Lines and Angles Questions

Lines and Angles Questions are given here to help the students get practice for their exams. As we know, lines and angles are the important concepts included in Class 7 and 9 of the NCERT curriculum. Lines and angles are the basic figures in geometry that we learn in our early geometry classes. In this article, you will get the questions and answers for lines and angles, based on the latest NCERT syllabus.

What are Lines and Angles?

In geometry, a line has no endpoints on either side, whereas a line segment has two endpoints, and a ray has only one endpoint (its initial point). An angle is formed when two lines (or rays or line segments) meet.

Also, visit: Lines and Angles

Lines and Angles Questions and Answers

1. An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?

Solution:

We know that complementary angles add up to 90°.

Let θ and 90° – θ be the complementary angles.

If θ = 45°, then the other angle will be: 90° – θ = 90° – 45° = 45°.

If θ > 45°, the other angle cannot be 45°.

Now consider, θ > 45°

Multiplying by -1 on both sides,

-θ < -45°

Adding 90° on both sides,

90° – θ < 90° – 45°

90° – θ < 45°

Therefore, if one angle is greater than 45°, its complementary angle must be less than 45°.

2. What is the type of another angle of a linear pair if

(a) one of its angles is acute?

(b) one of its angles is obtuse?

(c) one of its angles is right?

Solution:

If the sum of angles is 180°, they form a linear pair.

(a) If we subtract an acute angle from 180°, we get an angle greater than 90°.

Therefore, if one angle is acute, the other angle will be obtuse.

(b) If we subtract an obtuse angle from 180°, we get an angle less than 90°.

Therefore, if one angle is obtuse, the other angle will be acute.

(c) If we subtract 90° from 180°, we get 90°.

Therefore, if one angle is a right angle, the other angle will also be a right angle.

3. In the figure, l || m and a line t intersect lines l and m at P and Q, respectively. Find the sum 2a + b.

Solution:

Given that, l || m and t is a transversal.

Also, from the given figure,

132° and a are corresponding angles.

Therefore, a = 132°.

Also, b and 132° are vertically opposite angles.

We know that the pair of vertically opposite angles are equal.

So, b = 132°

Now,

2a + b = 2 x 132° + 132°

= 264° + 132°

= 396°

4. In the figure, AB||EF, ED||CB and ∠APE is 39°. Find∠CQF.

Solution:

Given,

ED||BC and AB is the transversal.

∠APE – 39°

∠QBP = ∠APE {corresponding angles}

⇒ ∠QBP = 39°

Now, AB||EF and BC is the transversal.

∠FQB = ∠QBP {alternate interior angles}

⇒ ∠FQB = 39°

Also,

∠CQF + ∠FQB = 180° {linear pair}

∠CQF + 39° = 180°

∠CQF = 180° – 39°

∠CQF = 141°

Therefore, ∠CQF = 141°.

5. In the figure, AB, CD and EF are three lines concurrent at O. Find the value of y.

Solution:

From the given figure,

∠AOE = ∠BOF = 5y {vertically opposite angles are equal}

Also, ∠COE + ∠AOE + ∠AOD = 180° {COD is a straight line and the corresponding angles form a linear pair}

2y + 5y + 2y = 180°

9y = 180°

y = 180°/9

y = 20°

6. If a transversal intersects two parallel lines, and the difference of two interior angles on the same side of a transversal is 20°, find the angles.

Solution:

Let a and b be the two interior angles on the same side of the transversal n for parallel lines l and m.

According to the given,

a – b = 20°

b = a – 20°

Also, a + b = 180° {since the sum of interior angles made by the transversal is 180°}

a + a – 20° = 180° {since b = a – 20°}

2a = 180° + 20°

2a = 200°

a = 200°/2

a = 100°

So, b = 100° – 20° = 80°

Hence, the required interior angles are 100° and 80°.

7. In the figure, if AB || DE, ∠ BAC = 35° and ∠ CDE = 53°, find ∠ DCE.

Solution:

We know that,

AE is a transversal since AB || DE.

Here, ∠BAC and ∠AED are alternate interior angles.

Hence, ∠BAC = ∠AED

∠BAC = 35° {given}

∠AED = 35°

Now consider the triangle CDE.

We know that the sum of the interior angles of a triangle is 180°.

∴ ∠DCE +∠CED +∠CDE = 180°

Substituting the values, we get;

∠DCE + 35° + 53° = 180°

∠DCE = 180° – 35° – 53° = 92°

8. The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.

Solution:

Given that the ratio of angles of a triangle is 2 : 3 : 4.

Let 2x, 3x, and 4x be the angles of a triangle.

We know that the sum of angles of a triangle is 180°.

So, 2x + 3x + 4x = 180°

9x = 180°

x = 180°/9 = 20°

Now,

2x = 2 × 20° = 40°

3x = 3 × 20° = 60°

4x = 4 × 20° = 80

Therefore, the angles of the triangle are 40°, 60°, and 80°, respectively.

9. In the figure, AB||CD, find the reflex ∠ EFG.

Solution:

Given,

AB || CD

From the given figure,

∠2 + ∠ FGD = 180°{Co-interior angles of a transversal}

∠2 = 180° – 135° = 45°

∠1 = ∠ AEF = 34°{alternate interior angles}

∠1 + ∠2 = 34° + 45° = 79°

Reflex ∠EFG = 360° -(∠ 1 + ∠ 2)

= 360° – 79°

= 281°

10. 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:

Let AD be the transversal that intersects two lines, PQ and RS, at points B and C, respectively, as shown in the figure.

Ray BE is the bisector of ∠ABQ, and ray CG is the bisector of ∠BCS, and BE || CG.

We need to prove that PQ || RS.

As we know, the ray BE is the bisector of ∠ABQ.

so, ∠ABE = (1/2)∠ ABQ….(1)

Similarly,

ray CG is the bisector of ∠BCS.

so, ∠BCG = (1/2)∠BCS….(2)

Also, BE || CG and AD is the transversal.

By the corresponding angles axiom,

Thus, ∠ABE = ∠BCG….(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. So they are equal.

By the converse of corresponding angles axiom,

PQ || RS

Hence proved.

Video Lesson on Intersecting and Parallel lines

Practice Questions on Lines and Angles

1. Measures (in degrees) of two supplementary angles are consecutive odd integers. Find the angles.
2. Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
3. Amisha makes a star with the help of line segments a, b, c, d, e and f, in which a || d, b || e and c || f. Chhaya marks an angle as 120° as shown in the figure, and asks Amisha to find the ∠x, ∠y and ∠z. Help Amisha in finding the angles.
4. In the figure, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28°, and ∠ QRT = 65°, find x and y.
5. Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.