Lines and angles Class 9 questions and solutions are available here to cover various problems on lines and angles. These questions are given as per Class 9 Maths Chapter 6 of the NCERT curriculum. Go through the lines and angles questions for Class 9 given below, along with their detailed solutions.
Definition of Lines and Angles
In mathematics and particularly in geometry, lines and angles are two related terms. Lines are straight one-dimensional figures extending endlessly in both directions, and do not have a thickness. The angles formed when lines meet at a common point is called the vertex.
Also, check: Lines and Angles Class 9 Notes
Lines and Angles Class 9 Questions and Answers
1. In the figure, lines PQ and RS intersect at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles.
Solution:
From the given figure,
∠POR +∠ROQ = 180° (linear pair of angles since PQ is a straight line)
Also, given that,
∠POR : ∠ROQ = 5 : 7
Therefore, ∠POR = (5/12) × 180° = 75°
Similarly, ∠ROQ = (7/12) × 180° = 105°
Now, ∠POS = ∠ROQ = 105° (vertically opposite angles)
∠SOQ = ∠POR = 75° (vertically opposite angles)
2. In the figure, if x + y = w + z, then prove that AOB is a straight line.
Solution:
To prove AOB is a straight line, we will have to prove x+y is a linear pair
i.e. x + y = 180°.
We know that the angles around a point are 360°.
So, x + y + w + z = 360°
x + y = w + z (given)
Thus, (x + y) + (x + y) = 360°
2(x + y) = 360°
∴ (x + y) = 180°
That means AOB is a straight line.
Hence proved.
3. If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
Solution:
Given:
PQ and RS are the two lines and a transversal AD intersects these two lines at points B and C, respectively.
Ray BE is the bisector of ∠ABQ and ray CG is the bisector of ∠BCS, and BE || CG.
To prove: PQ || RS
Proof:
Given that ray, BE is the bisector of ∠ABQ.
Therefore, ∠ABE = (1/2) ∠ABQ….(1)
Similarly, ray CG is the bisector of ∠BCS.
Therefore, ∠BCG = (1/2) ∠BCS….(2)
As we know, BE || CG and AD is the transversal.
Therefore, ∠ABE = ∠BCG (corresponding angles axiom)…..(3)
Substituting (1) and (2) in (3), we get;
(1/2) ∠ABQ = (1/2) ∠BCS
⇒ ∠ABQ = ∠BCS
These are the corresponding angles formed by transversal AD with PQ and RS; they are equal.
Thus, by the converse of corresponding angles axiom, we have;
PQ || RS
Hence proved.
4. In the below figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Solution:
In the given figure, we can see that ∠APQ and ∠PQR are alternate interior angles.
That means, ∠APQ = ∠PQR = 50°
i.e., ∠PQR = x = 50°
Also,
∠APR = ∠PRD (alternate interior angles)
∠PRD = 127° (given)
⇒ ∠APR = 127°
We know that
∠APR = ∠APQ +∠QPR
⇒ 127° = 50°+ y
⇒ y = 127° – 50° = 77°
Therefore, x = 50° and y = 77°
5. In the figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
In the given figure, AE is a transversal and AB || DE.
Here, ∠BAC and ∠AED are alternate interior angles.
⇒ ∠BAC = ∠AED
⇒ ∠AED = 35° (since it is given that ∠BAC = 35°)
In the triangle CDE,
∠DCE + ∠CED + ∠CDE = 180° (angle sum property of the triangle)
Substituting the values, we get;
∠DCE + 35° + 53° = 180°
∠DCE + 88° = 180°
∠DCE = 180° – 88° = 92°
Therefore, ∠DCE = 92°
6. In the figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)].
Solution:
Construction:
Extend DE to intersect BC at point P.
Given that EF||BC and DP is the transversal,
∠DEF = ∠DPC …(1) [corresponding angles]
Also given, AB||DP and BC is the transversal,
∠DPC = ∠ABC …(2) [corresponding angles]
From (1) and (2), we get;
∠ABC = ∠DEF
Hence proved.
7. An exterior angle of a triangle is 105°, and its two interior opposite angles are equal. Find the measure of each of these equal angles.
Solution:
Given,
The exterior angle of triangle = 105°
Let x be the measure of two interior opposite angles of the triangle.
We know that the exterior angle of a triangle is equal to the sum of interior opposite angles.
105° = x + x
⇒ 2x = 105°
⇒ x = 105°/2
⇒ x = 52½° = 52.5°
8. Bisectors of interior ∠B and exterior ∠ACD of a Δ ABC intersect at point T. Prove that ∠BTC = ½ ∠BAC.
Solution:
Given:
△ABC, produce BC to D, and the bisectors of ∠ABC and ∠ACD meet at point T.
To prove: ∠BTC = ½ ∠BAC
Proof:
In △ABC, ∠ACD is an exterior angle.
We know that,
The exterior angle of a triangle is equal to the sum of two opposite angles,
Then,
∠ACD = ∠ABC + ∠CAB
Dividing both sides of the equation by 2, we get;
⇒ ½ ∠ACD = ½ ∠CAB + ½ ∠ABC
⇒ ∠TCD = ½ ∠CAB + ½ ∠ABC .…(1)
[∵CT is a bisector of ∠ACD ⇒ ½ ∠ACD = ∠TCD]
As we know, the exterior angle of a triangle is equal to the sum of two opposite angles.
In â–³ BTC,
∠TCD = ∠BTC + ∠CBT
⇒ ∠TCD = ∠BTC + ½ ∠ABC ….(2)
[∵BT is bisector of △ ABC ⇒∠CBT = ½ ∠ABC ]
From (1) and (2), we get;
½ ∠CAB + ½ ∠ABC = ∠BTC + ½ ∠ABC
⇒ ½ ∠CAB = ∠BTC
or
½ ∠BAC = ∠BTC
Hence proved.
9. In the below figure, AB, CD and EF are three concurrent lines intersecting at O. Find the value of y.
Solution:
From the given figure,
∠AOE = ∠BOF = 5y (vertically opposite angles)
CD is a straight line.
Thus, ∠COE + ∠AOE + ∠AOD = 180°
2y + 5y + 2y = 180°
9y = 180°
y = 180°/9
y = 20°
Hence, the value of y is 20°.
10. In the figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = ½ ∠QPR.
Solution:
Consider the ΔPQR. ∠PRS is the exterior angle, and ∠QPR and ∠PQR are interior angles.
As we know, the exterior angle of a triangle is equal to the sum of two opposite angles.
∠PRS = ∠QPR + ∠PQR
⇒ ∠PRS – ∠PQR = ∠QPR ———–(i)
In the ΔQRT,
∠TRS = ∠TQR + ∠QTR
⇒ ∠QTR = ∠TRS – ∠TQR
We know that QT and RT bisect ∠PQR and ∠PRS, respectively.
⇒ ∠PRS = 2 ∠TRS
Similarly, ∠PQR = 2∠TQR
Now,
∠QTR = ½ ∠PRS – ½∠PQR
⇒ ∠QTR = ½ (∠PRS -∠PQR)
⇒ ∠QTR = ½ ∠QPR [From (i)]
Hence proved.
Practice Problems on Lines and Angles Class 9
- AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (below figure). Show that AP || BQ.
- If AB || CD, CD || EF and y : z = 3 : 7, find x from the below figure.
- Prove that two lines that are perpendicular to two intersecting lines intersect each other.
- If the ratio of angles of a triangle is 2 : 4 : 3, find the angles.
- If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2 : 3, find the greater of the two angles.
Comments