Chapter 11.2, Problem 25WE

To determine

To Find: The area.

Answer to Problem 25WE

  $\Delta DFE\sim \Delta DGF\sim \Delta FGE$

  $20,4,16$

Explanation of Solution

Given: $\overline{FG}$ is the altitude of the hypotenuse of the triangle $DEF$ . Name three similar triangles.

  

According to the theorem $8-1$ if the altitude drawn to the hypotenuse of a right triangle then the two triangles formed are similar to the original triangle.

Thus

  $\Delta DFE\sim \Delta DGF\sim \Delta FGE$

Now

In $\Delta DFE$ , $GEandFG$ are the segment and altitude.

Thus

  $\begin{array}{l}FG=\sqrt{DG\times GE}\\ FG=\sqrt{2(8)}\\ FG=4\end{array}$

Now in $\Delta DFE$

  $\begin{array}{l}b=DE=2+8=10\\ h=FG=4\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(10)(4)\\ A=20\end{array}$

Now in $\Delta DGF$

  $\begin{array}{l}b=DG=2\\ h=FG=4\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(2)(4)\\ A=4\end{array}$

Now in $\Delta FGE$

  $\begin{array}{l}b=GE=8\\ h=FG=4\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(8)(4)\\ A=16\end{array}$

Hence, the area of the three similar triangles are $20,4,16$