Class 9

Jharkhand Board JAC Class 9 Maths Solutions Chapter 1 Number Systems Ex 1.3 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 1 Number Systems Exercise 1.3

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has:
(i) \(\frac{36}{100}\)
(ii) \(\frac{1}{11}\)
(iii) \(4 \frac{1}{8}\)
(iv) \(\frac{3}{13}\)
(v) \(\frac{2}{11}\)
(vi) \(\frac{329}{400}\)
Answer:
(i) \(\frac{36}{100}\) = 0.36 (Terminating)

(ii) \(\frac{1}{11}\) = 0.09090909… = \(0 . \overline{09}\) (Non-terminating and repeating)

(iii) \(4 \frac{1}{8}\) = \(\frac{33}{8}\) =4.125 (Terminating)

(iv) \(\frac{3}{13}\) = 0.230769230769… = \(0 . \overline{230769}\) (Non-terminating and repeating)

(v) \(\frac{2}{11}\) = 0.181818181818… = \(0 . \overline{18}\) (Non-terminating and repeating)

(vi) \(\frac{329}{400}\) = 0.8225 (Terminating)

Question 2.
You know that \(\frac{1}{7}\) = 0.142857. Can you predict what the decimal expansion of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) are without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of \(\frac{1}{7}\) carefully.]
Answer:
Yes, We can do this by:

\(\frac{2}{7}=2 \times \frac{1}{7}=2 \times 0 . \overline{142857}=0 . \overline{285714}\)
\(\frac{3}{7}=3 \times \frac{1}{7}=3 \times 0 . \overline{142857}=0 . \overline{428571}\)
\(\frac{4}{7}=4 \times \frac{1}{7}=4 \times 0 . \overline{142857}=0 . \overline{571428}\)
\(\frac{5}{7}=5 \times \frac{1}{7}=5 \times 0 . \overline{142857}=0 . \overline{714285}\)
\(\frac{6}{7}=6 \times \frac{1}{7}=6 \times 0 . \overline{142857}=0 . \overline{857142}\)

Question 3.
Express the following in the form p/q, where p and q are integers and q ≠ 0.
(i) \(0 . \overline{6}\)
(ii) \(0 . 4 \overline{7}\)
(iii) \(0 . \overline{001}\)
Answer:
(i) \(0 . \overline{6}\) = 0.666…
Let x = 0.666…
∴ 10x = 6.66…
∴ 10x = 6 + 0.66…
∴ 10x = 6 + 0.666…
∴ 10x = 6 + x
∴ 9x = 6
x = \(\frac{2}{3}\)

(ii) \(0 . 4 \overline{7}\) = 0.4777… = \(\frac{4}{10}\) + \(\frac{0.777}{10}\)
Let x = 0.777…
∴ 10x = 7.77…
∴ 10x = 7.777…
∴ 10x = 7 + 0.777…
∴ 10x = 7 + x
∴ x = \(\frac{7}{9}\)
\(0 . 4 \overline{7}\) = \(\frac{4}{10}\) + \(\frac{0.777}{10}\) = \(\frac{4}{10}\) + \(\frac{7}{90}\)
= \(\frac{36}{90}\) + \(\frac{7}{90}\) = \(\frac{43}{90}\)

(iii) \(0 . \overline{001}\) =0.001001…
Let x = 0.001001…
∴ 1000x = 1.001….
∴ 1000x = 1.001001…
∴ 1000x = 1 + 0.001001 …
∴ l000x= 1 + X
∴ 999x = 1
x = \(\frac{1}{999}\)

Question 4.
Express 0.99999…in the form \(\frac{p}{q}\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Answer:
Let x = 0.9999…
10x = 9.999…
∴ 10x = 9.9999…
∴ 10x = 9 + 0.9999…
∴ 10x = 9 + x
∴ 9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999… is too much near 1. Therefore, the answer 1 can be justified.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.
Answer:
\(\frac{1}{17}\) = 0.05882352941176470588…
= \(0 . \overline{0588235294117647}\)
There are 16 digits in the repeating block of the decimal expansion of \(\frac{1}{17}\)

Question 6.
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Answer:
We observe that when q is 2, 4, 5, 8, 10,… then the decimal expansion is terminating. For example:
\(\frac{1}{2}\) = 0.5, denominator q = 2¹
\(\frac{4}{5}\) = 0.8, denominator q = 5¹
We can observe that terminating decimal may be obtained in the situation where prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Answer:
Three numbers whose decimal expansions are non-terminating non-recurring are:
(i) 0.303003000300003…
(ii) 0.505005000500005…
(iii) 0.7207200720007200007200000…

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).
Answer:
\(\frac{5}{7}\) = \(0 . \overline{714285}\)

\(\frac{9}{11}\) = \(0 . \overline{81}\)
Three different irrational numbers are:
0.73073007300073000073…
0.75075007500075000075…
0.76076007600076000076…

Question 9.
Classify the following numbers as rational or irrational:
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478…
(v) 1.101001000100001…
Answer:
(i) \(\sqrt{23}\) = 4.79583152331…
Since the decimal expansion is non-terminating and non-recurring therefore, it is an irrational number.

(ii) \(\sqrt{225}\) = 15 = \(\frac{15}{1}\)
The number is rational number as it can represented in \(\frac{p}{q}\) form where p, q ∈ Z and q ≠ 0.

(iii) 0.3796
Since the decimal expansion is terminating therefore, it is a rational number.

(iv) 7.478478… = \(7 . \overline{478}\)
Since, this decimal expansion is non-terminating recurring, therefore, it is a rational number.

(v) 1.101001000100001…
Since the decimal expansion is non-terminating and non-repeating, therefore, it is an irrational number.

Students should go through these JAC Class 9 Maths Notes Chapter 8 Quadrilaterals will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 8 Quadrilaterals

Quadrilateral
A quadrilateral is a closed figure obtained by joining four points (with no three points collinear) in an order.
→ Since, ‘quad’ means ‘four’ and ‘lateral’ is for ‘sides therefore quadrilateral means a figure bounded by four sides’
→ Every quadrilateral has:
(A) Four vertices
(B) Four sides
(C) Four angles and
(D) Two diagonals.
→ A diagonal is a line segment obtained on joining the opposite vertices.

Sum of the Angles of a Quadrilateral:

Consider a quadrilateral ABCD as shown in figure. Join A and C to get the diagonal AC which divides the quadrilateral ABCD into two triangles ABC and ADC.
We know the sum of the angles of each triangle is 180°
∴ In ΔABC; ∠CAB + ∠B + ∠BCA = 180°
and In ΔADC; ∠DAC + ∠D + ∠DCA = 180°
On adding, we get:
(∠CAB + ∠DAC) + ∠B + ∠D + (∠BCA + ∠DCA) = 180° + 180°
⇒ ∠A + ∠B + ∠D + ∠C = 360°
Thus, the sum of the angles of a quadrilateral is 360°.

Types of Quadrilaterals:
→ Trapezium: It is a quadrilateral in which one pair of opposite sides are parallel and one pair is unparallel. In the quadrilateral ABCD, drawn alongside, sides AB and DC are parallel and AD and BC are unparallel therefore it is a trapezium

→ Parallelogram: It is a quadrilateral in which both the pairs of opposite sides are equal and parallel. The figure shows a quadrilateral ABCD in which AB is parallel and equal to DC and AD is parallel and equal to BC, therefore ABCD is a parallelogram.

Here, (A) ∠A = ∠C and ∠B = ∠D
(B) AB = CD and AD = BC
(C) AB || CD and AD || BC
→ Rectangle: It is a parallelogram whose each angle is 90°.
(a) ∠A + ∠B = 90° + 90° = 180°
⇒ AD || BC, also AD = BC.
(b) ∠B + ∠C = 90° + 90° = 180°
⇒ AB || DC, also AB = DC.
(c) Diagonals AC and BD are equal.

Rectangle ABCD is also a parallelogram.
→ Rhombus: It is a also parallelogram whose all the sides are equal and diagonals are perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA; AC ⊥ BD.; therefore it is a rhombus.

→ Square: It is a parallelogram whose all the sides are equal and each angle is 90°. Also, diagonals are equal and perpendicular to each other. The figure shows a parallelogram ABCD in which AB = BC = CD = DA, ∠A = ∠B = ∠C = ∠D = 90°, AC ⊥ BD and AC = BD, therefore ABCD is a square.

→ Kite: It is not parallelogram in which two pairs of adjacent sides are equal The figure shows a quadrilateral ABCD in which adjacent sides AB and AD are equal i.e. AB = AD and also the other pair of adjacent sides are equal i.e., BC = CD; therefore it is a kite or kite shaped figure.

