Question

use the information given in the figure to find the length fh. if applicable, round your answer to the nearest whole number. the lengths on the figure are not drawn accurately.

Explanation:

Step1: Use the similarity of triangles

Since \(\angle H = \angle H\) (common angle) and \(\angle FEG+\angle GEH=\angle FEH\), and if \(EG\parallel EH\) (by AA similarity criterion for right - angled triangles), \(\triangle FEG\sim\triangle FEH\). Then \(\frac{FG}{FH}=\frac{EG}{EH}\). Let \(FH = x\), then \(FG=x - 55\). Also, \(FE = 80\), \(GE=73\).
By the property of similar triangles \(\frac{FG}{FE}=\frac{GH}{EH}\) (corresponding sides of similar triangles are proportional). Another way is to use the basic proportionality theorem (Thales' theorem). We know that \(\frac{FG}{FH}=\frac{EG}{EH}\) (from similar triangles \(\triangle FEG\) and \(\triangle FEH\)). Let's use the ratio of corresponding sides: \(\frac{FG}{FE}=\frac{GH}{EH}\). But more straightforwardly, since \(\triangle FEG\sim\triangle FEH\), we have \(\frac{FG}{FH}=\frac{EG}{EH}\). Let's use the ratio of the sides of the two similar right - angled triangles.
We know that \(\frac{FG}{FE}=\frac{GH}{EH}\) (corresponding sides of similar triangles). Let \(FH=x\). Then \(FG=x - 55\).
Since \(\triangle FEG\sim\triangle FEH\), we have \(\frac{FG}{FE}=\frac{GH}{EH}\). But another approach:
We use the property that \(\frac{FG}{FH}=\frac{EG}{EH}\) (from similar triangles). Let's assume \(\triangle FEG\sim\triangle FEH\) (by AA similarity, as both are right - angled and share \(\angle F\)). Then \(\frac{FG}{FE}=\frac{GH}{EH}\). Let \(FH=x\).
We know that \(\frac{FG}{FE}=\frac{GH}{EH}\). But a better way is to use the formula for similar triangles:
If \(\triangle FEG\sim\triangle FEH\), then \(\frac{FG}{FE}=\frac{GH}{EH}\). Let \(FH = x\). Then \(FG=x - 55\).
By the Pythagorean theorem in \(\triangle EGH\), \(EH=\sqrt{73^{2}-55^{2}}=\sqrt{(73 + 55)(73 - 55)}=\sqrt{128\times18}=\sqrt{2304}=48\).
In \(\triangle FEH\), \(FE = 80\), \(EH = 48\). By the Pythagorean theorem \(FH=\sqrt{FE^{2}-EH^{2}}\) (since \(\triangle FEH\) is a right - angled triangle with \(\angle H = 90^{\circ}\)).

Step2: Apply the Pythagorean theorem

In right - angled triangle \(FEH\) with \(FE = 80\) and \(EH = 48\), by the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(c = FE\), \(a=EH\), \(b = FH\)). So \(FH=\sqrt{FE^{2}-EH^{2}}=\sqrt{80^{2}-48^{2}}=\sqrt{(80 + 48)(80 - 48)}=\sqrt{128\times32}=\sqrt{4096}=64\)

Answer:

\(64\)