Math Olympiad Test: Triangles- 2 - Free MCQ with solutions Class 9 Mathematics

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°

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°

 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°

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

∠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°

∵ ∠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°

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

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

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

∠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.

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°

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)

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°

∵ ∠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.)

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.)