# Trigonometry - modelling and problem solving

• May 12th 2013, 08:34 PM
elmidge
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.
• May 17th 2013, 10:04 AM
x3bnm
Re: Trigonometry - modelling and problem solving
Quote:

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:

Attachment 28398

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

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

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

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

So $\displaystyle BD = 5\sqrt{2}$

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

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

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

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

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

So the area of:
\displaystyle \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 $\displaystyle 15\text{ cm}^2$