# Thread: Trigonometry - modelling and problem solving

1. ## Trigonometry - modelling and problem solving

I am having trouble finding the length of a sloping edge in the following problem:

A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the legth of a sloping edge and the area of a triangular face.

2. ## Re: Trigonometry - modelling and problem solving

Originally Posted by elmidge
I am having trouble finding the length of a sloping edge in the following problem:

A right pyramid, ht 6 cm stands on a square base of side 5cm. Find the length of a sloping edge and the area of a triangular face.
Suppose the right pyramid is like the one in the following image:

Right pyramid $ABCDE$ has a square base $BCDE$. This square base $BCDE$ has all equal sides where each side has length $5 \text{ cm}$. We also know that the height of the right pyramid is $6\text{ cm}$ which is $AO$

Now in sqare $BCDE$, $\triangle BED$ is a right triangle with $\angle BED$ a right angle.
So by Pythogorean theorem:

$BD^2 = BE^2 + ED^2$

But $BE = ED = 5 \text{ cm}$

So $BD = 5\sqrt{2}$

$BO = \frac{1}{2}BD = \frac{5}{\sqrt{2}} \text{ cm}$

We know that $\text{height} = AO = 6\text{ cm}$

Now $\triangle AOB$ is a right triangle with $\angle AOB$ a right angle.
So:

$AB^2 = AO^2 + BO^2$
$\therefore AB = \frac{\sqrt{97}}{\sqrt{2}}......\text{[because AO = } 6 \text{ cm and BO }= \frac{5}{\sqrt{2}}\text{ cm]}$

That's the length of the slopping edge which is $\frac{\sqrt{97}}{\sqrt{2}}\text{ cm}$

So the area of:
\begin{align*}\triangle ABC =& \frac{1}{2} \times BC \times AO\\=& 15\text{ cm}^2...\text{[because BC} = 5 \text{ cm and AO} = 6 \text{ cm]}\end{align*}

So the area of the triangular face is $15\text{ cm}^2$