The answer is 114sqrt{6} in³
A regular hexahedron is actually a cube.
Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²
Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²
D² = f² + a²
f² = 2a²
D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3
Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3
V = a³ = (D/√3)³
We have:
D = 8√2 in
Answer:
The answer I believe is 20
Step-by-step explanation:
Just add -34 + 54
Answer:
The width of the scale model is 33 inches.
Step-by-step explanation:
This question is solved making a relation with the scale model.
In the scale, 3 inches are worth 11 real feet.
The actual width of the building is 121 feet, so we find it's scale by a rule of three.
3 inches - 11 feet
x inches - 121 feet

Simplifying by 3, both sides

The width of the scale model is 33 inches.
The height of the tank must be at least 1 foot, or 12 inches. We know the floor area (which is length x width) must be at least 400 inches. Therefore these minimum dimensions already tell us that the minimum volume is 400 x 12 = 4800 cubic inches. Since we have a maximum of 5000 cubic inches, the volume must be within the range of 4800 - 5000 cubic inches.
We can set the height at exactly 1 ft (or 12 inches). Then we can select length and width that multiply to 400 square inches, for example, L = 40 inches and W = 10 in. This gives us a tank of dimensions 40 x 10 x 12 = 4800 cubic inches, which fits all the criteria.