Lines and Angles Class 9 Worksheet have been designed as per the latest pattern for CBSE, NCERT and KVS for Grade 9. Students are always suggested to solve printable worksheets for Mathematics Lines and Angles Grade 9 as they can be really helpful to clear their concepts and improve problem solving skills. We at worksheetsbag.com have provided here free PDF worksheets for students in standard 9 so that you can easily take print of these test sheets and use them daily for practice. All worksheets are easy to download and have been designed by teachers of Class 9 for benefit of students and is available for free download.
Mathematics Lines And Angles Worksheets for Class 9
We have provided chapter-wise worksheets for class 9 Mathematics Lines And Angles which the students can download in Pdf format for free. This is the best collection of Mathematics Lines And Angles standard 9th worksheets with important questions and answers for each grade 9th Mathematics Lines And Angles chapter so that the students are able to properly practice and gain more marks in Class 9 Mathematics Lines And Angles class tests and exams.
Chapter-wise Class 9 Mathematics Lines And Angles Worksheets Pdf Download
1. Basic Terms and Definitions
2. Intersecting Lines and Non-Intersecting Lines
3. Pairs of Angles
4. Parallel Lines and a Transversal
5. Lines Parallel to the same Line
6. Angle Sum Property of a Triangle
(1) Point- We often represent a point by a fine dot made with a fine sharpened pencil on a piece of paper.
(2) Line- A line is completely known if we are given any two distinct points. Line AB is
(3) Line segment- A part (or portion) of a line with two end points is called a line segment
(4) Ray- A part of line with one end point is called a ray.
(5) Collinear points- If three or more points lie on the same line, they are called collinear points otherwise they are called non-collinear points.
Types of Angles-
(1) Acute angle – An acute angle measure between 0o and 90o
(2) Right angle – A right angle is exactly equal to 90o
(3) Obtuse angle – An angle greater than 90o but less than 180o
(4) Straight angle – A straight angle is equal to 180o
(5) Reflex angle – An angle which is greater than 180o but less than 360o is called a reflex angle.
(6) Complementary angles – Two angles whose sum is 90o are called complementary angles.
(7) Supplementary angle – Two angles whose sum is 180o are called supplementary angles.
(8) Adjacent angles – Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of common arm
(9) Linear pair- Two angles form a linear pair, if their non-common arms form a line.
(10) Vertically opposite angles- Vertically opposite angles are formed when two lines intersect each other at a point.
(a) Corresponding angles
(b) Alternate interior angles
(c) Alternate exterior angles
(d) Interior angles on the same side of the transversal
• If a transversal intersects two parallel lines, then
(i) each pair of corresponding angles is equal.
(ii) each pair of alternate interior angles is equal.
(iii) each pair of interior angle on the same side of the transversal is supplementary.
• If a transversal interacts two lines such that, either
(i) any one pair of corresponding angles is equal, or
(ii) any one pair of alternate interior angles is equal or
(iii) any one pair of interior angles on the same side of the transversal is supplementary then the lines are parallel.
• Lines which are parallel to a given line are parallel to each other.
• The sum of the three angles of a triangle is 180o
• If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
Question. If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle
Question. The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(a) an acute angled triangle
(b) an obtuse angled triangle
(c) a right triangle
(d) an isosceles triangle
Question. An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is
(a) 37 (10/2)
(b) 52 (10/2)
(c) 72 (10/2)
Question. In Fig. 6.3, if OP||RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to
Question. In Fig. 6.1, if AB || CD || EF, PQ || RS, ∠RQD= 25° and ∠CQP = 60°, then ∠QRS is equal to
Question. If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
Question. Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
Question. In Fig. 6.2, POQ is a line. The value of x is
Question. Can a triangle have two obtuse angles? Give reason for your answer.
Answer. An angle whose measure is more than 90o but less than 180o is called an obtuse angle. A triangle cannot have two obtuse angles because the sum of all the angles of it cannot be more than 180o. It is always equal to 180o.
Question. For what value of x + y in Fig. 6.4 will ABC be a line? Justify your answer.
Answer. In the given figure, x and y are two adjacent angles. For ABC to be a straight line, the sum of two adjacent angles x and y must be 1800 .
Question. How many triangles can be drawn having is angles as 53°, 64° and 63°? Give reason for your answer.
Answer. Sum of these angles = 53° + 64° + 63° = 180°. So, we can draw infinitely many triangles, sum of the angles of every triangle having its angles as 53°, 64° and 63° is 180°.
Question. Can a triangle have all angles less than 60°? Give reason for your answer.
