20. The figure shows a cube \( A B C D E F G H \) of side \( 8 \mathrm{~cm} . E A \perp A B \) and \( E C \perp B C \). (a) Find the volume of the pyramid \( A B C D E \). (b) Find the total surface area of the pyramid \( A B C D E \).
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To solve the problem, let's start with the calculations step-by-step. (a) To find the volume of the pyramid \( A B C D E \), we can use the formula for the volume of a pyramid, which is: \[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \] The base of the pyramid is the square \( A B C D \). Since it's a cube with side length \( 8 \) cm, the area of the base \( A B C D \) is: \[ \text{Base Area} = \text{side}^2 = 8 \times 8 = 64 \, \text{cm}^2 \] The height of the pyramid \( E \) from the base \( A B C D \) is equal to the side of the cube, which is \( 8 \) cm. Now, substitute these values into the volume formula: \[ V = \frac{1}{3} \times 64 \times 8 = \frac{512}{3} \approx 170.67 \, \text{cm}^3 \] (b) To find the total surface area of the pyramid \( A B C D E \), we need to calculate the area of the base and the areas of the triangular faces. The area of the base \( A B C D \) is \( 64 \, \text{cm}^2 \). Now, we will find the area of the triangular faces \( ABE, BCE, CDE, DAE \). Each triangular face has a base of \( 8 \, \text{cm} \) and a height (the slant height from \( E \) to the base) that we will calculate. The height from \( E \) to the midpoint of any side of the square base is given by the Pythagorean theorem. The distance from \( E \) to any vertex of the base (like \( A \)) is: \[ \text{distance} = \sqrt{(4)^2 + (4)^2 + (8)^2} = \sqrt{16 + 16 + 64} = \sqrt{96} = 4\sqrt{6} \, \text{cm} \] Each triangle has a height from \( E \) to the base \( A B \) of: \[ \text{slant height} = \frac{E \text{ to } A + E \text{ to } B}{2} \] The area of each triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \), so for one triangle: \[ \text{Area} = \frac{1}{2} \times 8 \times h \] To find \( h \): \[ h = \sqrt{(8^2) - (\frac{8}{2})^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3} \] The area of one triangular face is: \[ \text{Area} = \frac{1}{2} \times 8 \times 4\sqrt{3} = 16\sqrt{3} \] Now, since there are four triangular faces: \[ \text{Total Area (Triangular faces)} = 4 \times 16\sqrt{3} = 64\sqrt{3} \] Finally, the total surface area of the pyramid is: \[ \text{Total Surface Area} = \text{Base Area} + \text{Total Area (Triangular faces)} = 64 + 64\sqrt{3} \approx 64 + 110.85 \approx 174.85 \, \text{cm}^2 \] So, the final answers are approximately: (a) \( \frac{512}{3} \, \text{cm}^3 \) or \( 170.67 \, \text{cm}^3 \), (b) \( 64 + 64\sqrt{3} \, \text{cm}^2 \) or \( 174.85 \, \text{cm}^2 \).
