CAT 2006 — QA Question 25
Basics of QuadrilateralsEasy
An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?
Answer & solution
Correct answer: 150
- A
75
- B
90
- C
120
- D
135
150
Solution
BP = PC = BC
m ∠BPC = m ∠PCB = m ∠PBC = 60°
Also, PC = CD = BP = AB
â ABP and â PCD are isosceles triangles.
m ∠ABP = m ∠PCD = 90 – 60 = 30°
∴ m ∠APB = m ∠DPC = (180 – 30)/2 = 75°
∴ m ∠APD = 360 – (m ∠APB + m ∠DPC + m ∠BPC) = 360 – (75 + 75 + 60) = 150°
Hence, option (e).