2007 Section 2 Question 20

20) A

This is a simple Pythagoras question. 

Let long side = A = (6 + √5) 

Let shorter side = B = (3 + 2√5)

Let third side = C 

A² = B² + C²

C² = A² – B²

C² = (6 + √5)² – (3 + 2√5)²

C² = [6² + (2 x 6 x √5 ) – √5²] + [3² + (2 x 3 x 2√5) + (2√5)²]

C² = (36 + 12√5 + 5) – (9 + 12√5 + 20) 

C² = (41 + 12√5) – (29 + 12√5)

C² = 41 – 29 + 12√5 – 12√5

C² = 12

C = √12

C = √(2 x 2 x 3)
C = √(2² x 3)

C = 2√3