Chapter 9, Problem 1RP

a.

To determine

To calculate: The side GF if side HG is 4 and side EG is 6.

Answer to Problem 1RP

The side GF is $9$ .

Explanation of Solution

Given information:

Side HG = 4,

Side EG = 6.

Formula used:

The below property is used:

The corresponding sides of similar triangles are congruent.

Calculation:

  

ΔHGE is similar to ΔEGF

The corresponding sides of similar triangles are congruent.

  $\begin{array}{l}\frac{HG}{EG}=\frac{EG}{FG}\\ \frac{4}{6}=\frac{6}{FG}\\ FG=\frac{36}{4}=9\end{array}$

b.

To determine

To find: The side EH if side GH is 4 and side GF is 12.

Answer to Problem 1RP

The side EH is $8$ .

Explanation of Solution

Given information:

Side GH = 4,

Side GF = 6.

Formula used:

The below property is used:

The corresponding sides of similar triangles are congruent.

Calculation:

  

ΔEGF is similar to ΔHEF

The corresponding sides of similar triangles are congruent.

  $\begin{array}{l}\frac{GH}{EH}=\frac{EH}{HF}\\ \frac{4}{EH}=\frac{EH}{16}\\ {\left(EH\right)}^2=64\\ EH=\sqrt{64}=8\end{array}$

c.

To determine

To find: The side HF if side EF is $2\sqrt{5}$ and side GF is 4.

Answer to Problem 1RP

The side HF is $5$ .

Explanation of Solution

Given information:

Side EF = $2\sqrt{5}$ ,

Side GF = 4.

Formula used:

The below property is used:

The corresponding sides of similar triangles are congruent.

Calculation:

  

ΔEGF is similar to ΔHEF

The corresponding sides of similar triangles are congruent.

  $\begin{array}{l}\frac{GF}{EF}=\frac{EF}{FH}\\ \frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{FH}\\ FH=\frac{20}{4}\\ FH=5\end{array}$

d.

To determine

To calculate: The side HF if side EH is 2 and side EF is 3.

Answer to Problem 1RP

The side HF is $3.60$ .

Explanation of Solution

Given information:

Side EH = 2,

Side EF = 3.

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:

  

Side HF can be calculated by applying Pythagoras Theorem.

In right angle triangle HEF , we get

  $\begin{array}{l}{\left(HF\right)}^2={\left(HE\right)}^2+{\left(EF\right)}^2\\ {\left(HF\right)}^2={\left(2\right)}^2+{\left(3\right)}^2\\ {\left(HF\right)}^2=4+9\\ \left(HF\right)=13\\ HF=\sqrt{13}=3.60\end{array}$