Question

The net below shows the dimensions of a triangular pyramid with equilateral base.What is the lateral surface area of this triangular pyramid?A: 36 cm²B: 54 cm²C: 64.8 cm²D: 72 cm²

Answer 1

$\begin{gathered} \text{area triangle base=}\frac{base\cdot high}{2} \\ \text{high}=\sqrt[]{5^2-2.5^2} \\ \text{high}=4.3\operatorname{cm} \\ \text{area triangle base=}\frac{5cm\cdot4.3\operatorname{cm}}{2} \\ \text{area triangle base=}10.75\operatorname{cm}^2 \\ \text{Surface area =10.75 cm}^2+(3)\cdot(\frac{5\operatorname{cm}\cdot7.2\operatorname{cm}}{2}) \\ \text{Surface area =10.75 cm}^2+(3)\cdot(18cm^2) \\ \text{Surface area =10.75 cm}^2+54\operatorname{cm} \\ \text{Surface area =}64.75\operatorname{cm}\approx64.8\operatorname{cm} \\ The\text{Surface area is }64.8\operatorname{cm} \end{gathered}$