Example 5
Completion requirements
| Example 5 |
Multimedia |
Write a simplified expression for the height of a right square-based pyramid with a volume of \(2x^2 + 7x + 6\) and a base length of \(x + 2\). State any non-permissible values. Note the formula for the volume of a pyramid is \(V = \frac{s^2h}{3}\), where \(s\) is the base length and \(h\) is the height from the apex to the centre of the base.
Step 1: Rewrite the formula to isolate \(h\).
Step 2: Substitute the expressions given for the volume and base length into the formula.
Step 3: State any NPVs, and simplify.
\[\begin{align}
V &= \frac{{s^2 h}}{3} \\
\frac{{3V}}{{s^2 }} &= h \\
\end{align}\]
V &= \frac{{s^2 h}}{3} \\
\frac{{3V}}{{s^2 }} &= h \\
\end{align}\]
Step 2: Substitute the expressions given for the volume and base length into the formula.
\[\begin{align}
h &= \frac{{3V}}{{s^2 }} \\
h &= \frac{{3\left( {2x^2 + 7x + 6} \right)}}{{\left( {x + 2} \right)^2 }} \\
\end{align}\]
h &= \frac{{3V}}{{s^2 }} \\
h &= \frac{{3\left( {2x^2 + 7x + 6} \right)}}{{\left( {x + 2} \right)^2 }} \\
\end{align}\]
Step 3: State any NPVs, and simplify.
\[\begin{align}
x + 2 &\ne 0 \\
x &\ne - 2 \\
\end{align}\]
\[\begin{align}
h &= \frac{{3\left( {2x^2 + 7x + 6} \right)}}{{\left( {x + 2} \right)^2 }} \\
h &= \frac{{3 {\color{red} \cancel {\color{#444} {\left( {x + 2} \right)}}} \left( {2x + 3} \right)}}{{{\color{red} \cancel {\color{#444}{\left( {x + 2} \right)}}}\left( {x + 2} \right)}} \\
h &= \frac{{3\left( {2x + 3} \right)}}{{x + 2}}, \thinspace x \ne - 2 \\
\end{align}\]
x + 2 &\ne 0 \\
x &\ne - 2 \\
\end{align}\]
\[\begin{align}
h &= \frac{{3\left( {2x^2 + 7x + 6} \right)}}{{\left( {x + 2} \right)^2 }} \\
h &= \frac{{3 {\color{red} \cancel {\color{#444} {\left( {x + 2} \right)}}} \left( {2x + 3} \right)}}{{{\color{red} \cancel {\color{#444}{\left( {x + 2} \right)}}}\left( {x + 2} \right)}} \\
h &= \frac{{3\left( {2x + 3} \right)}}{{x + 2}}, \thinspace x \ne - 2 \\
\end{align}\]