Chapter 15, Problem 26CR

To determine

To find: the lengths of the segment joining the mid-points of two opposite sides.

Answer to Problem 26CR

  $PG=7.6$ .

Explanation of Solution

Given:

Diagonals of the kite are $6\text{and}14$ .

Calculation:

According to the question, the diagonals of the kite are $6\text{and}14$ .

Let us consider, $Q,P,G$ be the mis-points of $\overline{DC},\overline{AD},\overline{BC}$ respectively.

Now, draw the pictorial representation of the kite as per the above conditions.

  

  $\begin{array}{l}BD=6\left\{given\right\}\\ AC=4\left\{given\right\}\end{array}$

Now, $Q,P,G$ be the mis-points of $\overline{DC},\overline{AD},\overline{BC}$ respectively.

  $\begin{array}{l}GQ=3\{\text{by Midline theorem}}\\ PQ=7\{\text{by Midline theorem}}\end{array}$

  $\begin{array}{l}\angle 1={90}^{\circ }\{\text{diagonals of the kite bisects}}\\ \angle 1=\angle 2\{\text{since, BD}\parallel \text{GQ}}\\ \angle 2=\angle 3\{\text{since, AC}\parallel \text{PQ}}\end{array}$

As $\angle GQP$ is a right triangle, then by Pythagoras theorem,

  $\begin{array}{l}P{G}^2=P{Q}^2+G{Q}^2\\ P{G}^2=7^2+3^2\\ P{G}^2=49+9=58\\ PG=\sqrt{58}\\ \Rightarrow PG=7.6\end{array}$

Therefore, the lengths of the segment joining the mid-points of two opposite sides is $PG=7.6$ .