Answer. A triangle cannot have all angle less than 60°. Then, sum of all the angles will be less than 180° whereas sum of all the angles of a triangle is always 180°.
Question. Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.
Answer. When two lines l and m are perpendicular to the same line n, each of the two corresponding angles formed by these lines l and m with the line n are equal (each is equal to 90o). Hence, the line l and m are parallel.
Question. How many triangles can be drawn having its angles as 45°, 64° and 72°? Give reason for your answer.
Answer. We cannot draw any triangle having its angles 45°, 64° and 72° because the sum of the angles (45° + 64° + 72° =181°) cannot be 181°.
Question. In Fig. 6.5, find the value of x for which the lines l and m are parallel.
Answer. If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary. Here, the two given lines l and m are parallel.
Angles x and 44o, are consecutive interior angles on the same side of the transversal.
Therefore, x + 440 =1800
Hence, x =1800 − 440 =1360
Question. If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.
Answer. If two intersect each other at a point, then four angles are formed. If one of these four angles is a right angle, then each of the other three angles will also be a right by linear pair axiom.
Question. In Fig.6.6, which of the two lines are parallel and why?
Answer. For fig(i), a transversal intersects two lines such that the sum of interior angles on the same side on the same side of the transversal is 132o + 48o = 180o.
Therefore, the line l and m are parallel.
For fig. (ii), a transversal intersects two line such that the sum of interior angles on the same sides of the transversal is 73o + 106o = 179o.
Therefore, the lines p and q are not parallel.
Question. Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.
Answer. No, each of these angles will be a right angle only when they form a linear pair, i.e., when the non-common arms of the given two adjacent angles are two opposite rays.
Question. In Fig. 6.10, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.
Answer. We have,
∠5 + ∠6 =1800 [Angles pf a linear pair]
⇒ ∠5 + 1200 = 1800 ⇒ ∠5 = 1800 −1200 = 600
Now, ∠1 = ∠5 [Each = 60o]
But, these are corresponding angles.
Therefore, the lines m and n are parallel.
Question. AP and BQ are the bisectors of the two alternate interior angles formed by intersection of a transversal t with parallel lines l and m (Fig. 6.1 1). Show that AP || BQ.
Answer. ∵ l || m and t is the transversal
∠MAB = ∠SBA
[Alt. ∠s ]
⇒ 1/2 ∠MAB = 1/2 ∠SBA ⇒ ∠2 = ∠3
But, ∠2 and ∠3 are alternate angles.
Hence, AP || BQ.
Question. If in Fig. 6.1 1, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.
AP is the bisector of ∠MAB and BQ is the bisector of ∠SBA. We are given that AP || BQ.
As AP || BQ, So ∠2 = ∠3 [Alt. ∠s ]
∴ 2∠2 = 2∠3
⇒ ∠2 + Ð2 = ∠3 + ∠3
⇒ ∠1+ ∠2 = ∠3 + ∠4 [∵ ∠1 = ∠2 and ∠3 = ∠4]
⇒ ∠MAB = ∠SBA
But, these are alternate angles. Hence, the lines l and m are parallel, i.e., l || m.
Question. In Fig. 6.9, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, O and B are collinear.
Answer. Given: In figure, OD ⊥ OE, OD and OE are the bisector of ∠AOC and ∠BOC.
To prove: points A, O and B are collinear i.e., AOB is a straight line.
Proof: Since, OD and OE bisect angles ∠AOC and ∠BOC respectively.
∴ ∠AOC = 2∠DOC …(1)
And ∠COB = 2∠COE …(2)
On adding equations (1) and (2), we get
∠AOC = ∠COB = 2∠DOC + 2∠COE
⇒ ∠AOC + ∠COB = 2(∠DOC + ∠COE)
⇒ ∠AOC + ∠COB = 2∠DOE
⇒ ∠AOC + ∠COB = 2 × 900 [∵ OD ⊥ OE]
⇒ ∠AOC + ∠COB =1800
∴ ∠AOB =1800
So, ∠AOC + ∠COB are forming linear pair or AOB is a straight line. Hence, points A, O and B are collinear.
Question. In Fig. 6.12, BA|| ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)].
Answer. Produce DE to intersect BC at P (say).
EF || BC and DP is the transversal,
∴ ∠DEF = ∠DPC …(1) [Corres. ∠s ]
Now, AB||DP and BC is the transversal,
∴ ∠DPC = ∠ABC …(2) [Corres. ∠s ]
From (1) and (2), we get
∠ABC = ∠DEF
Question. Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Two lines p and n are respectively perpendicular to two parallel line l and m, i.e., p ⊥ l and n ⊥ m.