Remarks:

• Square, rectangle and rhombus are all parallelograms.
• Kite and trapezium are not parallelograms.
• A square is a rectangle.
• A square is a rhombus.
• A parallelogram is a trapezium.

Parallelogram Theorems
A parallelogram is a quadrilateral in which both the pairs of opposite sides are equal and parallel.

Theorem 1.
A diagonal of a parallelogram divides the parallelogram into two congruent triangles.
Proof:
Given: A parallelogram ABCD.
To Prove: A diagonal divides the parallelogram
into two congruent triangles i.e., if diagonal AC is drawn then ΔABC ≅ ΔCDA and if diagonal BD is drawn
then ΔABD ≅ ΔCDB.
Construction: Join A and C
Proof: Since, ABCD is a parallelogram
AB || DC and AD || BC
In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
And, AC = AC [Common side]
∴ ΔABC ≅ ΔCDA [By ASA]
Similarly, we can prove that
ΔABD ≅ ΔCDB

Theorem 2.
In a parallelogram, opposite sides are equal.

Proof:
Given: A parallelogram ABCD in which
AB || DC and AD || BC.
To Prove: Opposite sides are equal i.e.. AB = DC and AD = BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ AB = DC and AD = BC [By CPCT]
Hence, proved.

Theorem 3.
If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AD = BC.
To Prove: ABCD is a parallelogram ie, AB || DC and AD || BC
Construction: Join A and C
Proof: In ΔABC and ΔCDA
AB = DC [Given]
AD = BC [Given]
And AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By SSS]
⇒ ∠1 = ∠3 [By CPCT]
And ∠2 = ∠4 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 4.
In a parallelogram, opposite angles are equal.

Proof:
Given: A parallelogram ABCD in which AB || DC and AD || BC.
To Prove: Opposite angles are equal i.e. ∠A = ∠C and ∠B = ∠D
Construction: Draw diagonal AC.
Proof: In ΔABC and ΔCDA
∠BAC = ∠DCA [Alternate angles]
∠BCA = ∠DAC [Alternate angles]
AC = AC [Common]
∴ ΔABC ≅ ΔCDA [By ASA]
⇒ ∠B = ∠D [By CPCT]
Similarly, we can prove that
∠A = ∠C Hence, proved.

Theorem 5.
If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof:
Given: A quadrilateral ABCD in which opposite angles are equal. i.e., ∠A = ∠C and ∠B = ∠D
To prove: ABCD is a parallelogram i.e.,
AB || DC and AD || BC
Proof: Since the sum of the angles of quadrilateral is 360°
⇒ ∠A + ∠B + ∠C+ ∠D = 360°
⇒ ∠A + ∠D + ∠A + ∠D = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠D = 360°
⇒ ∠A + ∠D = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
⇒ AB || DC
Similarly, ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B + ∠A + ∠B = 360°
[∵ ∠A = ∠C and ∠B = ∠D]
⇒ 2∠A + 2∠B = 360°
⇒ ∠A + ∠B = 180°
[∵ The sum of interior angles on the same side of transversal AB is 180°]
∴ AD || BC
So, AB || DC and AD || BC
⇒ ABCD is a parallelogram.
Hence, proved.

Theorem 6.
The diagonal of a parallelogram bisect each other.

Proof:
Given: A parallelogram ABCD. Its diagonals AC and BD intersect each other at point O..
To Prove: Diagonals AC and BD bisect each other i.e., OA = OC and OB = OD.
Proof: In ΔAOB and ΔCOD
∵ AB || DC and BD is a transversal.
∴ ∠ABO = ∠CDO [Alternate angles]
∵ AB || DC and AC is a transversal line.
∴ ∠BAO = ∠DCO [Alternate angles]
And, AB = DC
⇒ ΔAOB ≅ ΔCOD [By ASA]
⇒ OA = OC and OB = OD [By CPCT]
Hence, proved.

Theorem 7.
If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Proof:
Given: A quadrilateral ABCD whose diagonals AC and BD bisect each other at point O.
i.e., OA = OC and OB = OD
To prove: ABCD is a parallelogram i.e..
AB || DC and AD || BC.
Proof: In ΔAOB and ΔCOD
OA = OC [Given]
OB = OD [Given]
And, ∠AOB = ∠COD [Vertically opposite angles]
⇒ ΔAOB ≅ ΔCOD [By SAS]
⇒ ∠1 = ∠2 [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
∴ AB is parallel to DC ie., AB || DC Similarly,
ΔAOD ≅ ΔCOB [By SAS]
⇒ ∠3 = ∠4
But these are also alternate angles
⇒ AD || BC
AB || DC and AD || BC
⇒ ABCD is parallelogram.
Hence, proved.

Theorem 8.
A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel.

Proof:
Given: A quadrilateral ABCD in which AB = DC and AB || DC.
To Prove: ABCD is a parallelogram, i.e. AB || DC and AD || BC.
Construction: Join A and C.
Proof: Since AB is parallel to DC and AC is transversal
∠BAC = ∠DCA [Alternate angles]
AB = DC [Given]
And AC = AC [Common]
⇒ ΔBAC ≅ ΔDCA [By SAS]
⇒ ∠BCA = ∠DAC [By CPCT]
But these are alternate angles and whenever alternate angles are equal, the lines are parallel.
⇒ AD || BC
Now, AB || DC (given) and AD || BC
[Proved above]
⇒ ABCD is a parallelogram
Hence, proved.

Remarks:
In order to prove that given quadrilateral is parallelogram, we can prove any one of the following.

• Opposite angles of the quadrilateral are equal, or
• Diagonals of the quadrilateral bisect each other, or
• A pair of opposite sides is parallel and is of equal length, or
• Opposite sides are equal.
• Every diagonal divides the parallelogram into two congruent triangles.

Mid-Point Theorem
Statement: In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it.

Given: A triangle ABC in which P is the mid-point of side AB and Q is the mid-point of side AC.
To Prove: PQ is parallel to BC and is half of it
i.e., PQ || BC and PQ = \(\frac{1}{2}\)BC
Construction: Produce PQ upto point such that PQ = QR. Join R and C.
Proof: In ΔAPQ and ΔCRQ
PQ = QR [By construction]
AQ = QC [Given]
And, ∠AQP = ∠CQR [Vertically opposite angles]
⇒ ΔAPQ ≅ ΔCRQ [By SAS]
⇒ AP = CR [By CPCT]
And, ∠APQ = ∠CRQ [By CPCT]
But, ∠APQ and ∠CRQ are alternate angles and we know, whenever the alternate angles are equal, the lines are parallel.
⇒ AP || CR
⇒ AB || CR
⇒ BP || CR
Given, P is mid-point of AB
⇒ AP = BP
⇒ CR = BP [As, AP = CR]
Now, BP = CR and BP || CR
⇒ BCRP is a parallelogram.
[When any pair of opposite sides are equal and parallel, the quadrilateral is a parallelogram]
BCRP is a parallelogram and opposite sides of a parallelogram are equal and parallel.
∴ PR = BC and PR || BC
Since, PQ = QR
⇒ PQ = \(\frac{1}{2}\)PR = \(\frac{1}{2}\)BC [AS, PR = BC]
Also, PQ || BC [As, PR || BC]
∴ PQ || BC and PQ = \(\frac{1}{2}\)BC
Hence, proved.

Converse of the Mid-Point Theorem
Statement: The line drawn through the midpoint of one side of a triangle parallel to the another side bisects the third side.

Given: A triangle ABC in which is the midpoint of side AB and PQ is parallel to BC.
To prove: PQ bisects the third side AC i.e., AQ = QC.
Construction: Through C, draw CR parallel to BA, which meets PQ produced at point R.
Proof: Since, PQ || BC i.e., PR || BC [Given]
CR || BA i.e., CR || BP [By construction]
∴ Opposite sides of quadrilateral PBCR are parallel.
⇒ PBCR is a parallelogram
⇒ BP = CR
Also, BP = AP [As Pis mid-point of AB]
∴ CR = AP (As CR = BP)
Now, AB || CR and AC is transversal, ∠PAQ = ∠ROQ [Alternate angles]
Also, AB || CR and PR is transversal, ∠APQ = ∠CRQ [Alternate angles]
In ΔAPQ and ΔCRQ
CR = AP, ∠PAQ = ∠RCQ and ∠APQ = ∠CRQ
⇒ ΔAPQ ≅ ΔCRO [By ASA]
⇒ AQ = QC Hence, proved.

