Chapter 5.4, Problem 37WE

To determine

To write: name of the quadrilateral formed by the bisectors of its angles

Answer to Problem 37WE

Rectangle

Explanation of Solution

Given information: A rectangle is given, and its angles are bisected to form a quadrilateral.

  

ABCD is a rectangle and each angle of a rectangle is equal to $90°$ .

So,

  $\angle A=\angle D=90°$

  $\angle A+\angle D=90°+90°=180°$

Multiply both sides by $\frac{1}{2}$

  $\begin{array}{l}\frac{1}{2}\left(\angle A+\angle D\right)=90°\\ \Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle D=90°\end{array}$

As its angles are bisected,

  $\angle 1=\frac{1}{2}\angle A,\angle 2=\frac{1}{2}\angle D$

So,

  $\angle 1+\angle 2=90°$

In $\Delta AQD$ ,

  $\begin{array}{l}\angle 1+\angle 2+\angle Q=180°\\ \Rightarrow 90°+\angle Q=180°\\ \Rightarrow \angle Q=180°-90°=90°\end{array}$

Similarly,

  $\angle P=\angle Q=\angle R=\angle S=90°$

Now,

  $\angle P+\angle S=90°+90°=180°$

If a transversal intersects two lines such that a pair of interior angles is supplementary, then the lines are parallel.

So,

  $PQ\parallel RS$

Also,

  $\angle P+\angle Q=90°+90°=180°$

If a transversal intersects two lines such that a pair of interior angles is supplementary, then the lines are parallel.

So,

  $PS\parallel RQ$

Therefore,

  $PQRS$ is a parallelogram as both sides are parallel.

Also, each of the angle in parallelogram is equal to $90°$ .

So,

  $PQRS$ is a rectangle.

(A parallelogram in which each angle is equal to $90°$ is a rectangle)