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All questions of Triangles for Class 9 Exam

P is a point equidistant from two lines l and m intersecting at point A, then
  • a)
    ∠BAP = ∠APC
  • b)
    ∠CAP = ∠BPA
  • c)
    ∠CAP = ∠BAP
  • d)
    None of there 
Correct answer is option 'C'. Can you explain this answer?

Mahima Shah answered
Understanding the Geometry of Point P
When a point P is equidistant from two intersecting lines l and m at point A, it creates specific angles that relate to point P and the lines.
Key Concepts
- Equidistant Definition: Point P being equidistant from lines l and m means that the perpendicular distances from P to both lines are equal.
- Intersection at Point A: The lines l and m intersect at point A, creating angles around A.
Angles Formed by Point P
- Angle Relationships: The angles formed at point A can be expressed in terms of P:
- Angle BAP is the angle between line l and line segment AP.
- Angle CAP is the angle between line m and line segment AP.
- Angle BPA is the angle between line l and line segment BP.
Reasoning for the Correct Answer
- Since P is equidistant from lines l and m, the angles formed with respect to A will have specific relationships:
- Angle CAP = Angle BAP: This is true because both angles are subtended by the same line segment AP from point P, making them equal.
- Angle CAP does not equal Angle BPA: These angles are not directly related as they involve different line segments and points.
Conclusion
Thus, the correct answer is option 'C':
Angle CAP = Angle BAP.
This equality arises due to the symmetry created by point P being equidistant from both lines.

Side QR of a triangle PQR is produced both ways and the measures of exterior angles formed are 86° and 124°. The measure of  ∠P is :
  • a)
    30°
  • b)
    40°
  • c)
    60°
  • d)
    80°
Correct answer is option 'A'. Can you explain this answer?

∠PQX and ∠PRY are exterior angles for ΔPQR

∴ ∠P + ∠PRY = 86° ...(i)
∠P + ∠PQX = 124° ...(ii)
Adding (i) and (ii)
 2∠P +∠PRX + ∠PQX = 210°
⇒∠P + (∠P +∠PRX + ∠PQX) = 210°
⇒ ∠P + 180° = 210°
⇒ ∠P = 30°

In a ΔABC,∠A = 50°, ∠B = 60°. The longest side of the triangle will be
  • a)
    AB
  • b)
    BC
  • c)
    CA
  • d)
    None of these 
Correct answer is option 'A'. Can you explain this answer?

Avani Shah answered
Understanding Triangle Properties
In triangle ABC, we are given the angles:
- Angle A = 50°
- Angle B = 60°
To determine the longest side, we need to utilize the property that states: the side opposite the largest angle is the longest side.
Calculating Angle C
First, we calculate the third angle (C) using the triangle angle sum property:
- Angle C = 180° - (Angle A + Angle B)
- Angle C = 180° - (50° + 60°) = 70°
Now we have all angles in triangle ABC:
- Angle A = 50°
- Angle B = 60°
- Angle C = 70°
Identifying the Longest Side
Next, we compare the angles to determine which is the largest:
- Angle A = 50°
- Angle B = 60°
- Angle C = 70° (Largest angle)
The side opposite the largest angle (Angle C) is side AB.
Conclusion
Since side AB is opposite the largest angle (70°), it is the longest side of triangle ABC.
Thus, the correct answer is:
- a) AB
This aligns with the triangle inequality property, confirming that side AB is indeed the longest side in triangle ABC.

If one angle of the triangle is equal to the sum of the other two angles then the triangle is
  • a)
    Acute angled triangle
  • b)
    Isosceles/ equilateral triangle  
  • c)
    Obtuse angled triangle
  • d)
    Right angled triangle 
Correct answer is option 'D'. Can you explain this answer?

Nilofer Singh answered
Answer:

Explanation:
In a triangle, the sum of all angles is always 180 degrees. Let's assume the three angles of the triangle are A, B, and C.

Given:
One angle (let's assume A) is equal to the sum of the other two angles (B and C).

To prove:
The triangle is a right-angled triangle.

Proof:
Let's assume that angle A is equal to the sum of angles B and C.
So, A = B + C

Now, let's assume that the triangle is not a right-angled triangle. In this case, the sum of angles B and C would be less than 90 degrees (acute-angled triangle) or greater than 90 degrees (obtuse-angled triangle).

Case 1: Acute-angled triangle
If the sum of angles B and C is less than 90 degrees, it means that angle A (which is equal to B + C) would also be less than 90 degrees. However, this contradicts the given condition that angle A is equal to the sum of angles B and C. Therefore, an acute-angled triangle cannot satisfy the given condition.

