Chapter 11.7, Problem 30WE

To determine

To Find: $LJ$

Answer to Problem 30WE

  $LJ=6\sqrt{2}cm$

Explanation of Solution

Given: In $\Delta LMN$ , $LK=12cm$ . Through point $J$ OF $LK$ a line is drawn parallel to $MN$ dividing the triangle into two regions of equal area.

Observe the figure.

  

Here

  $\Delta LQP\sim \Delta LMN$

Thus

  $\begin{array}{l}\frac{ar(LQP)}{ar(LMN)}=\frac{1}{2}\\ therefore\\ \frac{LJ}{LK}=\frac{1}{\sqrt{2}}\\ LK=12\\ thus\\ \frac{LJ}{12}=\frac{1}{\sqrt{2}}\\ LJ=6\sqrt{2}cm\end{array}$