Chapter 12.5, Problem 1AE

To determine

Toshow:The diagram of the triangle and show that the area is $A=\frac{3x\left(4-x\right)}{2}$ .

Explanation of Solution

Given information:The original sides of triangle are $5$ , $5$ , $8$ instead of $5$ , $5$ , $6$ .

Proof:

The diagram according to given dimensions is shown in below.

  

As the area of the rectangle $\text{CDEF}$ is equal to the product of length of side $\text{CD}$ and $\text{DE}$ .

So, the area of the rectangle is equal to $2xy$ .

In right-angle $\Delta \text{VOB}$ , use the Pythagoras theorem for the length of side $\text{VO}$ .

  $\begin{array}{c}{\text{P}}^2+{\text{B}}^2={\text{H}}^2\\ {\text{VO}}^2+{\text{OB}}^2={\text{VB}}^2\\ {\text{VO}}^2+4^2=5^2\\ {\text{VO}}^2+16=25\end{array}$

Further simplify,

  $\begin{array}{c}{\text{VO}}^2+16=25\\ {\text{VO}}^2=9\\ \text{VO}=\sqrt{9}\\ \text{VO}=3\end{array}$

So, by the Pythagoras theorem the length of the side $\text{VO}$ is equal to $3$ .

Now assume that $\angle \text{B}$ is equal to $c$ and as the sum of interior angles of triangle is $180°$ .So, the third angle of $\Delta \text{VOB}$ and $\Delta \text{EDB}$ is:

  $180°-90°-c=90°-c$

As all the three angles of $\Delta \text{VOB}$ and $\Delta \text{EDB}$ are equal.So, the triangles are similar $\Delta \text{VOB}\sim \Delta \text{EDB}$ .

So, by the similarity of two triangles the corresponding sides of the triangles are in ratio that is $\frac{\text{ED}}{\text{VO}}=\frac{\text{DB}}{\text{OB}}$ .

Now simplify the proportion.

  $\begin{array}{c}\frac{\text{ED}}{\text{VO}}=\frac{\text{DB}}{\text{OB}}\\ \frac{y}{3}=\frac{4-x}{4}\\ y=\frac{3\left(4-x\right)}{4}\end{array}$

Now substitute $\frac{3\left(4-x\right)}{4}$ for $y$ in area formula.

  $\begin{array}{c}\text{A}=2xy\\ =2x\left(\frac{3\left(4-x\right)}{4}\right)\\ =\frac{3x\left(4-x\right)}{2}\end{array}$

Hence, the area is given by $\text{A}=\frac{3x\left(4-x\right)}{2}$ .