Justification:
Drop a perpendicular CM to AB from C.
Now in ΔAMC and ΔBMC
∠AMC = ∠BMC [Right angle]
∠A = ∠B [Construction]
CM = CM (Common)
∴ ΔAMC ≅ ΔBMC
⇒ AC = BC [CPCT]

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.5 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.5

Question 1.
Find the values of x and y in the figures given below.
i)

Solution:
From the figure x = y [ ∵ angles opp. to opp. to equal sides]
But x + y + 30° = 180°
∴ x + y = 180° – 30° = 150°
⇒ x + y = \(\frac { 150° }{ 2 }\) = 75°

ii)

Solution:
From the figure x° + 110° = 180°
[ ∵ Opp. angles of a cyclic quad, are supplementary]
y + 85°= 180°
∴ x= 180° – 110°; y = 180° – 85°
x = 70°; y = 95°

iii)

Solution:
From the figure x = 90° [angle in a semi-circle]
∴ y = 90° – 50° [∵ angle sum property]
= 40°

Question 2.
Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle.
Solution:
Given : ∠A + ∠C = 180°
∴ ∠B + ∠D = 360° – 180°
[ ∵ sum of the four angles of a quad. is 360 ].
Now in □ABCD, sum of the pairs of opp. angles is 180°.
∴ □ABCD must be a cyclic quadrilateral, i.e., D also lie on the same circle on which the vertices A, B and C lie. Hence proved.

Question 3.
Prove that a cyclic rhombus is a square.
Solution:

Let □ABCD be a cyclic rhombus,
i.e., AB = BC = CD = DA and
∠A + ∠C = ∠B + ∠D = 180°
But a rhombus is basically a parallelo-gram.

Question 4.
For each of the following, draw a circle and inscribe the figure given. If a poly¬gon of the given type can’t be in-scribed, write not possible.
a) Rectangle
b) Trapezium .
c) Obtuse triangle
d) Non-rectangular parallelogram
e) Acute isosceles triangle
f) A quadrilateral PQRS with PR as diameter.
Solution:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.4

Question 1.
In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:
’O’ is the centre
∠AOB = 100°

Thus ∠ACB = \(\frac{1}{2}\) ∠AOB
[∵ angle made by an arc at the centre is twice the angle made by it on the remaining part]]
= \(\frac{1}{2}\) x 100° = 50°
∠ACB and ∠ADB are supplementary
[ ∵ Opp. angles of a cyclic quadrilateral]
∴ ∠ADB = 180°-50° = 130°
[OR]
∠ADB is the angle made by the major arc \(\widehat{\mathrm{ACB}}\) at D.
∴ ∠ADB = \(\frac{1}{2}\)∠AOB [where ∠AOB is the angle; made by \(\widehat{\mathrm{ACB}}\) at the centre]
= \(\frac{1}{2}\) [360° – 100°] [from the figure]
= \(\frac{1}{2}\) x 260° = 130°

Question 2.
In the figure, ∠BAD = 40° then find ∠BCD.

Solution:
‘O’ is the centre of the circle.
∴ In ΔOAB; OA = OB (radii)
∴ ∠OAB = ∠OBA = 40°
(∵ angles opp. to equal sides)
Now ∠AOB = 180° – (40° + 40°)
(∵ angle sum property of ΔOAB)
= 180°-80° = 100°
But ∠AOB = ∠COD = 100°
Also ∠OCD = ∠ODC [OC = OD]
= 40° as in ΔOAB
∴ ∠BCD = 40°
(OR)
In ΔOAB and ΔOCD
OA = OD (radii)
OB = OC (radii)
∠AOB = ∠COD (vertically opp. angles)
∴ ΔOAB ≅ ΔOCD
∴ ∠BCD = ∠OBA = 40°
[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

Question 3.
In the figure, ‘O’ is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR.

Solution:
‘O’ is the centre; ∠POR = 120°
∠PQR = \(\frac{1}{2}\)∠POR [∵ angle made by an arc at the centre is, twice the angle made by it on the remaining part]
∠PSR = \(\frac{1}{2}\) [Angle made by \(\widehat{\mathrm{PQR}}\) at the centre]
∠PSR = \(\frac{1}{2}\) [360° – 120°] from the fig.
= \(\frac{1}{2}\) x 240 = 120°

Question 4.
If a parallelogram is cyclic, then it is a rectangle. Justify.
Solution:

