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Math Olympiad Test: Triangles- 2 - Class 9 MCQ


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15 Questions MCQ Test - Math Olympiad Test: Triangles- 2

Math Olympiad Test: Triangles- 2 for Class 9 2024 is part of Class 9 preparation. The Math Olympiad Test: Triangles- 2 questions and answers have been prepared according to the Class 9 exam syllabus.The Math Olympiad Test: Triangles- 2 MCQs are made for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Triangles- 2 below.
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Math Olympiad Test: Triangles- 2 - Question 1

An exterior angle of a triangle is 100° and the interior opposite angles are in ratio 1 : 4. The measure of the smallest angle of the triangle is 

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 1

Let the measure of interior opposite angles be x and 4x respectively.
∴ A/Q,
x + 4x = 100°
⇒ 5x = 100° ⇒ x = 20°
∴ The measure of the smallest angle = x = 20°

Math Olympiad Test: Triangles- 2 - Question 2

AN is the bisector of ∠A and AM ⊥ BC. Then a measure of ∠MAN is:

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 2

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°

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Math Olympiad Test: Triangles- 2 - Question 3

The sum of all the exterior angles of a triangle is

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 3

 In ΔABC

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

Math Olympiad Test: Triangles- 2 - Question 4

In an isosceles triangle AB = AC. Side AB is extended to P such that ∠CAP = 108°. The measure of ∠ABC is:

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 4


∵ AB = AC
∴ ∠ABC = ∠ACB
∵ ∠CAP is an exterior angle for ∠ABC,
∵ ∠CAP = ∠ABC + ∠ACB [using (i)]
⇒ 108° = 2∠ABC

Math Olympiad Test: Triangles- 2 - Question 5

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:

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 5

∠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°

Math Olympiad Test: Triangles- 2 - Question 6

The value of x in the adjoining figure will be:

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 6

∵ ∠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°

Math Olympiad Test: Triangles- 2 - Question 7

ABC is an isosceles such that AB = AC and AD is the median to base BC. Then, ∠BAD =

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 7

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

Math Olympiad Test: Triangles- 2 - Question 8

O is any point in the interior of ΔABC, then 

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 8

AB + AC > BC ...(i)
OB + OC > BC ...(ii)
Using (i) and (ii)
AB + AC > OB + OC

Math Olympiad Test: Triangles- 2 - Question 9

In ΔPQR, S is any point on the side QR. Then 

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 9


In ΔPQS,
PQ + QS > PS ...(i)
In ΔPSR,
PR + RS > PS ...(ii)
Using (i) and (ii)
PQ + PR + (QS + RS) > 2PS
⇒ PQ + PR + (QR + QR) > 2PS

Math Olympiad Test: Triangles- 2 - Question 10

In a ΔABC, ∠A = 50°, ∠B = 60°. The longest side of the triangle will be

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 10

∠A = 50°, ∠B = 60° 
∠C = 180° - (∠A + ∠B)
= 180° - 110°
= 70°
∵ ∠C is the largest angle of ΔABC
∴ AB is the largest side of ΔABC.

Math Olympiad Test: Triangles- 2 - Question 11

If two isosceles triangles have a common base, then the line joining their vertices will

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 11


Given: ΔABC and ΔBCP are isosceles ΔS with common base BC.
In ΔABP and ΔACP
AB = AC (given)
BP = PC (given)
AP = AP (common)
∴ ΔABP ≅ ΔACP (S.S.S. criterion)
∴ BO = OC and
∠AOB = ∠AOC = 180°
⇒ 2∠AOC = 180°
⇒ ∠AOB = 90°

Math Olympiad Test: Triangles- 2 - Question 12

If the length of three of the altitudes of a triangle are equal, then the triangle must be a/an

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 12

In ΔBEC and ΔCFB

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 ΔADC and C + A 
⇒ ∠A = ∠C
∴ AB = BC ...(ii)
Using (i) and (ii) 
AB = BC = AC  (Δ should be equilateral)

Math Olympiad Test: Triangles- 2 - Question 13

PQRS is a square and SRT is an equilateral triangle. The measure of ∠TQR is:

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 13

In ΔPTS and ΔQTR
(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°

Math Olympiad Test: Triangles- 2 - Question 14

In ΔABC, AB = AC, and the bisect are of angles B and C intersect at point O, then the ray AO

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 14


∵ ∠B = ∠C

⇒ OB = OC ...(i)
(sides opposite to equal angles are equal)
Now,
In ΔABO and ΔACO
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.)

Math Olympiad Test: Triangles- 2 - Question 15

P is a point equidistant from two lines l and m intersecting at point A, then

Detailed Solution for Math Olympiad Test: Triangles- 2 - Question 15

In ΔABP and ΔACP

PC = PB (Given)
∠ACP = ∠ABP = 90°
AP = AP (common)
∴ ΔABP ≅ ΔACP
(By R.H.S. congruence criterion)
⇒ ∠BAP = ∠CAP (C.P.C.T.)

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