Chapter 5.5, Problem 29WE

a.

To determine

To determine the new quadrilateral formed by joining the midpoints of sides of a convex kite.

Answer to Problem 29WE

Rectangle

Explanation of Solution

Given:

A convex kite

  

Consider the kite  $ABCD$  as shown below.

  $P,Q,R,S$ are the midpoints of sides  $AB,AD,BC$ and $CD$ respectively.

Join  $BD$  and  $AC.$

  

$\Delta ABD$ is an isosceles triangle. [ $\because \overrightarrow{AD}=\overrightarrow{DC}$ ]

Now, as $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $AD$ ,

  $PQ=\frac{1}{2}BD$ and $PQ$ is parallel to $BD$ [Midpoint theorem]

Similarly in $\Delta BCD,RS=\frac{1}{2}BD$ and $RS$ is parallel to $BD$ .

Hence, $PQ$ is parallel to $RS$ and $PQ=RS$ .

Similarly, $PR=QS$ and $PR$ is parallel to $QS$ .

Now in $\Delta ABC$ and $\Delta ACD,$

  $AB=AC$

  $BC=CD$

  $AC=AC$

Hence by side-side-side (SSS) congruency criteria,

  $\Rightarrow \angle BAC=\angle DAC$ [Corresponding parts of congruent triangles are congruent (equal)]

Therefore, $AC$ is the bisector of isosceles $\Delta ABC$ and hence $AC\perp BD$ and $PQ\perp PR$ .

Hence the quadrilateral $PQRS$ is a rectangle.

Conclusion:

Therefore, quadrilateral formed by joining the midpoints of sides of a convex kite is a rectangle.

b.

To determine

To determine the new quadrilateral formed by joining the midpoints of sides of a nonconvex kite.

Answer to Problem 29WE

A parallelogram

Explanation of Solution

Given:

A nonconvex kite

  

$P,Q,R,$ and $S$ are the midpoints of $AB,BC,CD,$ and $DA$ respectively.

Now join $DB$

In $\Delta ABD$ , as $S$ and $P$ are midpoints of $AD$ and $AB$ respectively.

By Midpoint Theorem,

  $SP$ is parallel to $DB$ ….(i) and $SP=\frac{1}{2}DB$ ….(ii).

In $\Delta CBD$ , as $R$ and $Q$ are midpoints of $CD$ and $CB$ respectively.

By Midpoint Theorem,

  $RQ$ is parallel to $DB$ ….(iii) and $RQ=\frac{1}{2}DB$ ….(iv).

  $\therefore$ From equation (i) and equation (ii),

  $SP$ is parallel to $DB$ is parallel to $RQ$ .

  $\Rightarrow SP$ is parallel to $RQ$ .

From equation (ii) and equation (iv),

  $SP=\frac{1}{2}DB=RQ$

  $\Rightarrow SP=RQ$

Similarly by joining $AC$ ,

  $PC=SR$ and $PC$ is parallel to $SR$ .

Therefore, $PQRS$ is a parallelogram.

Conclusion:

Therefore, $PQRS$ is a parallelogram.