Two ladders AB and CD are inclined on two floors of a building as shown below, such that BC = √2 m. If the height of a floor is 4√2 m, how much minimum distance is to be covered to walk from point A to point D?
(8 + 9√2) m
Given:
BC = √2 m
Height of floor = 4√2 m
Formula Used:
sin θ = perpendicular/hypotenuse
Calculation:
BX = DY = 4√2 m
Considering ∆AXB,
⇒ sin 30° = perpendicular/hypotenuse = BX/AB
⇒ ½ = 4√2/AB
⇒ AB = 8√2 m
Similarly,
Considering ∆CYD,
⇒ sin 45° = perpendicular/hypotenuse = DY/CD
⇒ 1/√2 = 4√2/CD
⇒ CD = 8 m
Now, minimum distance to be covered to walk from point A to point D = AB + BC + CD
∴ Required distance = 8√2 + √2 + 8 = (8 + 9√2) mSimilar Questions
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