Question

2. A right square pyramid is inverted so it is pointing down. The pyramid’s base has an area of 100 square centimeters and the area of one face is 60 square centimeters.
a. What is the volume of the pyramid?


b. The pyramid was filled with a solution. Some of the solution was drained out the vertex of the pyramid into a cylindrical dish with a radius of four centimeters. The level of liquid in the pyramid dropped three centimeters. How deep is the solution in the dish?

Answer 1

Answer:

2. a. The volume of the pyramid is approximately 363.62 cm³

b. The depth of the solution in the dish is approximately 0.15046 cm

Step-by-step explanation:

2. a. A right square pyramid is a pyramid with the vertex at the top directly above the center of the base

The area of the base of the right square pyramid = 100 cm²

The area of one of the face = 60 cm²

∴ The length of the side of the base, s = √(100 cm²) = 10 cm

The area of the face = 1/2 × s × l

Where;

l = The slant height

∴ 1/2 × 10 cm × l = 60 cm²

l = 60 cm²/(5 cm) = 12 cm

l = 12 cm

The height of the pyramid, h = √(l² - (s/2)²)

∴ h = √(12² - 5²) = √(119)

The volume of the pyramid = 1/3 × Base Area × Height

∴ The volume of the pyramid, V = 1/3 × 100 cm² × √(119) cm ≈ 363.62 cm³

The volume of the pyramid, V ≈ 363.62 cm³

b. The initial level of the solution in the pyramid = The pyramid was filled with the solution

∴ The initial level of the solution in the pyramid = The height of the pyramid

The radius of the cylindrical dish into which the solution is poured, r = 4 cm

The level by which the liquid drops in the pyramid after pouring = 3 cm

Therefore, by proportion of volume, we have;

Let V₁ and 'V' represent the volume that was drained out the vertex into the cylindrical dish and the volume of the whole pyramid respectively

Therefore;

V₁/V = (3/(√119))³

V₁ = V × (3/(√119))³

V = 1/3 × 100 cm² × √(119) cm

∴ V₁ = 1/3 × 100 cm² × √(119) cm × (3/(√119))³ =   100 cm² × 1 cm × 3²/(119) = 900/119 cm³

The volume that was drained out the vertex, V₁ = 900/119 cm³

The volume of the cylindrical dish, $V_d$ = π·r²·h

Where h = The depth of the solution in the dish

∴ When $V_d$ = V₁, and r = 4 cm, we have;

π × 4² × h = 900/119

h = (900/119)/(π × 4²) ≈ 0.15046

The depth of the solution in the dish, h ≈ 0.15046 cm.