Chapter 9.4, Problem 25PSC

a.

To determine

To calculate:Thelength of a side HF.

Answer to Problem 25PSC

The length of a side HF is 8.

Explanation of Solution

Given information:

Side $GH=10,$

Side $HE=17,$

Base of a triangle $(GE)=21,$

  $\angle GFH=90°.$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Let

  $\begin{array}{l}GF=x,HF=y\\ FE=21-x\end{array}$

Apply Pythagoras Theorem in triangles GHF and HFE.

In right angle triangle GHF , we get

  $\begin{array}{l}{\left(10\right)}^2={\left(x\right)}^2+{\left(y\right)}^2\\ 100-x^2=y^2----Equation1\end{array}$

In right angle triangle HFE , we get

  $\begin{array}{l}{\left(17\right)}^2={\left(21-x\right)}^2+{\left(y\right)}^2\\ {\left(17\right)}^2-{\left(21-x\right)}^2=y^2----Equation2\end{array}$

From Equation 1 and Equation 2, we get

  $\begin{array}{l}100-x^2={\left(17\right)}^2-{\left(21-x\right)}^2\\ 100-x^2=289-441+42x-x^2\\ 100=-152+42x\\ 252=42x\\ 6=x\\ \therefore x=6\\ \therefore GF=6\end{array}$

In right angled triangle GHF , we get

  $\begin{array}{l}{\left(6\right)}^2+{\left(HF\right)}^2={\left(10\right)}^2\\ 36+{\left(HF\right)}^2=100\\ HF=\sqrt{64}\\ \therefore HF=8\end{array}$

b.

To determine

To show:Triangles $\Delta EHF$ and $\Delta HGF$ are similar.

Explanation of Solution

Given information:

Side $GH=10,$

Side $HE=17,$

Base of a triangle $(GE)=21,$

  $\angle GFH=90°.$

Formula used:

The below property is used:

If two triangles are similar then their corresponding sides must be in proportion

Proof:

  

If $\Delta EHF~\Delta HGF$ then their corresponding sides must be in proportion.

  $\begin{array}{l}\frac{EH}{HF}?\frac{HG}{GF}\\ \frac{17}{8}?\frac{10}{6}\\ 102\ne 80\end{array}$

Their corresponding sides are not in proportion, so the triangles are not congruent.

Thus $\Delta EHF$ and $\Delta HGF$ are not congruent.