Let □ABCD be a parallelogram such
that A, B, C and D lie on the circle.
∴∠A + ∠C = 180° and ∠B + ∠D = 180°
[Opp. angles of a cyclic quadri lateral are supplementary]
But ∠A = ∠C and ∠B = ∠D
[∵ Opp. angles of a ||gm are equal]
∴∠A = ∠C =∠B =∠D = \(\frac{180}{2}\) = 90°
Hence □ABCD is a rectangle

Question 5.
In the figure, ‘O’ is the centr of the circle. 0M = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.
OM bisects AB.
∴ AM = \(\frac{\mathrm{AB}}{2}=\frac{8}{2}\) = 4 cm
OA² = OM² + AM² [ ∵ Pythagoras theorem]
OA \(\begin{array}{l}
=\sqrt{3^{2}+4^{2}} \\
=\sqrt{9+16}=\sqrt{25}
\end{array}\)
= 5cm

Question 6.
In the figure, ‘O’ is the centre of the circle and OM, ON are the perpen-diculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6 cm. Find RS.

Solution:
‘O’ is the centre of the circle.
OM = ON and 0M ⊥ PQ; ON ⊥ RS
Thus the chords FQ and RS are equal.
[ ∵ chords which are equidistant from the centre are equal in length]
∴ RS = PQ = 6cm

Question 7.
A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the
radius of the circle.

Solution:

A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle.
∴ Radius = 4 cm.

Question 8.
Draw a circle with any radius and then draw two chords equidistant
from the centre.
Solution:

1. Draw a circle with centre P.
2. Draw any two radii.
3. Mark off two points M and N oh these radii. Such that PM = PN.
4. Draw perpendicular through M and N to these radii.

Question 9.
In the given figure, ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of ΔOCD.

Solution:
‘O’ is the centre of the circle.
AB, CD are equal chords
⇒ They subtend equal angles at the centre.
∴ ∠AOB =∠COD = 70°
Now in ΔOCD
∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]
∴ ∠OCD + ∠ODC + 70° = 180°
= ∠OCD +∠ODC = 180° – 70° = 110°
∴ ∠OCD + ∠ODC = 110° = 55°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.3

Question 1.
Draw the following triangles and construct circumcircles for them.
(OR)
In ΔABC, AB = 6 cm, BC = 7 cm and ∠A = 60°.
Construct a circumcircle to the triangle XYZ given XY = 6cm, YZ = 7cm and ∠Y = 60°. Also, write steps of construction.
Solution:

Steps of construction :

1. Draw the triangle with given mea-sures.
2. Draw perpendicular bisectors to the sides.
3. The point of concurrence of per-pendicular bisectors be S’.
4. With centre S; SA as radius, draw a circle which also passes through B and C.
5. This is the required circumcircle.

ii) In ΔPQR; PQ = 5 cm, QR = 6 cm and RP = 8.2 cm.

Steps of construction:

1. Draw ΔPQR with given measures.
2. Draw perpendicular to PQ, QR and RS; let they meet at ‘S’.
3. With S as centre and SP as radius draw a circle.
4. This is the required circumcircle.

iii) In ΔXYZ, XY = 4.8 cm, ∠X = 60°and ∠Y = 70°.

Steps of construction:

1. Draw ΔXYZ with given measures.
2. Draw perpendicular bisectors to the sides of ΔXYZ, let the point of con-currence be S’.
3. Draw the circle (S, \(\overline{\mathrm{SX}}\)).
4. This is the required circumcircle.

Question 2.
Draw two circles passing A, B where AB = 5.4 cm.
(OR)
Draw a line segment AB with 5.4 cm. length and draw two different circles that passes through both A and B.
Solution:

Steps of construction:

1. Draw a line segment AB = 5.4 cm.
2. Draw the perpendicular bisector \(\stackrel{\leftrightarrow}{X Y}\) of AB.
3. Take any point P on \(\stackrel{\leftrightarrow}{X Y}\).
4. With P as centre and PA as radius draw a circle.
5. Let Q be another point on XY.
6. Draw the circle (Q, \(\overline{\mathrm{QA}}\)).

Question 3.
If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.
Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.
Join A, B to form the common chord
\(\overline{\mathrm{AB}}\). Let ‘O’ be the midpoint of AB.
Join ‘O’ with P and Q.
Now in ΔAPO and ΔBPO
AP = BP (radii)
PO = PO (common)
AO = BO (∵ O is the midpoint)
∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)
Also ∠AOP = ∠BOP (CPCT)
But these are linear pair of angles.
∴ ∠AOP = ∠BOP = 90°
Similarly in ΔAOQ and ΔBOQ
AQ = BQ (radii)
AO = BO (∵ O is the midpoint of AB)
OQ = OQ (common)
∴ AAOQ ≅ ABOQ
Also ∠AOQ = ∠BOQ (CPCT)
Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)
∴ ∠AOQ = ∠BOQ = \(\frac{180^{\circ}}{2}\) = 90°
Now ∠AOP + ∠AOQ = 180°
∴ PQ is a line.
Hence the proof.

