Geometric Mean
You are familiar with the Arithmetic Mean of two numbers \(a\) and \(b\) by calculating \(\Large\frac{a+b}{2}\). There is another type of mean called the Geometric Mean. You worked with the geometric mean in the Square Puzzles Investigation. The geometric mean of two positive numbers is calculated by \(\sqrt{a\cdot b}\). For example: the geometric mean of \(7\) and \(11\) is \(\sqrt{(7)(11)} = \sqrt{77}\)
Altitude
Altitude - perpendicular distance from a vertex to the opposite side of a triangle (must have a right angle!). Move the vertex around too.
You are familiar with the Arithmetic Mean of two numbers \(a\) and \(b\) by calculating \(\Large\frac{a+b}{2}\). There is another type of mean called the Geometric Mean. You worked with the geometric mean in the Square Puzzles Investigation. The geometric mean of two positive numbers is calculated by \(\sqrt{a\cdot b}\). For example: the geometric mean of \(7\) and \(11\) is \(\sqrt{(7)(11)} = \sqrt{77}\)
Altitude
Altitude - perpendicular distance from a vertex to the opposite side of a triangle (must have a right angle!). Move the vertex around too.
The altitude from point \(Q\) to \(\overline{TY}\) is drawn here:
An altitude to the hypotenuse is a line segment drawn from point \(B\) (the right angle) perpendicular to the hypotenuse, \(\overline{AC}\).
We can mark \(\angle A\) and \(\angle C\) to make it easier for us to see the similar triangles.
\(\angle A \cong \angle A\) is always true by the reflexive property, and the same is true for \(\angle C\). Therefore, we can draw three right triangles, using transformations, that are similar by \(AA\sim\). Use the app below to see the transformations
Notice that the triangles are drawn with the same orientation to easily show correspondence.
Since we have similar triangles, we can set up proportions to solve for the missing lengths. For example,
\[\frac{AB}{AC}=\frac{BD}{BC}=\frac{AD}{AB}\]
Since we have similar triangles, we can set up proportions to solve for the missing lengths. For example,
\[\frac{AB}{AC}=\frac{BD}{BC}=\frac{AD}{AB}\]
Using transformations we can re-draw the triangles with the same orientation.
The triangles are similar by \(AA\sim\). Since there is only one variable to solve for here, we only need to set up one proportion using corresponding sides.
\(\Large\frac{2}{x} = \frac{x}{17}\)
\((\Large\frac{17}{17})(\frac{2}{x}) = (\frac{x}{17})(\frac{x}{x})\)
\(\Large\frac{34}{17x} = \frac{x^2}{17x}\)
Since the denominators are equal, the numerators are equal.
Therefore: \(x^2 = (2)(17)\)
Notice that the value of \(x\) is the geometric mean of \(2\) and \(17\).
\(x = \sqrt{34}\)
\(\Large\frac{2}{x} = \frac{x}{17}\)
\((\Large\frac{17}{17})(\frac{2}{x}) = (\frac{x}{17})(\frac{x}{x})\)
\(\Large\frac{34}{17x} = \frac{x^2}{17x}\)
Since the denominators are equal, the numerators are equal.
Therefore: \(x^2 = (2)(17)\)
Notice that the value of \(x\) is the geometric mean of \(2\) and \(17\).
\(x = \sqrt{34}\)