Question

V- ABCD is a regular square pyramid. Find the following numerical answers.

$Areaof\Delta VBC=\_\_\_$

Short answer

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

Step-by-step solution

Step 1. Given information.

The given$V-ABCD$ is a regular square pyramid.

Also,$AB=12$ and$BC=12$ .

Step 2. Determine the value.

The edge length of square base is 12 and height of pyramid is 8.

The length of OMcan be calculated as:

$\begin{array}{c}OM=\frac{1}{2}AB\\ =\frac{1}{2}\left(12\right)\\ =6\end{array}$

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

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

Further simplify.

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

The height cannot be negative. Neglect the negative value of l.

Therefore, the value of the slant height l is 10 unit.

Step 3. Determine the area.

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 $60{\text{unit}}^2$