Jharkhand Board JAC Class 9 Maths Solutions Chapter 1 Number Systems Ex 1.2 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 1 Number Systems Exercise 1.2

Question 1.
State whether the following statements are true or false. Justify your Answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form \(\sqrt{m}\) , where m is a natural number.
(iii) Every real number is an irrational number.
Answer:
(i) True, since the collection of real numbers is made up of rational and irrational numbers.
(ii) False, since negative numbers cannot be expressed as square roots.
(iii) False, as real numbers include both rational and irrational numbers. Therefore, every real number cannot be an irrational number, e.g. 2 is a real number but not an irrational number.

Question 2.
Are the square roots of all positive inte-gers irrational? If not, give an example of the square root of a number that is a rational number.
Answer:
No, the square roots of all positive integers are not irrational. For example \(\sqrt{4}\) = 2, which is rational number.

Question 3.
Show how \(\sqrt{5}\) can be represented on the number line.
Answer:
Step 1: Let AB be a line segment of length 2 units on number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit. Join CA.
Step 3: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
AB² + BC² = CA²
⇒ 2² + 1² = CA²
⇒ CA² = 5
⇒ CA = \(\sqrt{5}\)
Thus, CA is a line of length \(\sqrt{5}\) units.

Step 4: Taking CA as a radius and A as a centre draw an arc touching the number line. The point D at which number line gets intersected by arc of length AC is at \(\sqrt{5}\) distance from 0.
Thus, \(\sqrt{5}\) is represented on the number line as shown in the figure.

Jharkhand Board JAC Class 9 Maths Solutions Chapter 1 Number Systems Ex 1.1 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 1 Number Systems Exercise 1.1

Question 1.
Is zero a rational number? Can you write it in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0?
Answer:
Yes, Zero is a rational number as it can be represented as \(\frac{0}{1}\) or \(\frac{0}{2}\) and so on.

Question 2.
Find six rational numbers between 3 and 4.
Answer:
There are infinitely many rational numbers between 3 and 4.
3 and 4 can be represented as \(\frac{24}{8}\) and \(\frac{32}{8}\) respectively.
Therefore, six rational numbers between 3 and 4 are: \(\frac{25}{8}\), \(\frac{26}{8}\), \(\frac{27}{8}\), \(\frac{28}{8}\), \(\frac{29}{8}\), \(\frac{30}{8}\).

Question 3.
Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\).
Answer:
There are infinitely many rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\)
\(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\)
\(\frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\)
Therefore, five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\) are \(\frac{19}{30}\), \(\frac{20}{30}\), \(\frac{21}{30}\), \(\frac{22}{30}\), \(\frac{23}{30}\).

Question 4.
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Answer:
(i) True, since the collection of whole numbers contains all natural numbers.
(ii) False, as integers may be negative but whole numbers are always positive, e.g. – 1 is an integer but not a whole number.
(iii) False, as rational numbers may be fractional but whole numbers are not fractional, e.g. \(\frac{2}{3}\) is a rational number but not a whole number.

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 1 Number Systems Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 1 Number Systems

Question 1.
Find 4 rational numbers between 2 and 3.
Solution :
Write 2 and 3 multiplying in numerator and denominator with (4 + 1).
i.e. 2 = \(\frac{2 \times 5}{1 \times 5}=\frac{10}{5}\)
or 3 = \(\frac{3 \times 5}{1 \times 5}=\frac{15}{5}\)
So, the four required rational numbers are \(\frac {11}{5}\), \(\frac {12}{5}\), \(\frac {13}{5}\), \(\frac {14}{5}\)

Question 2.
Find three rational numbers between a and b (a < b).
Solution :
a < b
⇒ a + a < b + a
⇒ 2a < a + b
⇒ a < \(\frac{a+b}{2}\)
Again, a < b
⇒ a + b < b + b
⇒ a + b < 2b
⇒ \(\frac{a+b}{2}\) < b
∴ a < \(\frac{a+b}{2}\) < b
i.e. \(\frac{a+b}{2}\) lies between a and b.
Hence, 1st rational number between a and b is \(\frac{a+b}{2}\)

Question 3.
Find 6 rational numbers between \(\frac {1}{3}\) and \(\frac {1}{2}\)
Solution :
LCM of 3 and 2 = 6
∴ \(\frac{1}{3}=\frac{2}{6}\) and \(\frac{1}{3}=\frac{2}{6}\)
Now, \(\frac{2}{6}=\frac{20}{60}\) and \(\frac{3}{6}=\frac{30}{60}\)
∴ \(\frac{21}{60}, \frac{22}{60}, \frac{23}{60}, \frac{24}{60}, \frac{25}{60}\) and \(\frac {26}{60}\) lie between \(\frac {1}{3}\) and \(\frac {1}{2}\).

Question 4.
Find 5 rational numbers between and \(\frac {3}{5}\) and \(\frac {4}{5}\)
Solution :
As \(\frac{3}{5}=\frac{3 \times 10}{5 \times 10}=\frac{30}{50}\) and
\(\frac{4}{5}=\frac{4 \times 10}{5 \times 10}=\frac{40}{50}\)
∴ 5 rationals between \(\frac {3}{5}\) and \(\frac {4}{5}\) are \(\frac{31}{50}, \frac{32}{50}, \frac{33}{50}, \frac{34}{50}\) and \(\frac {35}{50}\).

Question 5.
Find two irrational numbers between 2 and 2.5.
Solution :
\(\sqrt{2 \times 2.5}=\sqrt{5}\)
Since there is no rational number whose square is 5. So, \(\sqrt{5}\) is irrational.
Also, \(\sqrt{2 \times \sqrt{5}}\) is an irrational number.
∴ Irrational numbers between 2 and 2.5 are \(\sqrt{5}\) and \(\sqrt{3 \sqrt{5}}\).

Question 6.
Find two irrational numbers between \(\sqrt{2}\) and \(\sqrt{3}\).
Solution :
\(\sqrt{\sqrt{2} \times \sqrt{3}}=\sqrt{\sqrt{6}}=\sqrt[4]{6}\)
Irrational number between \(\sqrt{2}\) and \(\sqrt{3}\) are \(\sqrt[4]{6}, \sqrt{\sqrt{2} \times \sqrt[4]{6}}=\sqrt[4]{2} \times \sqrt[8]{6}\)

Question 7.
Find two irrational numbers between 0.12 and 0.13.
Solution :
0.1201001000100001…….,
0.12101001000100001……

Question 8.
Find two irrational numbers between 0.3010010001……. and 0.3030030003…..
Solution :
0.302020020002……, 0.302030030003….

Question 9.
Arrange \(\sqrt{2}\), \(\sqrt[3]{3}\) and \(\sqrt[4]{5}\) in ascending order.
Solution :
L.C.M. of 2, 3, 4 is 12.

Question 10.
Which is greater \(\sqrt{7}\) – \(\sqrt{3}\) or \(\sqrt{5}\) – 1 ?
Solution :

⇒ \(\sqrt{7}\) – \(\sqrt{3}\) < \(\sqrt{5}\) – 1 ⇒ \(\sqrt{5}\) – 1 > \(\sqrt{7}\) – \(\sqrt{3}\)
So, \(\sqrt{5}\) – 1 is greater.

Question 11.
Find the R.F. (rationalizing factor) of the following :
(i) \(\sqrt{12}\)
(ii) \(\sqrt{162}\)
(iii) \(\sqrt[3]{4}\)
(iv) \(\sqrt[3]{16}\)
(v) \(\sqrt[4]{162}\)
(vi) 2 + \(\sqrt{3}\)
(vii) 7 – 4\(\sqrt{3}\)
(vii) 3\(\sqrt{3}\) + 2\(\sqrt{2}\)
(ix) \(\sqrt[3]{3}\) + \(\sqrt[3]{2}\)
Solution :
(i) \(\sqrt{12}\)
First write its simplest form i.e. 2\(\sqrt{3}\).
Now R.F. of \(\sqrt{3}\) is \(\sqrt{3}\)
∴ R.F. of \(\sqrt{12}\) is \(\sqrt{3}\).

(ii) \(\sqrt{162}\)
Simplest form of \(\sqrt{162}\) is 9\(\sqrt{2}\).
R.F. of \(\sqrt{2}\) is \(\sqrt{2}\).
∴ R.F. of \(\sqrt{162}\) is \(\sqrt{2}\)

(iii) \(\sqrt[3]{4}\)
\(\sqrt[3]{4} \times \sqrt[3]{4^2}=\sqrt[3]{4^3}\) = 4
∴ R.F. of \(\sqrt[3]{4}\) is \(\sqrt[3]{4^2}\) or \(\sqrt[3]{16}\).

