Q25. - Geometry

Question

$\overline{FG}$ is the altitude to the hypotenuse of$△DEF$. Name three similar triangles and find their areas. (Hint: See Theorem $8-1$ and Corollary 1 on pages ).$285-286$

Step-by-Step Solution

Answer

The name of three similar triangle are $\Delta DFE,\Delta DGF,\Delta FGE$ and the area of the three similar triangles are $20,4,16$respectively.

Step 1. Concept Used.

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,

$\Delta DFE~\Delta DGF~\Delta FGE$

Now, In$\Delta DFE,GE$ and $FG$ are the segment and altitude. Thus 

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

Step 2. Write the formula to find the area of triangle.

Area of triangle can be found using the formula

$A=\frac{1}{2}bh$

Whereb= Base of triangle and h= Height of triangle.

Step 3. Find the area of similar triangles.

In$△DFE$

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

In$△DGF$

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

In$△FGE$

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

 

 

Therefore, the name of three similar triangle are $\Delta DFE,\Delta DGF,\Delta FGE$ and the area of the three similar triangles are$20,4,16$ respectively.