Case 2: Obtuse-angled triangle
If the sum of angles B and C is greater than 90 degrees, it means that angle A (which is equal to B + C) would also be greater than 90 degrees. However, this again contradicts the given condition that angle A is equal to the sum of angles B and C. Therefore, an obtuse-angled triangle cannot satisfy the given condition.

Hence, the only possibility left is that the triangle is a right-angled triangle, where the sum of angles B and C is equal to 90 degrees. In this case, angle A (which is equal to B + C) would also be equal to 90 degrees.

Therefore, the correct answer is option 'D' - Right-angled triangle.

If two isosceles triangles have a common base, then the line joining their vertices will
  • a)
    Bisect them at acute angle
  • b)
    Bisect them at obtuse angle
  • c)
    Bisect them at right angle
  • d)
    NOT
Correct answer is option 'C'. Can you explain this answer?

Rajesh Yadav answered
Explanation:

To understand why the line joining the vertices of two isosceles triangles with a common base bisects them at a right angle, let's consider the properties of isosceles triangles.

Properties of Isosceles Triangles:
1. In an isosceles triangle, the two sides opposite the equal angles are of equal length.
2. The angles opposite the equal sides of an isosceles triangle are equal.

Given:
We have two isosceles triangles with a common base.

To Prove:
The line joining their vertices bisects the triangles at a right angle.

Proof:
Let's label the triangles as ABC and ABD, where AB is the common base and AC = BC and AD = BD.

Now, let's draw the line joining the vertices C and D.

Step 1:
Since AC = BC and AD = BD, we can conclude that triangles ABC and ABD are congruent by the Side-Side-Side (SSS) congruence criterion.

Step 2:
Using the congruence of the triangles, we can say that angle BAC is congruent to angle BAD and angle ABC is congruent to angle ABD.

Step 3:
Since angles BAC and BAD are congruent, and angles ABC and ABD are congruent, we can conclude that angle CAB is congruent to angle DAB.

Step 4:
When two angles are congruent, they are called vertically opposite angles. Vertically opposite angles are always equal.

Step 5:
Since angle CAB and angle DAB are vertically opposite angles and congruent, we can conclude that the line joining the vertices C and D bisects the triangles ABC and ABD at a right angle.

Hence, the correct answer is option 'C' - The line joining the vertices of two isosceles triangles with a common base bisects them at a right angle.

PQRS is a square and SRT is an equilateral triangle. The measure of ∠TQR is:
  • a)
    25°
  • b)
    55°
  • c)
    15°
  • d)
    35°
Correct answer is option 'C'. Can you explain this answer?

In ΔS PTS and QTR
(TR = TS = SR = PQ = QR = PS)
TR = TR (given)
PS = QR (given)
∠PST = ∠QRT
= 90°  +  60° = 150°
(Square)  (equilateral Δ)
∴ ΔPTS = ΔQTR
(By R-H-S congruence criterion)
TP = TQ (C.P.C.T)
∴ ∠TPS = ∠TQR (C.P.C.T)
Now,
In ΔTQR
TR = RQ
∴ ∠RTQ = ∠RQT, and
∠RTQ + ∠RQT +∠R = 180°
⇒ 2∠RQT + 90° + 60° = 180°

AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If ∠APR = 25°, ∠RQC = 30° and ∠CQF = 65° then
  • a)
    p = 55°, q = 40°
  • b)
    p = 50°, q = 45°
  • c)
    p = 35°, q = 60°
  • d)
    p = 60°, q = 35°
Correct answer is option 'A'. Can you explain this answer?

∠APR + ∠CQF = 65°
(corresponding ∠s)
∠APR + ∠OPQ = 65°
⇒ 25° + Q = 65°
⇒ Q = 40°
∵ ∠OQF is on exterior angle for DPOQ
∴ q + DPOQ = 65° + 30°
⇒ p + q = 65°
⇒ p = 95° - 40° = 55°

In ΔABC, AC >AB and AD is the bisector of ∠A. Then 
  • a)
    ∠ADC < 2∠ADB
  • b)
    ∠ADC < ∠ADB
  • c)
    ∠ADC > ∠ADB
  • d)
    ∠ADC = ∠ADB
Correct answer is option 'C'. Can you explain this answer?

In ΔABC, A

∵ AC > AB
∴ ∠ABC = ∠ACB
⇒ - ∠ABC < -∠ACB
⇒ 180° - ∠ABC < 180° - ∠ACB
⇒ 180° - ∠ABC - x  < 180° - ∠ACB - x 
⇒ ∠ADB < ∠ADC.

In ΔABC, AB = AC, and the bisect are of angles B and C intersect at point O, then the ray AO
  • a)
    will bisect ∠A
  • b)
    will not bisect ∠A
  • c)
    AO = CO
  • d)
    AO = BO
Correct answer is option 'A'. Can you explain this answer?


