Two hallways, one 8 feet wide and the other 6 feet wide, meet at right angles. Determine the length of the longest ladder that can be carried horizontally from one hallway into the other.
Make a sketch showing the ladder going around the corner and touching the side walls.
let the part of the ladder in the wider hallway be L1, and the other part of the ladder L2
You should see two similar right-angled triangles.
Let the angle be Ø so that in the larger triangle
sinØ = 8/L1 , and in the smaller
cosØ = 6/L2
L1 = 8/sinØ = 8(sinØ)^-1
L2 = 6/cosØ = 6(cosØ)^-1
L = L2 + L1
= 6(cosØ)^-1 + 8(sinØ)^-1
dL/dØ = -6(cosØ)^-2(-sinØ) - 8sinØ)^-2 (cosØ)
= 6sinØ/cos^2 Ø - 8cosØ/sin^2 Ø
= 0 for a max of L
6sinØ/cos^2 Ø = 8cosØ/sin^2 Ø
sin^3 Ø/cos^3 Ø = 8/6
tan^3 Ø ‚ = 4/3
tanØ = (4/3)^(1/3) = 1.10064...
Ø = 47.7429...°
L = 8/sinØ + 6/cosØ
= appr 19.73 ft
To find the length of the longest ladder that can be carried horizontally from one hallway into the other, we can use the Pythagorean theorem.
Let's consider the 8-foot wide hallway as the base of a right-angled triangle and the 6-foot wide hallway as the height.
Using the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides, we can calculate the length of the ladder.
The equation for the Pythagorean theorem is: c² = a² + b², where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.
In this case, the hypotenuse is the length of the ladder, and the other two sides are the 8-foot and 6-foot hallways.
Substituting the values into the equation:
c² = 8² + 6²
c² = 64 + 36
c² = 100
Taking the square root of both sides:
√c² = √100
c = 10 feet
Therefore, the longest ladder that can be carried horizontally from one hallway into the other is 10 feet long.