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DanielleElmas [232]
3 years ago
14

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 a

rea 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?
Mathematics
1 answer:
-Dominant- [34]3 years ago
4 0

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.

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