Chapter 7.6, Problem 24WE

To determine

To find: The nearest tenth of the meter from the frontages of the lots.

Answer to Problem 24WE

The value of $x,y$ and $z$ is $x=53.33m,y=40m$ and $z=46.67m$ .

Explanation of Solution

Given information:

Three lots with parallel side boundaries extend from the avenue to the boulevard as shown.

  

According to proportionality theorm ratio of two corresponding sides of two figure are same.

Consider that the frontages are x,y and z.

Apply proportionality theorm in first two quadrilaterals.

  $\frac{x}{40}=\frac{y}{30}$

Cross-multiply and solve for x,

  $\begin{array}{l}x=\frac{40y}{30}\hfill \\ x=\frac{4}{3}\hfill \\ \hfill \end{array}$

Apply proportionality theorem in first two quadrilaterals,

  $\begin{array}{cc}\frac{y}{30}& =\frac{z}{35}\\ z& =\frac{35y}{30}\\ z& =\frac{7y}{6}\end{array}$

It is clear from the figure that sum of all the frontages is 140m,

x + y +z =140

Put the values and solve for y,

  $\begin{array}{cc}\frac{4y}{3}+y+\frac{7y}{6}& =140\\ \frac{8y+6y+7y}{6}& =140\\ 21y& =140\times 6\\ y& =40\text{m}\end{array}$

Thus, the Value of x is,

  $\begin{array}{cc}x& =\frac{4y}{3}\\ & =\frac{4}{3}\times 40\\ & =53.33\text{m}\end{array}$

And the value of z is,

  $\begin{array}{cc}z& =\frac{7y}{6}\\ & =\frac{7}{6}\times 40\\ & =46.67\text{m}\end{array}$

Therefore, the value of $x,y$ and $z$ is $x=53.33m,y=40m$ and $z=46.67m$ .