Can anybody help mewith these problems...?

Question

1.) A regular triangular pyramid with a slant height of 9m has a volume equal to 50 m^3. Find the lateral area.
the answer must be 84.65m^2

2.) A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude of 25ft and base edge 10 ft. How much surface will rise if 6779 liters of water is added?
the answer must be 12 ft.

please help me with these problems...

Answer 1

2.)

Convert everything to SI units: 25 ft = 7.62 m, 10 ft = 3.05 m, 6779 liters = 6.779 m³.

The height of the pyramid is h = 7.62 m.

Let x is the height of „water pyramid“ (how much surface will rise).

Calculate the area of the base of the pyramid: A = 24.14 m²

Calculate the volume of the pyramid: V₁ = Ah/3 = 61.308 m³

The volume of the „water pyramid“ is V₂ = 6.779 m³.

Calculate the x from the equation

x³/h³ = V₂/V₁ = 0.1106

Solution: x = 3.657 m = 12 feet

Is every step clear?

The first question is easier, try to solve it similarly.

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