Question 4.
If two intersecting chords of a circle make equal angles with diameter pass¬ing through their point of intersection, prove that the chords are equal.

Solution:
Let ‘O’ be the centre of the circle.
PQ is a drametre.
\(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are two chords meeting at E, a point on the diameter.
∠AEO = ∠DEO
Drop two perpendiculars OL and OM from ‘O’ to AB and CD;
Now in ΔLEO and ΔMEO
∠LEO = ∠MEO [given]
EO = EO [Common]
∠ELO = ∠EMO [construction 90°]
∴ ΔLEO ≅ ΔMEO
[ ∵ A.A.S. congruence]
∴ OL = OM [CPCT]
i.e., The two chords \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are at equidistant from the centre ‘O’.
∴ AB = CD
[∵ Chords which are equi-distant from the centre are equal]
Hence proved.

Question 5.
In the given figure, AB is a chord of circle with centre ‘O’. CD is the diam-eter perpendicular to AB. Show that AD = BD.

Solution:
CD is diameter, O is the centre.
CD ⊥ AB; Let M be the point of inter-section.
Now in ΔAMD and ΔBMD
AM = BM [ ∵ radius perpendicular to a chord bisects it]
∠AMD =∠BMD [given]
DM = DM (common)
∴ ΔAMD ≅ ΔBMD
⇒ AD = BD [C.P.C.T]

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.2

Question 1.
In the figure, if AB = CD and ∠AOB = 90° find ∠COD.

Solution:
‘O’ is the centre of the circle.
AB = CD (equal chords from the figure)
∴ ∠AOB = ∠COD
[ ∵ equal chords make equal angles at the centre]
∴ ∠COD = 90° [ ∵ ∠AOB = 90° given]

Question 2.
In the figure, PQ = RS and ∠ORS = 48°.
Find ∠OPQ and ∠ROS.

Solution:
‘O’ is the centre of the circle.
PQ = RS [given, equal chords]
∴∠POQ = ∠ROS [ ∵ equal chords make equal angles at the centre]
∴ In ΔROS
∠ORS + ∠OSR + ∠ROS = 180°
[angle sum property]
∴ 48° + 48° + ∠ROS = 180°
[ ∵ OR = OS(radii); ΔORS is isosceles]
∴ ∠ROS = 180° – 96° = 84°
Also ∠POQ = ∠ROS = 84°
∴ ∠OPQ = ∠OQP
[∵ OP = OQ; radii]
= \(\frac { 1 }{ 2 }\)[180°-84°] = 48°

Question 3.
In the figure, PR and QS are two diameters. Is PQ = RS ?

Solution:
‘O’ is the centre of the circle.
[ ∵ PR, QS are diameters]
OP = OR (∵ radii) .
OQ = OS (∵ radii)
∠POQ = ∠ROS [vertically opp. angles]
∴ ΔOPQ ≅ ΔORS [SAS congruence]
∴ PQ = RS [CPCT]

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.1

Question 1.
Name the following from the given figure where ’O’ is the centre of the circle.
i) \(\overline{\mathbf{A O}}\)
ii) \(\overline{\mathbf{A B}}\)
iii) \(\widehat{\mathrm{BC}}\)
iv) \(\overline{\mathbf{A C}}\)
v) \(\widehat{\mathrm{DCB}}\)
vi) \(\widehat{\mathrm{ACB }}\)
vii) \(\overline{\mathbf{A D}}\)
viii) Shaded region

Solution:
i) \(\overline{\mathbf{A O}}\) – radius
ii) \(\overline{\mathbf{A B}}\) – diameter
iii) \(\widehat{\mathrm{BC}}\) – minor arc
iv) \(\overline{\mathbf{A C}}\) – chord
v) \(\widehat{\mathrm{DCB}}\) – major arc
vi) \(\widehat{\mathrm{ACB }}\) – semi circle
vii) \(\overline{\mathbf{A D}}\) – chord
viii) Shaded region – Minor segment

Question 2.
State true or false.
i) A circle divides the plane on which it lies into three parts. ( )
ii) The area enclosed by a chord and the minor arc is minor segment. ( )
iii) The area enclosed by a chord and the major arc is major segment. ( )
iv) A diameter divides the circle into two unequal parts. ( )
v) A sector is the area enclosed by two radii and a chord. ( )
vi) The longest of all chords of a circle is called a diameter. ( )
vii) The mid point of any diameter of a circle is the centre. ( )
Solution:
i) True
ii) True
iii) True
iv) False
v) False
vi)True
vii) True

