Math Olympiad Test: Triangles- 1 - Class 9 MCQ

Let the angles of triangle be x, y and
180° - ( x + y)
∴ 180° - (x + y) = (x + y)
⇒ 2(x + y) = 180°
⇒ x + y = 90°
∴ (180°- (x + y))  = 180° - 90° = 90°
∴ The triangle will be right angled triangle

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

∠AOC = 90º​​​​​​​ + 1/2∠​​​​​​​B
= 90º + 1/2 x 90º
= 90º +​​​​​​​ 45º
= 135º

x = 90° - 1/2∠A
= 90°​​​​​​​ - 1/2 x 70°
= 90°​​​​​​​ - 35°​​​​​​​ = 55°

x = 90° + 1/2∠A
= 90°​​​​​​​ + 1/2 x 72°
= 90°​​​​​​​ + 36°
= 126°

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

∠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°

In Δs ABC and DCE
∠ABE  = ∠DCE = ∠B/2
AB = CD (given)
BE = CE (∵ ∠ABE =∠DCE = ∠B/2)
∴ ΔABE ≅ ΔDCE [ By S-S-S congruence]

In ΔABC
AB + BC > AC ...(i)
In ΔADC,
AD + DC > AC ...(ii)
Using (i) and (ii)
CD + AD + AB + BC > 2AC

In ΔABC,

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