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