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 11 Areas Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas Exercise 11.3

Question 1.
In a triangle ABC, E is the midpoint of median AD. Show that

i) ar ΔABE = ar ΔACE
Solution:
In ΔABC; AD is a median.
∴ ΔABD = ΔACD …………….. (1)
(∵ Median divides a triangle in two equal triangles)
Also in ΔABD; BE is a median.
∴ ΔABE = ΔBED = \(\frac{1}{2}\)ΔABD …………..(2)
Also in ΔACD; CE is a median.
∴ ΔACE = ΔCDE = \(\frac{1}{2}\)ΔACD …………….(3)
From (1), (2) and (3);
ΔABE = ΔACE

(OR)

ΔABD = ΔACD [∵ AD is median in ΔABC]
\(\frac{1}{2}\) ΔABD = \(\frac{1}{2}\) ΔACD
[Dividing both sides by 2]
ΔABE = ΔAEC
[∵ BE is median of ΔABD, CE is median of ΔACD]
Hence proved.

ii) arΔABE = \(\frac{1}{2}\) ar(ΔABC)
Solution:
ΔABE = \(\frac{1}{2}\) (ΔABD)
[From (i); BE is median of ΔABD]
ΔABE = \(\frac{1}{2}\) [\(\frac{1}{2}\) ΔABC]
[∵ AD is median of ΔABC]
= \(\frac{1}{4}\) ΔABC
Hence,proved.

Question 2.
Show that the diagonals of a paral¬lelogram divide it into four triangles of equal area.
Solution:
□ABCD is a parallelogram.
The diagonals AC and BD bisect each other at ‘O’.
ΔABC and □ABCD lie on the same base AB and between the same parallels AB//CD.

∴ ΔABC = \(\frac{1}{4}\) □ABCD
Now in ΔABC; BO is a median
[∵ O is the midpoint of both diagonals AC and BD]
∴ ΔAOB = ΔBOC ………….(1)
[ ∵ Median divides a triangle into two triangles of equal area]
Similarly ∵ABD and □ABCD lie on the same base AB and between the same parallels AB//CD.]
∴ ΔABD = \(\frac{1}{2}\)□ ABCD
Also ΔAOB = ΔAOD …………..(2)
[ ∵ AO is the median of AABD]
From (1) & (2)
ΔAOB = ΔBOC = ΔAOD
Similarly we can prove that
ΔAOD = ΔCOD [ ∵ OD is the median of ΔACD]
∴ ΔAOB = ΔBOC = ΔCOD = ΔAOD
Hence proved.

Question 3.
In the figure, ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by \(\overline{\mathbf{A B}}\) at O, show that ar (ΔABC) = ar (ΔABD).

Solution:
From the figure, in ΔAOC; ΔBOD
OA = OB [ ∵ given]
∠AOC = ∠BOD [Vertically opp. angles]
∴ ΔAOC ≅ ΔBOD (SAS congruence)
Thus AC = BD (CPCT)
∠OAC = ∠OBD (CPCT) .
But these are alternate interior angles for the lines AC, BD.
∴ AC // BD
As AC = BD and AC // BD;
□ABCD is a parallelogram.
AB is a diagonal of oABCD
⇒ ΔABC ≅ ΔABD
(∵ diagonal divides a parallelogram into two congruent triangles)
∴ ar(ΔABC) = ar (ΔABD)

Question 4.
In the figure ΔABC; D, E and F are the midpoints of sides BC, CA and AB respectively. Show that
i)BDEF is a parallelogram
ii) ar (ΔDEF) = \(\frac { 1 }{ 4 }\) ar(ΔABC)
iii) ar (BDEF) = \(\frac { 1 }{ 2 }\) ar(ΔABC)

Solution:
i) In ΔABC; D, E and F are the mid¬points of the sides.
∴ EF//BC FD//AC ED//AB
EF = \(\frac { 1 }{ 2 }\)BC FD = \(\frac { 1 }{ 2 }\)AC ED = \(\frac { 1 }{ 2 }\)AB
[ ∵ line joining the mid points of any two sides of a triangle is parallel to third side and equal to half of it]
∴ In □BDEF
BD = EF [ ∵ D is mid point of BC and \(\frac { 1 }{ 2 }\) BC = EF]
DE = BF
∴ □BDEF is a parallelogram.

ii) □BDEF is a parallelogram (from (i))
Thus ΔBDF = ΔDEF
Similarly □CDFE; □AEDF are also parallelograms.
∴ ΔDEF = ΔCDE =ΔAEF
∴ ΔABC = ΔAEF+ ΔBDF + ΔCDF + ΔDEF
= 4ΔDEF
⇒ ΔDEF = \(\frac { 1 }{ 4 }\)ΔABC

iii) □BDEF = 2 ΔDEF …………..(1)
(from (ii))
ΔABC = 4 ΔDEF (2)
(from (ii))
From (1) & (2);
ΔABC = 2 (2ΔDEF) = 2 □BDEF
Hence proved.

