Arthur Pythagoras rules! 12 x 12 = 3 x 3 + x2, so x2 = 144 - 9 and x = sqrt 135 = 11.62 so to the nearest foot it reaches 12 feet (actually 11.62)
30 feet. And you don't have to round it to the nearest foot. It's exactly 30 feet.
10.9 [11]
The ladder forms a right angle with the building: the ground and the building forming the right angle and the ladder forming the hypotenuse. If the length of the ladder is L metres, then sin(49) = 12/L So L = 12/sin(49) = 15.9 = 16 metres.
30 feet. And you don't have to round it to the nearest foot. It's exactly 30 feet.
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10.9 [11]
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The ladder forms a right angle with the building: the ground and the building forming the right angle and the ladder forming the hypotenuse. If the length of the ladder is L metres, then sin(49) = 12/L So L = 12/sin(49) = 15.9 = 16 metres.
18
Using Pythagoras' theorem it is 20 feet
The height-to-base ratio for a ladder is determined by the relationship between the vertical height the ladder reaches and the distance from the base of the ladder to the wall or structure it leans against. A common guideline is to maintain a ratio of 4:1, meaning that for every four feet of height, the base should be one foot away from the wall. This helps ensure stability and safety while using the ladder.
Brian makes a ladder by cutting down saplings and tying them together with the help of his shoelaces and pieces of his outer shirt. He arranges the saplings like rungs and leans the ladder against the cliff to climb up.
If you are asking, what's the distance (x) from the bottom of the ladder to the wall, then... x squared + 2 squared = 3 squared x squared + 4 = 9 x squared = 5 x = the square root of 5, or approx 2.24 m
Assuming the wall is vertical, the wall, the ground and the ladder form an isosceles right-angled triangle. Pythagoras tells us that the square of the length of the ladder, in this case 225 equals the sum of the squares of the other two lengths, ie the height where the ladder touches the wall and the bottom of the ladder's distance from the wall. As these distances are equal in an isosceles triangle each must be the square root of (225/2) ie sqrt 112.5 which is 10.6066, as near as makes no difference to 10 ft 71/4 inches