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a regular square pyramid has base whose area is 250 cm^2. a section parallel to the base and 31.8 cm above it has an area of 40 cm^2 . find the ratio of the volume of the frustum to the volume of the pyramid.

Sagot :

We are given that the area of the base of a pyramid is 250 cm^2. We are asked to determine the ratio of the volumes of the frustum and the volume of the pyramid. To do that, let's remember that the volume of a pyramid is given by:

[tex]V=\frac{1}{3}A_bh[/tex]

Where:

[tex]\begin{gathered} A_b=\text{ area of the base} \\ h=\text{ height} \end{gathered}[/tex]

Now, the volume of the frustum is given by:

[tex]V_f=\frac{1}{3}h_f(A_b+A_f+\sqrt[]{A_bA_f})[/tex]

Where:

[tex]\begin{gathered} h_f=\text{ height of the frustum} \\ A_f=\text{ area of the base of the frustum} \end{gathered}[/tex]

Now, the ratio of between the volume of the frustum and the volume of the pyramid is:

[tex]\frac{V_f}{V_{}}=\frac{\frac{1}{3}h_f(A_b+A_f+\sqrt[]{A_bA_f})}{\frac{1}{3}A_bh_{}}[/tex]

We can cancel out the 1/3 and we get:

[tex]\frac{V_f}{V_{}}=\frac{h_f(A_b+A_f+\sqrt[]{A_bA_f})}{A_bh}[/tex]

Now, we need to determine the heights. To do that we will use the fact that the ratio of the squares of the height of the pyramid and the height to the top of the pyramid is equivalent to the ratio of the areas, therefore, we have:

[tex](\frac{h}{h_t})^2=\frac{A_b}{A_f}[/tex]

Now we substitute the areas:

[tex](\frac{h}{h_t})^2=\frac{250}{40}[/tex]

Taking square root we get:

[tex]\frac{h}{h_t}=\sqrt[]{\frac{250}{40}}[/tex]

Solving the operations:

[tex]\frac{h}{h_f}=2.5[/tex]

Now we multiply by the height of the frustum on both sides:

[tex]h=2.5h_t[/tex]

Now, let's look at the following diagram:

This shows us that the height of the frustum plus the height to the top must be equal to the height of the pyramid, therefore:

[tex]h=h_t+31.8[/tex]

Substituting the relationship we determined for the height to the top we get:

[tex]h=\frac{h}{2.5}+31.8[/tex]

Now we solve form the height "h" first by subtracting h/2.5 from both sides:

[tex]\begin{gathered} h-\frac{h}{2.5}=31.8 \\ \end{gathered}[/tex]

Solving the operations:

[tex]\frac{1.5h}{2.5}=31.8[/tex]

Now we multiply both sides by 2.5:

[tex]\begin{gathered} 1.5h=31.8\times2.5 \\ 1.5h=79.5 \end{gathered}[/tex]

Now we divide by 1.5:

[tex]h=\frac{79.5}{1.5}=53[/tex]

Therefore, the height of the pyramid is 53 cm. Now we substitute in the ratio of the volumes and we get:

[tex]\frac{V_f}{V_{}}=\frac{h_f(A_b+A_f+\sqrt[]{A_bA_f})}{A_bh}[/tex]

Substituting the values:

[tex]\frac{V_f}{V_{}}=\frac{(31.8)(250+40+\sqrt[]{(250)(40)})}{(250)(53)}[/tex]

Solving the operations:

[tex]\frac{V_f}{V}=0.936_{}[/tex]

Therefore, the ratio is 0.936

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