Question 5.
In the figure D, E are points on the sides AB and AC respectively of ΔABC such that ar (ΔDBC) = ar (ΔEBC). Prove that DE // BC.

Solution:
ΔDBC = ΔEBC
The two triangles are on the same base BC and between the same pair of lines BC and DE.
As they are equal in area.
∴ BC // DE.

Question 6.
In the figure, XY is a line parallel to BC is drawn through A. If BE // CA and CF // BA are drawn to meet XY at E and F respectively. Show that ar (ΔABE) = ar (ΔACF).

Solution:
Given that XY//BC; BE//CA; CF//BA
In quad ABCF; AB//CF and BC//AF
Hence □ABCF is a parallelogram.
Also in □ACBE ; BC//AE and AC//BE
Hence □ACBE is a parallelogram.
Now in □ABCF and □ACBE
ΔABC = ΔACF …………..(1);
ΔABC = ΔABE …………..(2)
[∵ Diagonal divides a parallelogram into two congruent triangles]
∴ ΔACF = ΔABE [from (1) & (2)]
Hence proved.

Question 7.
In the figure, diagonals AC and BD of a trapezium ABCD with AB//DC inter¬sect each other at ‘O’. Prove that ar (ΔAOD) = ar (ΔBOC)

Solution:
Given that AB // CD
Now ΔADC and ΔBCD are on the same base and between the same parallels AB // CD.
∴ ΔADC = ΔBCD
⇒ ΔADC – ΔCOD = ΔBCD – ΔCOD
⇒ ΔAOD = ΔBOC [from the figure]

Question 8.
In the figure ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ΔACB) = ar (ΔACF)
(ii) ar (AEDF) = ar (ABCDE)

Solution:
ABCDE is a pentagon.
AC//BF
i) ΔACB and ΔACF are on the same base AC and between the same parallels AC//BF.
∴ ΔACB = ΔACF

ii) □AEDF = □AEDC + ΔACF
= □AEDC + ΔABC
[ ∵ ΔACF = ΔACB]
= area (ABCDE)
Hence proved.

Question 9.
In the figure, if ar (ΔRAS) = ar (ΔRBS) and ar (ΔQRB) = ar (ΔPAS) then show that both the quadrilaterals PQSR and RSBA are trapeziums.

ΔRAS = ΔRBS ………….. (1)
Both the triangles are on the same base RS and between the same pair of lines RS and AB.
As their areas are equal RS must be parallel to AB.
⇒ RS//AB
∴ □ABRS is a quadrilateral in which AB//RS.
∴ □ABRS (or) □RSBA is a trapezium.
Now AQRB = APAS (given)
⇒ ΔQRB – ΔRBS = ΔPAS – ΔRAS
[from (1) ΔRBS = ΔRAS]
⇒ ΔQRS = ΔPRS
These two triangles are on the same base RS and between the same pair of lines RS and PQ.
As these two triangles have same area RS must be parallel to PQ.
⇒ RS // PQ
Now in quad PQRS; PQ//RS.
Hence □PQRS is a trapezium.

Question 10.
A villager Ramayya has a plot of land in the shape of a quadrilateral. The grampanchayat of the village decided to take over some portion of his plot from one of the comers to construct a school. Ramayya agrees to the above proposal with the condition that he should be given equal amount of land in exchange of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will implemented. (Draw a rough sketch of the plot.)

Solution:
Let □ABCD is the plot of Ramayya.
School be constructed in the region ΔMCD where M is a point on BC such that □ABCD ≅ ΔADE
Draw the diagonal BD.
Draw a line parallel to BD through C which meets AB produced at E.
Join D, E
ΔADE is the required triangle.

Analysis:
ΔCED and ΔCEB are on the same base CE and between the same parallels CE and DB.
∴ ΔCED = ΔCEB [also from the figure]
ΔCEM + ΔCMD = ΔCEM + ΔBME
∴ ΔCMD = ΔBME
∴ ΔADE = □ABCD