We have to show that p is parallel to n.
As n ⊥ m, so ∠1 = 900 …(1)
Again, p ⊥ l, So ∠2 = 900.
But, l is parallel to m, so
∠2 = ∠3 [corres. ∠s ]
∴ ∠2 = ∠90o …(2) [∵ ∠2 = 900 ]
From (1) and (2), we get
⇒ ∠1 = ∠3 [Each = 90o]
But, these are corresponding angles.
Hence, p || n.
Question. In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.
Answer. DE || QR and the line n is the transversal line.
∴ ∠EAB + ∠RBA = 1800 …(1)
[∵ If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary]
⇒ ∠PAB + ∠PBA = 900
[∵ AP is the bisector of ∠EAB and BP is the bisector of ∠RBA]
Now, from ∠APB, we have
∠APB = 1800 − (∠PAB + ∠PBA)
⇒ ∠APB = 1800 − 900 = 900
Question. The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Answer. Given: Ratio of angles is 2 : 3 : 4.
To find: Angles of triangle.
Proof: The ratio of angles of a triangle is 2 : 3 : 4.
Let the angles of a triangle be ∠A, ∠B and ∠C
Therefore, ∠A = 2x, then ∠B = 3x and ∠C = 4x.
In ΔABC, ∠A + ∠B + ∠C = 1800 [∵Sum of angles of a triangle is 180o]
∴ 2x + 3x + 4x =1800
⇒ 9x =1800 ⇒ x = 1800 / 9 = 200
∴ ∠A = 2x = 2 × 200 = 400
∠B = 3x = 3 × 200 = 600
And ∠C = 4x = 4 × 200 = 800
Hence, the angles of the triangles are 40o, 60o and 80o.
Question. In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180°
Answer. Produce ED to meet BC at P(say)
Now, EF || BC and EP is the transversal.
∴ ∠DEF = ∠EPC = 1800 …(1)
Again, EP || AB and BC is the transversal.
∴ ∠EPC = ∠ABC …(2) [ corresponding ∠s ]
From (1) and (2), we get
∠DEF = ∠ABC =1800
⇒ ∠ABC + ∠DEF =1800
Question. A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL = ∠ACB.
Given: In ΔABC
∠A = 900 and AL ⊥ BC.
To prove: ∠BAL = ∠ACB.
Proof: In ΔABC and ΔLAC,
∠BAC = ∠ALC …(1)
[Each = 90o]
And ∠ABC = ∠ABL. …(2) [Common angle]
Adding equations (1) and (2), we get
∠BAC + ∠ABC = ∠ALC + ∠ABC …(3)
In ΔABC, ∠BAC + ∠ACB + ∠ABC =1800
[Sum of angles of triangle is 180o]
⇒ ∠BAC + ∠ABC = 1800 − ∠ACB …(4)
In ΔABL, ∠ABL + ∠ALB + ∠BAL =1800
[Sum of the angles of triangle is 180o]
⇒ ∠ABL + ∠ALC = 1800 −∠BAL …(5)
[∠ALC = ∠ALB = 900]
Substituting the value from equation (4) and (5) in equation (3), we get
1800 − ∠ACB = 1800 − ∠BAL ⇒ ∠ACB = ∠BAL
Question. If two lines intersect, prove that the vertically opposite angles are equal.
Given: Two lines AB and CD intersect at point O.
To prove: (i) ∠AOC = ∠BOD
(ii) ∠AOD = ∠BOC
Proof: (i) Since, ray OA stands on line CD.
∴ ∠AOC = ∠AOD = 1800 …(1)
[Linear pair axiom]
Similarly, ray OD stands on line AB.
∴ ∠AOD = ∠BOD = 1800 …(2)
From equations (1) and (2), we get
∠AOC = ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD
(ii) Since, ray OD stands on line AB.
∴ ∠AOD + ∠BOD = 1800 …(3) [Linear pair axiom]
Similarly, ray OB stands on line CD.
∴ ∠DOB +∠BOC =1800 …(4)
From equations (3) and (4), we get
∠AOD +∠BOD = ∠DOB + ∠BOC ⇒ ∠AOD = ∠BOC
Question. Prove that through a given point, we can draw only one perpendicular to a given line.
[Hint: Use proof by contradiction].
Answer. From the point P, a perpendicular PM is drawn to the given line AB.
∴ ∠PMB = 900
Let if possible, we can draw another perpendicular PN to the line AB. Then,
∠PMB = 900
∴ ∠PMB = ∠PNB, which is possible only when PM and PN coincide with each other.
