Triangles — Chapter Test | Class 9 IMO

Q1

A ladder is leaning against a vertical wall of a shop in Chandni Chowk. The foot of the ladder is on the ground. The wall, the ground, and the ladder form a right-angled triangle. If the angle between the ladder and the ground is 60°, what is the angle between the ladder and the wall?

The sum of angles in a triangle is 180°. One angle is 90° (wall and ground) and another is 60°. The third angle = 180° − 90° − 60° = 30°.

Q2

If the exterior angle of a triangle is 115° and one of the interior opposite angles is 45°, then the other interior opposite angle is:

By the exterior angle property, the exterior angle is equal to the sum of the two interior opposite angles. So, 115° = 45° + x → x = 115° − 45° = 70°.

Q3

Anil is facing North. He turns 135° clockwise, then 45° anti-clockwise, and finally 90° clockwise. He walks 5 meters straight, turns left and walks 5 meters. His path forms what shape with his starting point?

Total angular turns: 135 - 45 + 90 = 180° clockwise (facing South). Walks 5m South, turns left (East) and walks 5m. The start point (0,0), intermediate point (0,-5), and final point (5,-5) form a right-angled triangle with two equal sides of 5m.

Q4

In ΔABC, if AB = AC and the bisectors of ∠B and ∠C intersect each other at O, then:

Since AB = AC, we have ∠B = ∠C, which means ½ ∠B = ½ ∠C, so ∠OBC = ∠OCB. In ΔOBC, since base angles are equal, the opposite sides must be equal, so OB = OC.

Q5

In ΔABC, AD is the perpendicular bisector of BC. Then ΔABC is a/an:

In ΔABD and ΔACD: BD = CD (since AD bisects BC), ∠ADB = ∠ADC = 90° (perpendicular), and AD = AD (common). By SAS, ΔABD ≅ ΔACD, which means AB = AC. Hence, it is an isosceles triangle.

Q6

Choose the correct alternative to fill in the blank series: SAS, ASA, AAS, RHS, ___.

The series consists of valid congruence criteria for triangles. Among the choices, only SSS is a valid congruence rule.

Q7

In ΔABC, if AB = AC, and side BA is produced to D such that AD = AB, then ∠BCD is equal to:

Let ∠ABC = ∠ACB = x. Since AD = AB = AC, in ΔACD, AD = AC → ∠ADC = ∠ACD = y. In ΔBCD, sum of angles is ∠B + ∠D + ∠BCD = 180° → x + y + (x + y) = 180° → 2(x + y) = 180° → x + y = 90°. Since ∠BCD = x + y, it equals 90°.

Q8

An agricultural field in Punjab is shaped like an isosceles triangle ABC with AB = AC. The farmer wants to lay an irrigation pipe from vertex A to the midpoint D of the boundary BC. If the angle ∠B is 48°, what angle does the irrigation pipe make with the boundary BC?

In an isosceles triangle with AB = AC, the line joining the vertex A to the midpoint of the base BC is perpendicular to BC. Thus, the angle is 90°.

Q9

If a triangle has vertices with coordinates that follow a specific pattern: (1,1), (2,3), (3,5), then these points are:

Slope of line joining (1,1) and (2,3) is (3-1)/(2-1) = 2. Slope joining (2,3) and (3,5) is (5-3)/(3-2) = 2. Since slopes are equal, points are collinear and do not form a triangle.

Q10

In ΔABC, the bisectors of ∠B and ∠C intersect at point O. Prove that ∠BOC = 90° + ½ ∠A. If ∠A = 70°, find ∠BOC.

In ΔBOC, ∠BOC = 180° − (½ ∠B + ½ ∠C) = 180° − ½ (∠B + ∠C). Since ∠B + ∠C = 180° − ∠A, we get ∠BOC = 180° − ½ (180° − ∠A) = 90° + ½ ∠A. Substituting ∠A = 70°: 90° + 35° = 125°.

Q11

In ΔABC, if AB = AC and ∠B = 50°, then find the value of ∠A.

Since AB = AC, the angles opposite to these sides are equal, so ∠C = ∠B = 50°. By angle sum property, ∠A + ∠B + ∠C = 180° → ∠A + 50° + 50° = 180° → ∠A = 80°.

Q12

Point A is to the North of Point B. Point B is to the West of Point C. If AB = BC = 6 km, then what is the direction of Point A with respect to Point C, and what is the shortest distance between them?

ABC forms a right-angled isosceles triangle with right angle at B. Distance AC = √(6² + 6²) = √72 = 6√2 km. Direction of A from C is North-West.

Q13

In ΔABC, AD is the median to side BC. Which of the following inequalities holds true for the sum of sides AB and AC relative to the median?

Produce AD to E such that DE = AD and join CE. In ΔABD and ΔECD, BD = CD, AD = ED, and ∠ADB = ∠EDC (vertically opposite). By SAS, ΔABD ≅ ΔECD → AB = EC. In ΔACE, AC + CE > AE → AC + AB > 2AD.

Q14

A traffic sign board in the shape of an isosceles triangle has its equal sides of length 20 cm each and the third side of length 30 cm. A technician wants to cut it exactly in half along its line of symmetry. What will be the base length of each new triangle?

The line of symmetry in an isosceles triangle originates from the vertex between the equal sides and acts as a perpendicular bisector to the base. The base of 30 cm is bisected into two parts of 30/2 = 15 cm each.

Q15

In ΔPQR, S is any point on side QR. Then which of the following inequalities is always valid?

In ΔPQS: PQ + QS > PS. In ΔPRS: PR + RS > PS. Adding these two inequalities: PQ + (QS + RS) + PR > 2PS → PQ + QR + PR > 2PS.