Question 254063
Both questions rely on the Pythagorean formula:  c^2 = a^2 + b^2.
.
The diagonal brace on the gate would be the hypotenuse of a right triangle with a=12 and b=5.
a^2 = 144
b^2 = 25
so
c^2 = 169
By inspection we see that is a perfect square:  sqrt(169) = 13.
That means the brace must be 13 ft.
.
The longest rod that can fit in a trunk will fit in three dimensions, for example, from the lower left front corner to the upper right back corner.
The height of the triangle will be one of known dimensions.
We are told 4x3x2, so we have to assume these are length by width by height.
a^2 = height^2 = 2^2 = 4
The other leg of the triangle the rod defines will be the diagonal of the base of the trunk.
The base of the trunk is a rectangle that is 4x3, so the diagonal will be 5.
b^2 = base^2 = 5^2 = 25
c^2 = 4 + 25 = 29
c = sqrt(29) = 5.385