Question

Geometry: VA > Chapter 11 > Section Exercises5 of anFind the lateral area and the surface area of the regularpyramid. Round your answers to the nearesthundredth, if necessary.8 in.405 in.

Answer 1

$\begin{gathered} \text{Lateral area of a triangular pyramid=3(}\frac{1}{2}\times base\times height) \\ =\frac{3}{2}bh \end{gathered}$$\begin{gathered} \text{where b=base=5in} \\ h=\text{height}=8in \end{gathered}$$\begin{gathered} \text{Lateral area=}\frac{\text{3}}{2}\times5\times8 \\ =3\times5\times4=60in^2 \end{gathered}$$\begin{gathered} \text{Surface of a triangular pyramid=}4(\frac{1}{2}\times b\times h) \\ =2bh \\ =2\times5\times8=80in^2 \end{gathered}$

Hence, the Total surface area= 80in²

Lateral area=60in².