(iv) \(\sqrt[3]{16}\)
Simplest form of \(\sqrt[3]{16}\) is 2\(\sqrt[3]{2}\)
Now, R.F. of \(\sqrt[3]{2}\) is \(\sqrt[3]{2^2}\)
∴ R.F. of \(\sqrt[3]{16}\) is \(\sqrt[3]{2^2}\) or \(\sqrt[3]{4}\).

(v) \(\sqrt[4]{162}\)
Simplest form of \(\sqrt[4]{162}\) is 3\(\sqrt[4]{2}\)
Now, R.F. of \(\sqrt[4]{2}\) is \(\sqrt[4]{2^3}\) or \(\sqrt[4]{8}\)
R.E. of \(\sqrt[4]{162}\) is \(\sqrt[4]{2^3}\) or \(\sqrt[4]{8}\)

(vi) 2 + \(\sqrt{3}\)
As (2 + \(\sqrt{3}\))(2 – \(\sqrt{3}\)) = (2)² – (\(\sqrt{3}\))²
= 4 – 3 = 1, which is rational.
∴ R.F. of 2 + \(\sqrt{3}\) is 2 – \(\sqrt{3}\).

(vii) 7 – 4\(\sqrt{3}\)
As (7 – 4\(\sqrt{3}\))(7 + 4\(\sqrt{3}\))
= (7)² – (4\(\sqrt{3}\))² = 49 – 48
= 1, which is rational
∴ R.E. of 7 – 4\(\sqrt{3}\) is 7 + 4\(\sqrt{3}\).

(viii) 3\(\sqrt{3}\) + 2\(\sqrt{2}\)
As (3\(\sqrt{3}\) + 2\(\sqrt{2}\))(3\(\sqrt{3}\) – 2\(\sqrt{2}\))
= (3\(\sqrt{3}\))² – (2\(\sqrt{2}\))² = 27 – 8
= 19, which is rational
∴ R.F. of 3\(\sqrt{3}\) + 2\(\sqrt{2}\) is 3\(\sqrt{3}\) – 2\(\sqrt{2}\)

(ix) \(\sqrt[3]{3}\) + \(\sqrt[3]{2}\)
As (\(\sqrt[3]{3}\) + \(\sqrt[3]{2}\))(\(\sqrt[3]{3^2}\) – \(\sqrt[3]{3}\) × \(\sqrt[3]{2}\) + \(\sqrt[3]{2^2}\))
= (\(\sqrt[3]{3}\))³ + (\(\sqrt[3]{2}\))³ = 3 + 2
= 5, which is rational
∴ R.F. of \(\sqrt[3]{3}\) + \(\sqrt[3]{2}\)
is (\(\sqrt[3]{3^2}\) – \(\sqrt[3]{3}\) × \(\sqrt[3]{2}\) + \(\sqrt[3]{2^2}\)).

Question 12.
Express the following surd with a rational denominator.
\(\frac{8}{\sqrt{15}+1-\sqrt{5}-\sqrt{3}}\)
Solution :

Question 13.
Rationalise the denominator of \(\frac{a^2}{\sqrt{a^2+b^2}+b}\)
Solution :

Question 14.
If \(\frac{3+2 \sqrt{2}}{3-\sqrt{2}}\) = a + b\(\sqrt{2}\), where a and b are rationals then find the values of a and b.
Solution :

Equating the rational and irrational parts, we get
a = \(\frac {13}{7}\) , b = \(\frac {9}{7}\)

Question 15.
If \(\sqrt{3}\) = 1.732, find the value of \(\frac{1}{\sqrt{3}-1}\).
Solution :

Question 16.
If \(\sqrt{5}\) = 2.236 and \(\sqrt{2}\) = 1.414, then evaluate : \(\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{4}{\sqrt{5}-\sqrt{2}}\)
Solution :

Question 17.
If x = \(\frac{1}{2+\sqrt{3}}\), find the value of x³ – x² – 11x + 3.
Solution :
As, x = \(\frac{1}{2+\sqrt{3}}\) = 2 – \(\sqrt{3}\)
⇒ x – 2 = –\(\sqrt{3}\)
⇒ (x – 2)² = (-\(\sqrt{3}\))²
[By Squaring both sides]
⇒ x² + 4 – 4x = 3
⇒ x² – 4x + 1 = 0
Now, x³ – x² – 11x + 3
= x³ – 4x² + x + 3x² – 12x + 3
= x (x² – 4x + 1) + 3 (x² – 4x + 1)
= x(0) + 3 (0) = 0 + 0 = 0

Question 18.
If x = 3 – \(\sqrt{8}\) , find the value of x³ + \(\frac{1}{x^3}\).
Solution :

Question 19.
If x = 1 + 2^1/3 + 2^2/3, show that x³ – 3x² – 3x – 1 = 0
Solution :
x = 1 + 2^1/3 + 2^2/3
⇒ x – 1 = (2^1/3 + 2^2/3)
⇒ (x – 1)³ = (2^1/3 + 2^2/3)³
⇒ (x – 1)³
⇒ (2^1/3)³ + (2^2/3)³ + 3.2^1/3 × 2^1/3(2^1/3 + 2^2/3)
⇒ (x – 1)³ = 2 + 2² + 3.2¹ (x – 1)
⇒ (x – 1)³ = 6 + 6(x – 1)
⇒ x³ – 3x² + 3x – 1 = 6x
⇒ x³ – 3x² – 3x – 1 = 0

Question 20.
Solve : \(\sqrt{x+3}+\sqrt{x-2}\) = 5
Solution :
⇒ \(\sqrt{x+3}+\sqrt{x-2}\) = 5
⇒ \(\sqrt{x+3}\) = 5 – \(\sqrt{x-2}\)
⇒ (\(\sqrt{x+3}\))² = (5 – \(\sqrt{x-2}\))²
[By squaring both sides]
⇒ x + 3 = 25 + (x – 2) – 10\(\sqrt{x-2}\)
⇒ x + 3 = 25 + x – 2 – 10\(\sqrt{x-2}\)
⇒ 3 – 23 = – 10\(\sqrt{x-2}\)
⇒ – 20 = – 10\(\sqrt{x-2}\)
⇒ 2 = \(\sqrt{x-2}\)
⇒ x – 2 = 4 [By squaring both sides]
⇒ x = 6

Question 21.
If x = 1 + \(\sqrt{2}\) + \(\sqrt{3}\), prove that x⁴ – 4x³ – 4x² + 16 – 8 = 0.
Solution :
x = 1 + \(\sqrt{2}\) + \(\sqrt{3}\)
⇒ x – 1 = \(\sqrt{2}\) + \(\sqrt{3}\)
⇒ (x – 1)² = (\(\sqrt{2}\) + \(\sqrt{3}\))²
[By squaring both sides]
⇒ x² + 1 – 2x = 2 + 3 + 2\(\sqrt{6}\)
⇒ x² – 2x – 4 = 2\(\sqrt{6}\)
⇒ (x² – 2x – 4)² = (2\(\sqrt{6}\))²
⇒ x⁴ + 4x² + 16 – 4x³ + 16x – 8x² = 24
⇒ x⁴ – 4x³ – 4x² + 16x + 16 – 24 = 0
⇒ x⁴ – 4x³ – 4x² + 16x – 8 = 0

Question 22.
Evaluate each of the following:
(i) 5² × 5⁴
(ii) 5⁸ ÷ 5³
(iii) (3²)³
(iv) (\(\frac {11}{12}\))³
(v) (\(\frac {3}{4}\))^-3
Solution :
Using the laws of indices, we have
(i) 5² × 5⁴ = 5²⁺⁴ = 5⁶ = 15625
[∵ a^m × aⁿ = a^m+n]

(ii) 5⁸ ÷ 5³ = \(\frac{5^8}{5^3}\) = 5^8-3 = 5⁵ = 3125
[∵ a^m ÷ aⁿ = a^m-n]

(iii) (3²)³ = 3^2×3 = 3⁶ = 729
[∵ (a^m)ⁿ = a^m×n]

Question 23.
Evaluate each of the following.
(i) (\(\frac {2}{11}\))⁴ × (\(\frac {11}{3}\))² × (\(\frac {3}{2}\))³
(ii) (\(\frac {1}{2}\))⁵ × (\(\frac {-2}{3}\))⁴ × (\(\frac {3}{5}\))^-1
(iii) 2⁵⁵ × 2⁶⁰ – 2⁹⁷ × 2¹⁸
(iv) (\(\frac {2}{3}\))³ × (\(\frac {2}{5}\))^-3 × (\(\frac {3}{5}\))²
Solution :
(i) We have,
(\(\frac {2}{11}\))⁴ × (\(\frac {11}{3}\))² × (\(\frac {3}{2}\))³

