Question

If △ABC is an isosceles right-triangle right angled at C such that AC = 5 cm. Then, AB =
(a) 2.5cm
(b) $5\sqrt{2}$ cm
(c) 10 cm
(d) 5 cm

Answer 1

Suppose BC is the ladder which is placed againts the wall OA. The foot of the ladder C is 15 m away from the foot O of the wall and its top reaches the window which is 20 m above the ground.


In right traingle ABC,
AB² = BC² + AC²
⇒ AB² = (5)² + (5)²
⇒ AB² = 25 + 25
⇒ AB² = 50
⇒ AB² = ${\left(5\sqrt{2}\right)}^2$
⇒ AB = $5\sqrt{2}$ cm
Hence, the correct answer is option (b).