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(}(1)/(2)* base* height) \\ =(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}*5*8 \\ =3*5*4=60in^2 \end{gathered}`
`\begin{gathered} \text{Surface of a triangular pyramid=}4((1)/(2)* b* h) \\ =2bh \\ =2*5*8=80in^2 \end{gathered}`

Hence, the Total surface area= 80in²

Lateral area=60in².