Question 24.
Simplify :
(i) \(\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}\)
(ii) \(\frac{16 \times 2^{n+1}-4 \times 2^n}{16 \times 2^{n+2}-2 \times 2^{n+2}}\)
Solution :
We have

Question 25.
Simplify :
\(\left(\frac{81}{16}\right)^{-3 / 4} \times\left[\left(\frac{25}{9}\right)^{-3 / 2} \div\left(\frac{5}{2}\right)^{-3}\right]\)
Solution :

Multiple Choice Questions

Question 1.
If x = 3 + \(\sqrt{8}\) and y = 3 – \(\sqrt{8}\) then \(\frac{1}{x^2}+\frac{1}{y^2}\) =
(a) – 34
(b) 34
(c) 12\(\sqrt{8}\)
(d) – 12\(\sqrt{8}\)
Solution :
(b) 34

Question 2.
If \(\frac{3+\sqrt{7}}{3-\sqrt{7}}\) = a + b\(\sqrt{7}\) then (a, b) =
(a) (8, -3)
(b) (-8, -3)
(c) (-8, 3)
(d) (8, 3)
Solution :
(d) (8, 3)

Question 3.
Which of the following is an irrational number ?
(a) 0.24
(b) \(0 . \overline{24}\)
(c) 0.5777….
(d) 0.242242224…
Solution :
(d) 0.242242224…

Question 4.
If x = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) and y = 1, then value of \(\frac{x-y}{x-3 y}\) is :
(a) \(\frac{5}{\sqrt{5}-4}\)
(b) \(\frac{5}{\sqrt{6}+4}\)
(c) \(\frac{\sqrt{6}-4}{5}\)
(d) \(\frac{\sqrt{6}+4}{5}\)
Solution :
(d) \(\frac{\sqrt{6}+4}{5}\)

Question 5.
\(\sqrt{2}\) is a/an _________ number.
(a) natural
(b) whole
(c) irrational
(d) integer
Solution :
(c) irrational

Question 6.
The value of \(\sqrt[5]{(32)^{-3}}\) is:
(a) 1/8
(b) 1/16
(c) 1/32
(d) None
Solution :
(a) 1/8

Question 7.
If x = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) and y = \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\) then value of x² + xy + y² is :
(a) 99
(b) 100
(c) 1
(d) 0
Solution :
(a) 99

Question 8.
Which of the following is not a rational number?
(a) 2 + \(\sqrt{3}\)
(b) – \(\frac {7}{11}\)
(c) 3.42
(d) \(0 .23 \overline{4}\)
Solution :
(a) 2 + \(\sqrt{3}\)

Question 9.
Which of the following is smallest?
(a) \(\sqrt[4]{5}\)
(b) \(\sqrt[5]{4}\)
(c) \(\sqrt{4}\)
(d) \(\sqrt{3}\)
Solution :
(b) \(\sqrt[5]{4}\)

Question 10.
The product of \(\sqrt{3}\) and \(\sqrt[3]{5}\) is:
(a) \(\sqrt[6]{375}\)
(b) \(\sqrt[6]{675}\)
(c) \(\sqrt[6]{575}\)
(d) \(\sqrt[6]{475}\)
Solution :
(b) \(\sqrt[6]{675}\)

Question 11.
The exponential form of \(\sqrt{\sqrt{2}} \times \sqrt{2} \times \sqrt{2}\) is :
(a) 2^1/16
(b) 8^3/4
(c) 2^3/4
(d) 8^1/2
Solution :
(c) 2^3/4

Question 12.
The value of x, if 5^x-3 × 3^2x-8 = 225, is:
(a) 1
(b) 2
(c) 3
(d) 5
Solution :
(d) 5

Question 13.
If 2^5x ÷ 2^x = \(\sqrt[5]{2^{20}}\) then x =
(a) 0
(b) – 1
(c) \(\frac {1}{2}\)
(d) 1
Solution :
(d) 1

Question 14.
If 10x = \(3 . \overline{3}\) = 3 + x, then x =
(a) \(\frac {1}{9}\)
(b) \(\frac {1}{3}\)
(c) 3
(d) 9
Solution :
(b) \(\frac {1}{3}\)

Question 15.
A rational number between \(\frac {1}{7}\) and \(\frac {1}{3}\) is
(a) \(\frac {29}{210}\)
(b) \(\frac {50}{210}\)
(c) \(\frac {81}{210}\)
(d) \(\frac {93}{210}\)
Solution :
(b) \(\frac {50}{210}\)

Students should go through these JAC Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 5 Introduction to Euclid’s Geometry

Introduction:
The credit for introducing geometrical concepts goes to the distinguished Greek mathematician ‘Euclid’ who is known as the “Father of Geometry” and the word geometry comes from the Geek words ‘geo’ which means “Earth’ and ‘Metreon’ which means ‘measure’.

Basic Concepts In Geometry:
A point, a ‘line’ and a plane are the basic concepts to be used in geometry.
→ Axioms:
The statement that is taken to be true without proof, to serve as a premise for further reasoning and arguments, are called axioms.

Euclid’s Definitions:

• A point is that which has no part.
• A line is breadthless length.
• The ends of a line segment are points.
• A straight line is that which has length only.
• A surface is that which has length and breadth only.
• The edges of surface are lines.
• A plane surface is a surface which lies evenly with the straight lines on itself.

Euclid’s Five Postulates:
→ A straight line may be drawn from any one point to any other point.
→ A terminated line or a line segment can be produced infinitely.

→ A circle can be drawn with any centre and of any radius.
→ All right angles are equal to one another.
→ If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced infinitely, meet on that side on which the sum of angles is less than two right angles.

Important Axioms:
→ A line is the collection of infinite number of points.
→ Through a given point, infinite lines can be drawn.

→ Given two distinct points, there is one and only one line that contains both the points.
→ If P is a point not lying on a line l, then one and only one line can be drawn through P which is parallel to l.

→ Two distinct lines cannot have more than one point in common.

→ Two lines which are both parallel to the same line, are parallel to each other. i.e. IF l || n, m || n ⇒ l || m.

Some Important Definitions:
→ Collinear points: Three or more points are said to be collinear if there is a line which contains all of them.

→ Concurrent lines: Three or more lines are said to be concurrent if there is a point which lies on all of them.

→ Intersecting lines: Two lines are intersecting if they have a common point. The common point is called the “point of intersection”.

→ Parallel lines: Two lines I and m in a plane are said to be parallel lines if they do not have a common point.

→ Line segment: Given two points A and B on a line l, the connected part (segment) of the line with end points at A and B is called the line segment AB.

→ Interior point of a line segment: A point R is called an interior point of a line segment PQ if R lies between Pand O but Ris neither P nor Q.

→ Congruence of line segment: Two line segments AB and CD are congruent if trace copy of one can be superposed on the other so as to cover it completely and exactly in this case we write AB ≅ CD. In other words we can say two lines are congruent if their lengths are same.
→ Distance between two points: The distance between two points P and Q is the length of the line segment PO.
→ Ray: Directed line segment is called a ray. If AB is a ray, then it is denoted by \(\overrightarrow{\mathrm{AB}}\). Point A is called initial point of ray.

→ Opposite rays: Two rays AB and AC are said to be opposite rays if they are collinear and point A is the only common point of the two rays and A lies in between B and C.

Theorem 1.
If l, m, n are lines in the same plane such that l intersects m and n || m, then l also intersects n.
Answer:
Given: Three lines l, m, n in the same plane such that intersects m and n || m.
To prove: Lines land n are intersecting lines.

Proof: Let l and n be non intersecting lines. Then. l || n. But, n || m [Given]
∴ l || n and n || m
⇒ l || m
⇒ l and m are non-intersecting lines.
This is a contradiction to the hypothesis that I and m are intersecting lines. So our supposition is wrong.
Hence, l intersects line n.

Theorem 2.
If lines AB, AC, AD and AE are parallel to a line l, then points A, B, C, D and E are collinear.
Answer:
Given: Lines AB, AC, AD and AE are parallel to a line l.
To prove: A, B, C, D, E are collinear.
Proof: Since AB, AC, AD and AE are all parallel to a line l. Therefore point A is outside l and lines AB, AC, AD, AE are drawn through A and each line is parallel to l.
But by parallel lines axiom, one and only one line can be drawn through the point A outside a line l and parallel to it.
This is possible only when A, B, C, D and E all lie on the same line. Hence, A, B, C, D and E are collinear.

