Chapter 12.5, Problem 6CK

To determine

To prove:The area of the rectangleis $A=\frac{8x\left(3-x\right)}{3}$ .

Explanation of Solution

Given information:Consider the given figure.

  

Proof:

In the given figure, $OB=3$ , $VB=5$ , $ED=y$ , $OD=x$ , $DB=3-x$ , and $CD=2x$

By the similarity property of triangles:

  $\begin{array}{c}\frac{y}{4}=\frac{3-x}{3}\\ y=\frac{4}{3}\left(3-x\right)\end{array}$

The formula used to find the area of rectangle is:

  $\text{Area}=\text{length}\times \text{breadth}$

The area of rectangle CDEF is:

  $\text{Area}=CD\times ED$

Substitute $2x$ for $CD$ and $y$ for $ED$ in the above formula.

  $\text{Area}=2xy$

Substitute $\frac{4}{3}\left(3-x\right)$ for $y$ in the above formula.

  $\begin{array}{c}\text{Area}=2x\left(\frac{4}{3}\left(3-x\right)\right)\\ =\frac{8x\left(3-x\right)}{3}\end{array}$

Hence it is proved thatthe area of the rectangle is $A=\frac{8x\left(3-x\right)}{3}$ .