∵ ∠B = ∠C

⇒ OB = OC ...(i)
(sides opposite to equal angles are equal)
Now,
In Δs ABO and ACO
AO = AO (common)
AB = AC (given)
OB = OC (From (i))
∴ ΔABO ≅ ΔACO
(By R-H-S congruence criterion)
∴ ∠BAO = ∠CAO = ∠A/2 (C.P.C.T)

AN is the bisector of ∠A and AM ⊥ BC. Then a measure of ∠MAN is:
  • a)
    35°
  • b)
    30°
  • c)
    20°
  • d)
    25°
Correct answer is option 'C'. Can you explain this answer?

Here ∠BAC = 180° -(75° +35°) = 70°

(∵ AN is the angle bisector of ∠A)
Now, in DANC,
∠ANC + ∠CAN + ∠NAC = 180°
⇒ ∠ANC +35° + 35° = 180°
⇒ ∠ANC = 110°
∵ ∠ANC is an exterior angle for ΔAMN 
∵ ∠ANC + ∠MAN = 110°
⇒ ∠MAN = 110°- ∠AMN 
= 110° - 90° = 20°

The sum of all the exterior angles of a triangle is
  • a)
    180°
  • b)
    360°
  • c)
    540°
  • d)
    270°
Correct answer is option 'B'. Can you explain this answer?

 In ΔABC

∠ABC + ∠ACB +∠BAC = 180°
Now,
using exterior angle theorem 
∠ACB + ∠ABC = ∠BAP ...(i)
∠ABC + ∠BAC = ∠ACQ ...(ii)
∠ACB + ∠BAC = ∠ABR ...(iii)
Adding Eqns (i), (ii) and (iii), we get
2 (∠ABC + ∠BAC +∠ABC)
= ∠BAP + ∠ACQ + ∠ABR
⇒ Sum of all exterior angle = 2 × 180° = 360°

Find x in the given figure
  • a)
    120°
  • b)
    135°
  • c)
    150°
  • d)
    110°
Correct answer is option 'C'. Can you explain this answer?


Constructing a line PQ || BC,
∠APQ = ∠ABC = 55°
∵ ∠AQL is an exterior angle for DAPQ
∴ ∠APQ + 25° = a
⇒ a = 25° + 55° = 80°
b = 70° (Alternate opposite ∠s)
∴ x = a + b = 70° + 80° = 150°

If the length of three of the altitudes of a triangle are equal, then the triangle must be a/an
  • a)
    Isosceles triangle
  • b)
    Equilateral triangle
  • c)
    Scalene triangle
  • d)
    Right triangle
Correct answer is option 'B'. Can you explain this answer?

In Δs BEC and CFB

BC = BC (Common)
BE = CF (Given)
∠BEC = ∠CFB = 90°
∴ ΔBEC = ∠CFB = 90°
(By R-H-S congruence criterion)
∠B =∠C (C-P-C-T)
∴ AB = AC ...(i)
Similarly in Δs ADC and C + A 
⇒ ∠A = ∠C
∴ AB = BC ...(ii)
Using (i) and (ii) 
AB = BC = AC  (Δ should be equilateral)

The value of  x from the adjoining figure will be:
  • a)
    41°
  • b)
    45°
  • c)
    42°
  • d)
    48°
Correct answer is option 'C'. Can you explain this answer?

∠ABC + ∠A = ∠ACD
⇒ ∠ACD = ∠ABC + 84°

⇒ ∠ECD = ∠EBC + 42° ...(i)
∵ ∠ECD is an exterior angle for ∠EBC
∴ ∠ECD =∠EBC = x ...(ii)
Comparing (i) and (ii), we get
x = 42°

ABC is an isosceles such that AB = AC and AD is the median to base BC. Then, ∠BAD =
  • a)
    40°
  • b)
    50°
  • c)
    60°
  • d)
    100°
Correct answer is option 'B'. Can you explain this answer?

In  Δs ABD and ACD,
AB = AC (Given)
AD = AD (Common)
BD = CD (∵ AC is median)
∴ ΔABD ≅ ΔACD
[By S-S-S congruence criterion]

The value of x in the adjoining figure will be:
  • a)
    120°
  • b)
    90°
  • c)
    65°
  • d)
    80°
Correct answer is option 'A'. Can you explain this answer?

∵ ∠DBC is the exterior angle for ∠DAB
∴ ∠ADB + ∠DAB = ∠DBC
⇒ ∠DBC = 25° + 55° = 80° 
∵ ∠x is an exterior angle for ∠EBC
∴ ∠EBC +∠ECB = x
⇒ 80° +  40° = x
⇒ x = 120°

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