Chapter 12.2, Problem 4CE

To determine

To find: The area of $\Delta \text{VBC}$ from the given figure.

Answer to Problem 4CE

The area of $\Delta VBC$ from the given figure is $60{\text{ unit}}^2.$

Explanation of Solution

Given information:

V-ABCD is a regular square pyramid.

  

Calculation:

Since, the pyramid has a square base ABCD .

Calculate the measure of $AB$ from the above figure.

  $\begin{array}{l}AB=2\left(OM\right)\\ OM=\frac{AB}{2}\\ OM=\frac{12}{2}\\ OM=6\end{array}$

In $\Delta \text{VOM}$ , apply Pythagoras Theorem to calculate the slant height.

  $\begin{array}{c}{\left(VO\right)}^2+{\left(OM\right)}^2={\left(VM\right)}^2\\ 8^2+6^2=l\\ l^2=64+36\\ l^2=100\end{array}$

Further simplify.

  $\begin{array}{c}l=\sqrt{100}\\ l=\pm 10\end{array}$

The height cannot be negative. Neglect the negative value of $l$ . Take the positive value of $l$ as 10.

In $\Delta VBC$ , calculate the area.

  $\begin{array}{c}\text{Area}\left(\Delta VBC\right)=\frac{1}{2}\times \text{base}\times \text{height}\\ \text{ =}\frac{1}{2}\times \left(12\right)\times \left(10\right)\\ =6\times 10\\ =60{\text{ unit}}^2\end{array}$

Therefore, the area of $\Delta VBC$ from the given figure is $60{\text{ unit}}^2.$