Chapter 11.2, Problem 39WE

To determine

To Find: The two possible values for the area.

Answer to Problem 39WE

  $936c{m}^2,504c{m}^2$

Explanation of Solution

Given: The diagonals of a parallelogram are $82and30$ and one altitude is $18$ cm long.

Consider the parallelogram $ABCD$

Draw an altitude $BF$ .

  

Using Pythagoras theorem in triangle $BFD$ .

  $\begin{array}{l}B{D}^2=B{F}^2+F{D}^2\\ 8^2={18}^2+F{D}^2\\ FD=80cm\end{array}$

Now in triangle $AEC$ .

  $\begin{array}{l}A{C}^2=A{E}^2+E{C}^2\\ {30}^2={18}^2+E{C}^2\\ EC=24cm\end{array}$

Now

Consider the right triangles $MNOP$

  $\begin{array}{l}AD\cong BC\\ AF=BN=18\\ i.e.\\ \Delta AED\cong \Delta BFC\\ Thus\\ DE=CF(byCPCT)\end{array}$

Now

  $\begin{array}{l}DF=DE+EC+CF\\ DE=28cm\\ DC=DE+EC\\ DC=52cm\\ now\\ A=bh\\ b=DC=52cm\\ h=18cm\\ A=(520(18)\\ A=936c{m}^2\end{array}$

Now consider another parallelogram.

  

Similarly

Area of parallelogram $MNOP$

  $\begin{array}{l}b=28cm\\ h=18\\ A=bh\\ A=(28)(18)\\ A=504c{m}^2\end{array}$

Hence, two possible values of area are $936c{m}^2,504c{m}^2$