Chapter 12, Problem 13MCQ

To determine

To find the surface area of the given regular pyramid.

Answer to Problem 13MCQ

  $16+8\sqrt{85}$ inches²

Explanation of Solution

Given:

  

Height of pyramid, $h=9$ inches

Side of base of pyramid, $a=4$ inches

Formula used:

Surface area of a regular pyramid $=A+\frac{1}{2}ps$

where $A=$ Area of base, $p=$ perimeter of base, $s=$ slant height of the pyramid.

Slant height $,s^2={\left(h\right)}^2+{\left(r\right)}^2\text{or}s=\sqrt{{\left(h\right)}^2+{\left(r\right)}^2}$

Calculation:

Since height of the pyramid is $h=9$ and edge of the base is $a=4$ .

Therefore, slant height of the pyramid is given as

  $\Rightarrow {\left(9\right)}^2+{\left(2\right)}^2=s^2$

  $\begin{array}{l}\Rightarrow 81+4=s^2\\ \Rightarrow 85=s^2\\ \Rightarrow s=\sqrt{85}\end{array}$

For the given pyramid,

Area of base, $A=a^2$

  $\begin{array}{l}=4^2\\ =16\end{array}$

Perimeter of base, $p=4a$

  $\begin{array}{l}=4\times 4\\ =16\end{array}$

Surface area $=A+\frac{1}{2}ps$

  $\begin{array}{l}=16+\frac{1}{2}\times 16\times \sqrt{85}\\ =16+8\sqrt{85}\end{array}$

Conclusion:

Therefore, surface area of the given pyramid is $16+8\sqrt{85}$ inches².