Hence, through a given point, we can draw only one perpendicular to a given line.
Question. A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
Answer. Given: Two lines DE and QR are parallel and are intersected by transversal at A and B
respectively. Also, BP and AF are the bisector of angles ∠ABR and ∠CAE respectively.
To prove: EP || FQ
Proof: Given, DE ||QR ⇒ ∠CAE = ∠ABR [Corresponding angles]
⇒ 1/2 ∠CAE = 1/2 ∠ABR [Dividing both sides by 2]
⇒ ∠CAE = ∠ABP
[∵ BP and AF are the bisector of angles ∠ABR and ∠CAE respectively.
As these are the corresponding angles on the transversal line n and are equal. Here, EP || FQ.
Question. Prove that a triangle must have at least two acute angles.
Answer. If the triangle is an acute angled triangle, then all its three angles are acute angle. Each of these angles is less than 90o, so they can make three angles sum equal to 180o.
If a triangle is a right triangle, then one angle which is right angle will be equal to 90o and the other two acute angles can make the three angles sum equal to 180o.
Hence, we can say that a triangle must have a least two acute angles.
Question. Bisectors of interior ∠B and exterior ∠ACD of a ΔABC intersect at the point T. Prove that 1/2 ∠BTC = ∠BAC
Answer. Given: ΔABC, produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.
∠BTC = 1/2 ∠BAC
Proof: In ΔABC, ∠ACD is an exterior angle.
∴ ∠ACD = ∠ABC + ∠CAB
[Exterior angle of a triangle is equal to the sum of two opposite angles]
⇒ 1/2 ∠ACD = 1/2 ∠CAB + 1/2 ∠ABC [Dividing both sides by 2]
⇒ ∠TCD = 1/2 ∠CAB + 1/2 ∠ABC …(1)
[∵ CT is a bisector of ∠ACD ⇒ 1/2 ∠ACD = ∠TCD]
In ΔBTC, ∠TCD = ∠BTC +∠CBT
[Exterior angle of the triangle is equal to the sum of two opposite angles]
⇒ ∠TCD = ∠BTC + 1/2 ∠ABC …(2)
[∵ BT is bisector of ΔABC ⇒ ∠CBT = 1/2 ∠ABC ]
From equation (1) and (2), we get
1/2 ∠CAB + 1/2 ∠ABC = ∠BTC + 1/2 ∠ABC
⇒ 1/2 ∠CAB = ∠BTC or 1/2 ∠BAC = ∠BTC
Question. Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.
[Hint: Use proof by contradiction].
Answer. Given: Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D. To prove: Two lines n and p intersecting at a point.
Proof: Let us consider lines n and p are not intersecting, then it means they are parallel to each other i.e., n || P. …(1)
Since, lines n and p are perpendicular to m and l respectively. But from equation (1), n || p, it implies that l and m. It is a contradiction Thus, our assumption is wrong.
Hence, lines n and p intersect at a point.
Question. In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = 1/2 (∠Q − ∠R)
Answer. Given: ΔPQR, ∠Q < ∠R, PA is the bisector of ∠QPR and PM ⊥ QR.
To prove: ∠APM = 1/2 (∠Q − ∠R)
Proof: Since, PA is the bisector of ∠QPR
∠QPA = ∠APR
In ∠ PQM, ∠Q + ∠PMQ + ∠QPM = 1800 …(1)
[Angle sum property of a triangle]
⇒ ∠Q + 900 + ∠QPM = 1800 [∴∠PMR = 900 ]
⇒ ∠Q = 900 − ∠QPM …(2)
In ΔPMR, ∠PMR +∠R +∠RPM = 1800
[Angle sum property of a triangle]
⇒ 900 + ∠R + ∠RPM = 1800 [∵ ∠PMR = 900 ]
⇒ ∠R = 1800 − 900 − ∠RPM
⇒ ∠Q = 900 −∠QPM
⇒ ∠PRM = 900 − ∠RPM …(3)
Subtracting equation (3) from equation (2), we get
∠Q − ∠R = (900 − ∠QPM) − (900 −∠RPM)
⇒ ∠Q − ∠R = ∠RPM −∠QPM
⇒ ∠Q − ∠R = (∠RPM + ∠APM) − (∠QPA − ∠APM) …(4)
⇒ ∠Q − ∠R = ∠QPA + ∠APM − ∠QPA + ∠APM [Using equation (1)]
⇒ ∠Q − ∠R = 2 ∠APM
⇒ ∠APM = 1/2 (∠Q −∠R)
Mathematics Lines And Angles Worksheets for Class 9 as per CBSE NCERT pattern
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