JAC Jharkhand Board Class 9th Maths Solutions in Hindi & English Medium

JAC Board Class 9th Maths Solutions in English Medium

JAC Class 9 Maths Chapter 1 Number Systems

• Chapter 1 Number Systems Ex 1.1
• Chapter 1 Number Systems Ex 1.2
• Chapter 1 Number Systems Ex 1.3
• Chapter 1 Number Systems Ex 1.4
• Chapter 1 Number Systems Ex 1.5
• Chapter 1 Number Systems Ex 1.6

JAC Class 9 Maths Chapter 2 Polynomials

• Chapter 2 Polynomials Ex 2.1
• Chapter 2 Polynomials Ex 2.2
• Chapter 2 Polynomials Ex 2.3
• Chapter 2 Polynomials Ex 2.4
• Chapter 2 Polynomials Ex 2.5

JAC Class 9 Maths Chapter 3 Coordinate Geometry

• Chapter 3 Coordinate Geometry Ex 3.1
• Chapter 3 Coordinate Geometry Ex 3.2
• Chapter 3 Coordinate Geometry Ex 3.3

JAC Class 9 Maths Chapter 4 Linear Equations in Two Variables

• Chapter 4 Linear Equations in Two Variables Ex 4.1
• Chapter 4 Linear Equations in Two Variables Ex 4.2
• Chapter 4 Linear Equations in Two Variables Ex 4.3
• Chapter 4 Linear Equations in Two Variables Ex 4.4

JAC Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry

• Chapter 5 Introduction to Euclid’s Geometry Ex 5.1
• Chapter 5 Introduction to Euclid’s Geometry Ex 5.2

JAC Class 9 Maths Chapter 6 Lines and Angles

• Chapter 6 Lines and Angles Ex 6.1
• Chapter 6 Lines and Angles Ex 6.2
• Chapter 6 Lines and Angles Ex 6.3

JAC Class 9 Maths Chapter 7 Triangles

• Chapter 7 Triangles Ex 7.1
• Chapter 7 Triangles Ex 7.2
• Chapter 7 Triangles Ex 7.3
• Chapter 7 Triangles Ex 7.4
• Chapter 7 Triangles Ex 7.5

JAC Class 9 Maths Chapter 8 Quadrilaterals

• Chapter 8 Quadrilaterals Ex 8.1
• Chapter 8 Quadrilaterals Ex 8.2

JAC Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

• Chapter 9 Areas of Parallelograms and Triangles Ex 9.1
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.2
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.3
• Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

JAC Class 9 Maths Chapter 10 Circles

• Chapter 10 Circles Ex 10.1
• Chapter 10 Circles Ex 10.2
• Chapter 10 Circles Ex 10.3
• Chapter 10 Circles Ex 10.4
• Chapter 10 Circles Ex 10.5
• Chapter 10 Circles Ex 10.6

JAC Class 9 Maths Chapter 11 Constructions

• Chapter 11 Constructions Ex 11.1
• Chapter 11 Constructions Ex 11.2

JAC Class 9 Maths Chapter 12 Heron’s Formula

• Chapter 12 Heron’s Formula Ex 12.1
• Chapter 12 Heron’s Formula Ex 12.2

JAC Class 9 Maths Chapter 13 Surface Areas and Volumes

• Chapter 13 Surface Areas and Volumes Ex 13.1
• Chapter 13 Surface Areas and Volumes Ex 13.2
• Chapter 13 Surface Areas and Volumes Ex 13.3
• Chapter 13 Surface Areas and Volumes Ex 13.4
• Chapter 13 Surface Areas and Volumes Ex 13.5
• Chapter 13 Surface Areas and Volumes Ex 13.6
• Chapter 13 Surface Areas and Volumes Ex 13.7
• Chapter 13 Surface Areas and Volumes Ex 13.8
• Chapter 13 Surface Areas and Volumes Ex 13.9

JAC Class 9 Maths Chapter 14 Statistics

• Chapter 14 Statistics Ex 14.1
• Chapter 14 Statistics Ex 14.2
• Chapter 14 Statistics Ex 14.3
• Chapter 14 Statistics Ex 14.4

JAC Class 9 Maths Chapter 15 Probability

• Chapter 15 Probability Ex 15.1

JAC Board Class 9th Maths Solutions in Hindi Medium

JAC Class 9 Maths Chapter 1 संख्या पद्धति

• Chapter 1 संख्या पद्धति Ex 1.1
• Chapter 1 संख्या पद्धति Ex 1.2
• Chapter 1 संख्या पद्धति Ex 1.3
• Chapter 1 संख्या पद्धति Ex 1.4
• Chapter 1 संख्या पद्धति Ex 1.5
• Chapter 1 संख्या पद्धति Ex 1.6

JAC Class 9 Maths Chapter 2 बहुपद

• Chapter 2 बहुपद Ex 2.1
• Chapter 2 बहुपद Ex 2.2
• Chapter 2 बहुपद Ex 2.3
• Chapter 2 बहुपद Ex 2.4
• Chapter 2 बहुपद Ex 2.5

JAC Class 9 Maths Chapter 3 निर्देशांक ज्यामिति

• Chapter 3 निर्देशांक ज्यामिति Ex 3.1
• Chapter 3 निर्देशांक ज्यामिति Ex 3.2
• Chapter 3 निर्देशांक ज्यामिति Ex 3.3

JAC Class 9 Maths Chapter 4 दो चरों वाले रैखिक समीकरण

• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.1
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.2
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.3
• Chapter 4 दो चरों वाले रैखिक समीकरण Ex 4.4

JAC Class 9 Maths Chapter 5 युक्लिड के ज्यामिति का परिचय

• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.1
• Chapter 5 युक्लिड के ज्यामिति का परिचय Ex 5.2

JAC Class 9 Maths Chapter 6 रेखाएँ और कोण

• Chapter 6 रेखाएँ और कोण Ex 6.1
• Chapter 6 रेखाएँ और कोण Ex 6.2
• Chapter 6 रेखाएँ और कोण Ex 6.3

JAC Class 9 Maths Chapter 7 त्रिभुज

• Chapter 7 त्रिभुज Ex 7.1
• Chapter 7 त्रिभुज Ex 7.2
• Chapter 7 त्रिभुज Ex 7.3
• Chapter 7 त्रिभुज Ex 7.4
• Chapter 7 त्रिभुज Ex 7.5

JAC Class 9 Maths Chapter 8 चतुर्भुज

• Chapter 8 चतुर्भुज Ex 8.1
• Chapter 8 चतुर्भुज Ex 8.2

JAC Class 9 Maths Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल

• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.1
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.2
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.3
• Chapter 9 समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल Ex 9.4

JAC Class 9 Maths Chapter 10 वृत्त

• Chapter 10 वृत्त Ex 10.1
• Chapter 10 वृत्त Ex 10.2
• Chapter 10 वृत्त Ex 10.3
• Chapter 10 वृत्त Ex 10.4
• Chapter 10 वृत्त Ex 10.5
• Chapter 10 वृत्त Ex 10.6

JAC Class 9 Maths Chapter 11 रचनाएँ

• Chapter 11 रचनाएँ Ex 11.1
• Chapter 11 रचनाएँ Ex 11.2

JAC Class 9 Maths Chapter 12 हीरोन का सूत्र

• Chapter 12 हीरोन का सूत्र Ex 12.1
• Chapter 12 हीरोन का सूत्र Ex 12.2

JAC Class 9 Maths Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.1
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.2
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.3
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.4
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.5
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.6
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.7
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.8
• Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.9

JAC Class 9 Maths Chapter 14 सांख्यिकी

• Chapter 14 सांख्यिकी Ex 14.1
• Chapter 14 सांख्यिकी Ex 14.2
• Chapter 14 सांख्यिकी Ex 14.3
• Chapter 14 सांख्यिकी Ex 14.4

JAC Class 9 Maths Chapter 15 प्रायिकता

• Chapter 15 प्रायिकता Ex 15.1

Students should go through these JAC Class 9 Maths Notes Chapter 7 Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 7 Triangles

Triangle
A plane figure bounded by three lines in a plane is called a triangle. Every triangle has three sides and three angels. If ABC is any triangle then AB, BC and CA are three sides and ∠A, ∠B and ∠C are three angles.

Types of Triangles
→ On the basis of sides we have three types of triangles:

• Scalene triangle – A triangle whose no two sides are equal is called a scalene triangle.
• Isosceles triangle – A triangle having two sides equal is called an isosceles triangle.
• Equilateral triangle – A triangle in which all sides are equal is called an equilateral triangle.

