Lines & Angles — Chapter Test | Class 9 IMO

Q1

Lines AB and CD intersect at O. If ∠AOC : ∠COB = 4:5, find the measure of ∠AOD.

∠AOC and ∠COB form a linear pair, so 4x + 5x = 180° → 9x = 180° → x = 20°. Thus, ∠COB = 5 × 20° = 100°. Since ∠AOD and ∠COB are vertically opposite angles, ∠AOD = ∠COB = 100°.

Q2

In the given figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Since AB and CD intersect at O, ∠AOC = ∠BOD = 40° (vertically opposite angles). Given ∠AOC + ∠BOE = 70°, we get 40° + ∠BOE = 70° → ∠BOE = 30°. Also, ∠AOC + ∠COE + ∠BOE = 180° (angles on a straight line AB) → 70° + ∠COE = 180° → ∠COE = 110°. Thus, reflex ∠COE = 360° − 110° = 250°.

Q3

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°. This statement is known as:

This is a basic geometric axiom known directly as the Linear Pair Axiom.

Q4

In the given figure, AB || CD. If ∠ABO = 118° and ∠CDO = 122°, find the measure of ∠BOD if O lies between the parallel lines.

Draw a line XY || AB through O. Then XY || CD as well. The angle ∠BOX = 180° − 118° = 62° (co-interior angles). The angle ∠DOX = 180° − 122° = 58° (co-interior angles). Thus, ∠BOD = ∠BOX + ∠DOX = 62° + 58° = 120°.

Q5

If two lines are intersected by a transversal such that the bisectors of a pair of corresponding angles are parallel, then the two lines are:

When the bisectors of corresponding angles are parallel, the corresponding angles themselves must be equal because they are exactly double the angles formed by the parallel bisectors. Since corresponding angles are equal, the original lines are parallel.

Q6

Statement 1: If two lines intersect, then the vertically opposite angles are equal. Statement 2: A linear pair of angles is always supplementary. Which of the following holds true?

Both statements are fundamental geometric theorems. Vertically opposite angles are always equal when lines intersect, and angles forming a linear pair always add up to 180° (supplementary).

Q7

If the side BC of a triangle ABC is produced on both sides, then the sum of the two exterior angles so formed is always:

Left exterior angle = ∠A + ∠C. Right exterior angle = ∠A + ∠B. Sum = 2∠A + ∠B + ∠C = ∠A + (∠A + ∠B + ∠C) = 180° + ∠A.

Q8

Rays OC and OE lie above a straight line AB through point O, dividing the straight angle into three parts: ∠AOC = 3y, ∠COE = y and ∠EOB = 2y. Find the value of y.

The three angles on the straight line AB add up to 180°: 3y + y + 2y = 180° → 6y = 180° → y = 30°.

Q9

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the greater of the two angles is:

Co-interior angles are supplementary. Let the angles be 2x and 3x. 2x + 3x = 180° → 5x = 180° → x = 36°. The greater angle is 3x = 3 × 36° = 108°.

Q10

In ΔABC, the bisectors of ∠B and ∠C intersect each other at a point O. Prove or find the measure of ∠BOC in terms of ∠A.

In ΔBOC, ∠BOC + ∠OBC + ∠OCB = 180°. Since OB and OC are bisectors, ∠OBC = ½∠B and ∠OCB = ½∠C. Thus, ∠BOC + ½(∠B + ∠C) = 180°. In ΔABC, ∠B + ∠C = 180° − ∠A. Substituting this: ∠BOC + ½(180° − ∠A) = 180° → ∠BOC + 90° − ½∠A = 180° → ∠BOC = 90° + ½∠A.

Q11

In ΔABC, side BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then the measure of ∠BAC is:

By the exterior angle theorem, the exterior angle of a triangle is equal to the sum of its two opposite interior angles. So, ∠ACD = ∠BAC + ∠ABC → 120° = ∠BAC + 40° → ∠BAC = 80°.

Q12

In ΔXYZ, the bisector of ∠X meets YZ at Q. If ∠XYZ = 60° and ∠XZY = 40°, find ∠XQY.

∠X = 180° − (60° + 40°) = 80°, so ∠YXQ = ½ × 80° = 40°. In ΔXQY, ∠XQY = 180° − (∠Y + ∠YXQ) = 180° − (60° + 40°) = 80°.

Q13

In a rectangular billiard table, a ball bounces off the cushion such that the angle of incidence equals the angle of reflection. If the path of the ball forms an angle of 50° with the cushion wall on entry, what is the angle between the incoming and outgoing path of the ball?

The angle with the cushion is 50° on both sides. Since the cushion is a straight line, the angle between the paths is 180° − (50° + 50°) = 80°.

Q14

An architect designs a window frame containing three structural bars meeting at a single point on a straight baseline. The angles between consecutive bars are in the ratio 1:2:3. If these three angles together form a straight angle on the baseline, find the measure of the largest angle.

Let the angles be x, 2x, and 3x. Since they form a straight angle, x + 2x + 3x = 180° → 6x = 180° → x = 30°. The largest angle is 3x = 3 × 30° = 90°.

Q15

In a town planning layout, Ashoka Road and Nehru Road are parallel to each other. Raman Street intersects both roads. If the consecutive interior angles formed by Raman Street with Ashoka Road and Nehru Road are (3x + 20)° and (2x − 40)° respectively, find the value of x.

Since the roads are parallel, the consecutive interior angles are supplementary. Therefore, (3x + 20) + (2x − 40) = 180 → 5x − 20 = 180 → 5x = 200 → x = 40.