Pyramid Problems with Solutions

Problem Statement: Find a formula for the total surface area of the pyramid shown above.

Solution:

The surface consists of four triangular faces (two pairs of congruent triangles) and a rectangular base.

Let \(S\) be the midpoint of diagonal \(AC\). For triangle \(DOC\):

\[\text{Altitude } H_1 = \sqrt{h^2 + \left(\dfrac{L}{2}\right)^2}\] \[\text{Area}(DOC) = \dfrac{1}{2} W \cdot \sqrt{h^2 + \left(\dfrac{L}{2}\right)^2}\]

For triangle \(AOD\):

\[\text{Altitude } H_2 = \sqrt{h^2 + \left(\dfrac{W}{2}\right)^2}\] \[\text{Area}(AOD) = \dfrac{1}{2} L \cdot \sqrt{h^2 + \left(\dfrac{W}{2}\right)^2}\]

Total lateral surface area:

\[A_{\text{lateral}} = 2 \times \text{Area}(DOC) + 2 \times \text{Area}(AOD)\] \[A_{\text{lateral}} = W \sqrt{h^2 + \left(\dfrac{L}{2}\right)^2} + L \sqrt{h^2 + \left(\dfrac{W}{2}\right)^2}\]

Total surface area including base:

\[A_{\text{total}} = W \sqrt{h^2 + \left(\dfrac{L}{2}\right)^2} + L \sqrt{h^2 + \left(\dfrac{W}{2}\right)^2} + LW\]

Problem Statement: A pyramid with a square base of side \(x\) and height \(h\) has a volume of \(1000 \text{ cm}^3\). Find the value of \(x\) that minimizes the surface area.

Solution:

Given volume:

\[\dfrac{1}{3} h x^2 = 1000 \quad \Rightarrow \quad h = \dfrac{3000}{x^2}\]

Using the formula from Problem 1 with \(L = W = x\):

\[S = 2x \sqrt{h^2 + \left(\dfrac{x}{2}\right)^2} + x^2\]

Substitute \(h\):

\[S = 2x \sqrt{\left(\dfrac{3000}{x^2}\right)^2 + \left(\dfrac{x}{2}\right)^2} + x^2\] \[S = 2x \sqrt{\dfrac{9,000,000}{x^4} + \dfrac{x^2}{4}} + x^2\]

To minimize \(S\), we can use calculus or graphical methods. The minimum occurs at approximately:

\[x \approx 12.9 \text{ cm}\]

Note: The exact solution can be found using calculus by setting \(\dfrac{dS}{dx} = 0\).