→ On the basis of angles we have three types of triangles:

• Right triangle – A triangle in which any one angle is a right angle (= 90°) is called right triangle.
• Acute triangle – A triangle in which all angles are acute (0° >angle >90°) is called an acute triangle.
• Obtuse (90° < angle < 180° ) triangle – A triangle in which any one angle is obtuse is called an obtuse triangle.

Congruent Figures
The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height.

In the above figures {Fig. (i) and Fig. (ii)} both are equal in length, width and height, so these are congruent figures.

Congruent Triangles
Two triangles are congruent if and only if one of them can be made to superimposed on the other, so as to cover it exactly.

If two triangles ΔABC and ΔDEF are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities.
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB = DE, BC = EF, AC = DF.
If ΔABC and ΔDEF are congruent under one to one correspondence A ↔ D, B ↔ E, C ↔ F then we write ΔABC ≅ ΔDEF We cannot write it as ΔABC ≅ ΔDFE or ΔABC ≅ ΔEDF or in other forms because ΔABC ≅ ΔDFE have following one-one correspondence A ↔ D, B ↔ F, C ↔ E.
Hence, we can say that ‘two triangles are congruent if and only if there exists a oneone correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

Sufficient Conditions for Congruence of two Triangles
→ SAS Congruence Criterion:

Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.

→ ASA Congruence Criterion:

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

→ AAS Congruence Criterion:
If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

→ SSS Congruence Criterion:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

→ RHS Congruence Criterion:
Two right angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

→ Congruence Relation in the Set of all Triangles:
By the definition of congruence of two triangles, we have following results.

• Every triangle is congruent to itself i.e., ΔABC ≅ ΔABC
• If ΔABC ≅ ΔDEF then ΔDEF ≅ ΔABC
• If ΔABC ≅ ΔDEF and ΔDEF ≅ ΔΡQR then ΔΑΒC ≅ ΔΡQR

NOTE: If two triangles are congruent then their corresponding sides and angles are also congruent by CPCT (corresponding parts of congruent triangles are also congruent).

Theorem 1.
Angles opposite to equal sides of an isosceles triangle are equal.

Given:
ΔABC in which AB = AC
To Prove: ∠B = ∠C
Construction: We draw the bisector AD of ∠A which meets BC in D.
Proof: In ΔABD and ΔACD, we have
AB = AC [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
And, AD = AD [Common side]
∴ By SAS criterion of congruence, we have
ΔΑΒD ≅ ΔΑCD
⇒ ∠B = ∠C [by CPCT]
Hence, proved.

Theorem 2.
If two angles of a triangle are equal, then sides opposite to them are also equal.
Given: ΔABC in which ∠B = ∠C
To Prove: AB = AC
Construction: We draw the bisector of ∠A

which meets BC in D.
Proof: In ΔABD and ΔACD, we have
∠B = ∠C [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
AD = AD [Common side]
∴ By AAS criterion of congruence, we get
ΔΑΒD ≅ ΔΑCD
⇒ AB = AC [By CPCT] Hence, proved.

Theorem 3.
If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.

Given: ΔABC in which AD is the bisector of ∠A meeting BC in D such that BD = CD
To Prove: ΔABC is an isosceles triangle.
Construction: We produce AD to E such that AD = DE and join EC
Proof: In ΔADB and ΔEDC, we have
AD = DE [By construction]
∠ADB = ∠CDE [Vertically opposite angles]
BD = DC [Given]
∴ By SAS criterion of congruence, we get
ΔADR ≅ ΔEDC ⇒ AB = EC ……(i)
And, ∠BAD = ∠CED [By CPCT]
But, ∠BAD = ∠CAD
∴ ∠CAD = ∠CED
⇒ AC = EC [Sides opposite to equal angles are equal]
⇒ AC = AB [By eq. (i)] Hence, proved

Some Inequality Relations In A Triangle
→ If two sides of a triangle are unequal, then the longer side has greater angle opposite to it, i.e., if in any ΔABC, AB > AC then ∠C > ∠B.
→ In a triangle the greater angle has the longer side opposite to it, ie, if in any ΔABC, ∠A > ∠B then BC > AC.
→ The sum of any two sides of a triangle is always greater than the third side, i.e., in any ΔABC, AB + BC > AC, BC + CA > AB and AC + AB > BC.
→ Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

P is any point not lying on line l, PM ⊥ l then PM < PN.
→ The difference of any two sides of a triangle is less than the third side, i.e., in any ΔABC, AB – BC < AC, BC – CA < AB and AC – AB < BC.

Students should go through these JAC Class 9 Maths Notes Chapter 3 Coordinate Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 3 Coordinate Geometry

Co-Ordinate System:
In two dimensional coordinate geometry, we generally use two types of coordinate systems.

• Cartesian or Rectangular coordinate system.
• Polar coordinate system.

In cartesian coordinate system we represent any point by ordered pair (x, y) where x and y are called x and y coordinate of that point respectively.
In polar coordinate system we represent any point by ordered pair (r, θ) where is called radius vector and ‘θ’ is called vectorial angle of that point, which will be studied in higher classes.

Cartesian Coordinate System:
→ Rectangular Coordinate Axes:
Let XX’ and YY’ are two lines such that XX’ is horizontal and YY’ is vertical lines in the same plane and they intersect each other at O. This intersecting point is called origin Now choose a convenient unit of length and starting from origin as zero, mark off a number scale on the horizontal line XX’, positive to the right of origin O and negative to the left of origin O. Also mark off the same scale on the vertical line YY’, positive upwards and negative downwards of the origin. The line XX’ is called X-axis and the line YY’ is known as Y-axis and the two lines taken together are called the coordinate axes.

→ Quadrants:
The coordinates axes XX’ and YY’ divide the plane of graph paper into four parts XY, X’Y, X’Y’ and XY’. These four parts are called the quadrants. The parts XY, X’Y, X’Y’ and XY’ are known as the first second, third and fourth quadrants respectively.

→ Cartesian Coordinates of a Point:
Let -axis and y-axis be the coordinate axes and P be any point in the plane. To find the position of P with respect of x-axis and y-axis, we draw two perpendicular line segment from P on both coordinate axes.

Let PM and PN be the perpendiculars on x-axis and y-axis resepectively. The length of the line segment OM is called the x-coordinate or abscissa of point P. Similarly the length of line segment ON is called the y-coordinate or ordinate of point P.

Let OM = x and ON = y. The position of the point P in the plane with respect to the coordinate axes is represented by the ordered pair (x, y). The ordered pair (x, y) is called the coordinates of point P. “Thus, for a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively”.

The above system of coordinating of ordered pair (x, y) with every point in plane is called the Rectangular or Cartesian coordinate system.

Cartesian coordinate system

→ Convention of Signs:
As discussed earlier that regions XOY, Χ’ΟΥ, Χ’ΟΥ’ and ΧΟΥ’ are known as the first second, third and fourth quadrants respectively. The ray OX is taken as positive X-axis, OX’ as negative x-axis, OY as positive y-axis and OY as negative y-axis. Thus we have.
In first quadrant: x > 0, y > 0
In second quadrant: x < 0, y > 0
In third quadrant: x < 0, y < 0
In fourth quadrant: x > 0, y < 0

→ Points on Axis:
If point P lies on x-axis then clearly its distance from x-axis will be zero, therefore we can say that its ordinate will be zero. In general, if any point lies on x-axis then its y-coordinate will be zero. Similarly if any point Q lies on y-axis, then its distance from y-axis will be zero therefore we can say its x-coordinate will be zero. In general, if any point lies on y-axis then its x-coordinate will be zero.

→ Plotting of Points:
In order to plot the points in a plane, we may use the following algorithm.
Step I: Draw two mutually perpendicular lines on the graph paper, one horizontal and other vertical.
Step II: Mark their intersection point as O (origin).
Step III: Choose a suitable scale on X-axis and Y-axis and mark the points on both the axes.
Step IV: Obtain the coordinates of the point which is to be plotted. Let the point be P(a, b). To plot this point start from the origin and |a| units move along OX, OX’ according as ‘a’ is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward |b| units according as ‘b’ is positive or negative respectively The point where we arrive finally is the required point P(a, b).

Distance Between Two Points:
→ If there are two points A (x₁, y₁) and B(x₂, y₂) on the XY plane, the distance between them is given by
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Jharkhand Board JAC Class 9 Maths Important Questions Chapter 4 Linear Equations in Two Variables Important Questions and Answers.

