Question

quiz 1: surface areas of prisms and pyramids
what is the surface area of a regular square pyramid with a slant height of 12 cm and a lateral edge of 13 cm?
100 cm²
240 cm²
340 cm²
640 cm²

Explanation:

Step1: Find base edge length

A regular square pyramid has lateral edge (slant height of the triangular face's right triangle) \( l = 13 \, \text{cm} \), slant height (height of triangular face) \( s = 12 \, \text{cm} \). The base edge half - length \( a/2 \) can be found by Pythagorean theorem: \( \frac{a}{2}=\sqrt{l^{2}-s^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169 - 144}=\sqrt{25} = 5 \, \text{cm} \). So base edge \( a = 10 \, \text{cm} \).

Step2: Calculate base area

Base is a square, so base area \( B=a^{2}=10^{2} = 100 \, \text{cm}^{2} \).

Step3: Calculate lateral surface area

Lateral surface area of a regular square pyramid \( LSA = 2\times a\times s\) (since there are 4 triangular faces, each with area \( \frac{1}{2}\times a\times s \), so total lateral surface area \( 4\times\frac{1}{2}\times a\times s=2as \)). Substituting \( a = 10 \, \text{cm} \) and \( s = 12 \, \text{cm} \), we get \( LSA=2\times10\times12 = 240 \, \text{cm}^{2} \).

Step4: Calculate total surface area

Total surface area \( TSA=B + LSA \). Substituting \( B = 100 \, \text{cm}^{2} \) and \( LSA = 240 \, \text{cm}^{2} \), we get \( TSA=100 + 240=340 \, \text{cm}^{2} \).

Answer:

\( 340 \, \text{cm}^2 \)