Chapter 59, Problem 1A

Plans call for triangle $\Delta BCD$to be sheared off a corner of a rectangle as shown in the figure. A decision is made to move one vertex of the triangle 1.5 in. from B to B' and the other vertex from D to D'. How much larger is the area of AB' D' than ABDEF ?

To determine

Difference of area between $A{B}^{\prime }{D}^{\prime }EF$ and $ABDEF$.

Answer to Problem 1A

The difference in area is $25.09{\text{in}}^{\text{2}}$.

Explanation of Solution

Given:

Rectangle with the dimensions is shown below:

Concept used:

Expression for the difference of area is given below:

  $A=area\left(\Delta BCD\right)-area\left(\Delta B^{\prime }C{D}^{\prime }\right)$  ......(1)

Here, difference of area is $A$.

Calculation:

Area of triangle $BCD$ is calculated as follows:

  $\begin{array}{l}area\left(\Delta BCD\right)=\frac{1}{2}BC\times CD\\ area\left(\Delta BCD\right)=\frac{1}{2}\times 18\times \left(17-6\right)\\ area\left(\Delta BCD\right)=99{\text{in}}^{\text{2}}\end{array}$

Area of triangle $B^{\prime }C{D}^{\prime }$ is calculated as follows:

  $\begin{array}{l}area\left(\Delta B^{\prime }C{D}^{\prime }\right)=\frac{1}{2}{B}^{\prime }C\times C{D}^{\prime }\\ area\left(\Delta B^{\prime }C{D}^{\prime }\right)=\frac{1}{2}\times 15.56\times \left(17-7.5\right)\\ area\left(\Delta B^{\prime }C{D}^{\prime }\right)=73.91{\text{in}}^{\text{2}}\end{array}$

Substitute $99{\text{unit}}^{\text{2}}$ for $area\left(\Delta BCD\right)$ and $73.91{\text{unit}}^{\text{2}}$ for $area\left(\Delta B^{\prime }C{D}^{\prime }\right)$ in equation (1).

  $\begin{array}{l}A=99-73.91\\ A=25.09{\text{in}}^{\text{2}}\end{array}$

Thus, the difference in area is $25.09{\text{in}}^{\text{2}}$.

Conclusion:

The difference in area is $25.09{\text{in}}^{\text{2}}$.