Question

finding the surface area of a three - dimensional figure continued
explain how your work in problem 2 would be different if the triangular bases were isosceles but not equilateral.

Explanation:

Step1: Find base - area of square - based pyramid

The base of the square - based pyramid is a square with side length $s = 8$ ft. The area of the base $B$ is given by the formula $B=s^{2}$. So, $B = 8^{2}=64$ square feet.

Step2: Find area of one triangular face

The formula for the area of a triangle is $A=\frac{1}{2}bh$, where the base $b = 8$ ft and the slant - height $h = 9.2$ ft. So, $A=\frac{1}{2}\times8\times9.2 = 36.8$ square feet.

Step3: Find total lateral area

Since a square - based pyramid has 4 triangular faces, the total lateral area $L$ is $L = 4A$. So, $L=4\times36.8 = 147.2$ square feet.

Step4: Find surface area

The surface area $S$ of a square - based pyramid is the sum of the base area and the lateral area, $S=B + L$. So, $S=64+147.2=211.2$ square feet.

For the second part:
If the triangular bases of the pyramid were isosceles but not equilateral:

• In the case of an equilateral - based pyramid (original problem), all triangular faces are congruent, so we can find the area of one face and multiply by 4.
• If the bases are isosceles non - equilateral, the non - base sides of the triangular faces may have different lengths. We would need to find the area of each non - congruent triangular face separately using Heron's formula (if we know the side lengths) or the formula $A=\frac{1}{2}bh$ (where we need to find the appropriate base and height for each face). The height of each non - congruent triangular face would be different, and we couldn't simply multiply the area of one face by 4 to get the lateral area.

Answer:

The surface area of the square - based pyramid is 211.2 square feet. If the triangular bases were isosceles but not equilateral, we would need to find the area of each non - congruent triangular face separately instead of multiplying the area of one triangular face by 4 to get the lateral area.