JAC Board Class 9th Maths Important Questions Chapter 4 Linear Equations in Two Variables

Question 1.
Prove that x = 3, y = 2 is a solution of 3x – 2y = 5.
Solution :
x = 3, y = 2 is a solution of 3x – 2y = 5, because L.H.S. = 3x – 2y = 3 × 3 – 2 × 2 = 9 – 4 = 5 = R.H.S.
i.e. x = 3, y = 2 satisfies the equation 3x – 2y = 5.
∴ It is a solution of the given equation.

Question 2.
Prove that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.
Solution :
Given eq, is 4x – y – 3 = 0 …(i)
First we put x = 1, y = 1 in L.H.S. of eq (i)
Here L.H.S. = 4x – y – 3 = 4 × 1 – 1 – 3 = 4 – 4 = 0 = R.H.S.
Now we put x = 2, y = 5 in eq. (i)
L.H.S. = 4x – y – 3 = 4 × 2 – 5 – 3 = 8 – 8 = 0 = R.H.S.
Since, x = 1, y = 1 and x = 2, y = 5, both pairs satisfied the given equation, therefore they are the solutions of given equation.

Question 3.
Determine whether x = 2, y = – 1 is a solution of equation 3x + 5y – 2 = 0.
Solution :
Given eq. is 3x + 5y – 2 = 0 ….(i)
Taking L.H.S. = 3x + 5y – 2 = 3 × 2 + 5 × (-1) – 2 = 6 – 5 – 2 = – 1 + 0 = R.H.S.
Here LH.S. ≠ R.H.S. therefore x = 2, y = – 1 is not a solution of given equation.

Question 4.
Draw the graph of
(i) 2x + 5 = 0
(ii) 3y – 15 = 0

Solution :
(i) Graph of 2x + 5 = 0
On simplifying it we get 2x = – 5
x = – \(\frac {5}{2}\)
First we plot point A₁ (-\(\frac {5}{2}\), 0) and then we plot any other point A₂ (-\(\frac {5}{2}\), 2) on the graph paper, then we join these two points we get required line l as shown in adjoining figure

(ii) Graph of 3y – 15 = 0
On simplifying it we get 3y = 15 y = 5.
⇒ y = \(\frac {15}{3}\) = 5
First we plot the point B₁(0, 5) and then we plot any other point B₂(3, 5) on the graph paper, then we join these two points we get required line m as shown in figure.
Note: A point which lies on the line is a solution of that equation. A point not lying on the line is not a solution of the equation.

Question 5.
Draw the graph of the line x – 2y = 3, from the graph find the coordinates of the point when
(i) x = – 5
(ii) y = 0

Solution :
Here given equation is x – 2y = 3.
Solving it for y we get 2y = x – 3
⇒ y = \(\frac {1}{2}\)x – \(\frac {3}{2}\)
Let x = 0, then y = \(\frac {1}{2}\)(0) – \(\frac {3}{2}\) = – \(\frac {3}{2}\)
x = 3, then y = \(\frac {1}{2}\)(3) – \(\frac {3}{2}\) = 0
x = – 2, then y Hence, we get y = \(\frac {1}{2}\)(-2) – \(\frac {3}{2}\) = – \(\frac {5}{2}\)
Hence, we get

x	0	3	– 2
Y	–\(\frac {3}{2}\)	0	–\(\frac {5}{2}\)

Clearly from the graph, when x = -5 then y = -4 so corresponding coordinates are (-5, -4) and when y = 0 then x = 3, so corresponding coordinates are (3, 0).

Question 6.
Draw the graphs of the lines represented by the equations x + y = 4 and 2x – y = 2 in the same graph. Also find the coordinates of the point where the two lines intersect.

Solution :
Given equations are x + y = 4 ………….(i)
and 2x – y = 2 ….. (ii)
(i) We have, y = 4 – x

x	0	2	4
Y	4	2	0

(ii) We have, y = 2x – 2

x	1	0	3
Y	0	– 2	4

By drawing the lines on a graph paper, clearly we can say that P is the point of intersection where coordinates are x = 2, y = 2, i.e., P(2, 2).

Question 7.
Solve : \(\frac {x}{2}\) = 3 + \(\frac {x}{3}\)
Solution :
Given \(\frac {x}{2}\) = 3 + \(\frac {x}{3}\)
⇒ \(\frac {x}{2}\) – \(\frac {x}{3}\) = 3
⇒ \(\frac{3 x-2 x}{6}\) = 3
⇒ \(\frac {x}{6}\) = 3
⇒ x = 18

Question 8.
Solve the following system of equations:
2x – 3y = 5
3x + 2y = 1
Solution :
Given eq. are 2x – 3y = 5 ………(i)
and 3x + 2y = 1 ……………(ii)
Multiplying eq. (i) by 3 and eq. (ii) by 2. we get 6x – 9y = 15 and 6x + 4y = 2 respectively.

⇒ -13y = 13
⇒ y = – 1
Putting the value of y in eq. (i) we get
2x – (3) × (-1) = 5
2x + 3 = 5
⇒ 2x = 5 – 3
⇒ 2x = 2
⇒ x = 1
∴ x = 1, y = – 1 is the solution of given system of linear equations.

Question 9.
Solve the following system of equations:
x + 4y = 14
7x – 3y = 5
Solution :
Let x + 4y = 14 ……..(i)
and 7x – 3y = 5 ……….(ii)
From equation (i)
x = 14 – 4y
Substitute the value of x in equation (ii)
⇒ 7(14 – 4y) – 3y = 5
⇒ 98 – 28y – 3y = 5
⇒ 98 – 31y = 5
⇒ 93 = 3ly
⇒ y = \(\frac {93}{31}\)
⇒ y = 3
x = 14 – 4y = 14 – 4 × 3 = 14 – 12 = 2
So, solution is x = 2 and y = 3.

Multiple Choice Questions

Question 1.
Which of the following equations is not a linear equation?
(a) 2x + 3 = 7x – 2
(b) \(\frac {2}{3}\)x + 5 = 3x – 4
(c) x² + 3 = 5x – 3
(d) (x – 2)² = x² + 8
Solution :
(c) x² + 3 = 5x – 3

Question 2.
Solution of equation \(\sqrt{3}\)x – 2 = 2\(\sqrt{3}\) + 4 is
(a) 2(\(\sqrt{3}\) – 1)
(b) 2(1 – \(\sqrt{3}\))
(c) 1 + \(\sqrt{3}\)
(d) 2(1 + \(\sqrt{3}\))
Solution :
(d) 2(1 + \(\sqrt{3}\))

Question 3.
The value of x which satisfies \(\frac{6 x+5}{4 x+7}=\frac{3 x+5}{2 x+6}\) is :
(a) -1
(b) 1
(c) 2
(d) -2
Solution :
(b) 1

Question 4.
Solution of \(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}\) = 3 is
(a) a + b – c
(b) a – b + c
(c) – a + b + c
(d) a + b + c
Solution :
(d) a + b + c

Question 5.
One-fourth of one-third of one-half of a number is 12, then number is
(a) 284
(b) 286
(c) 288
(d) 290
Solution :
(c) 288

Question 6.
A linear equation in two variables has maximum
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) None of these
Solution :
(c) infinitely many solutions

Question 7.
Solutions of the equation x – 2y = 2 is/are
(a) x = 4, y = 1
(b) x = 2, y = 0
(c) x = 6, y = 2
(d) All of these
Solution :
(d) All of these

Question 8.
The graph of line 5x + 3y = 4 cuts y-axis at the point
(a) (0, \(\frac {4}{3}\))
(b) (0, \(\frac {3}{4}\))
(c) (\(\frac {4}{3}\), 0)
(d) (\(\frac {4}{3}\), 0)
Solution :
(a) (0, \(\frac {4}{3}\))

Question 9.
If x = 1, y = 1 is a solution of equation 9ax + 12ay = 63, then the value of a is
(a) – 3
(b) 3
(c) 7
(d) 5
Solution :
(b) 3

Jharkhand Board JAC Class 9 Maths Solutions Chapter 3 Coordinate Geometry Ex 3.3 Textbook Exercise Questions and Answers.

JAC Board Class 9th Maths Solutions Chapter 3 Coordinate Geometry Ex 3.3

Page-65

Question 1.
In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1,2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
Answer:

(-2, 4) → Second quadrant
(3,-1) → Fourth quadrant
(-1,0) → x-axis
(1,2) → First quadrant
(-3, -5)→ Third quadrant

Question 2.
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

x	-2	-1	0	1	3
y	8	7	-1.25	3	-1

Ans. Points (x, y) on the plane, 1 unit = 1 cm