Chapter 8, Problem 16CUR

To determine

To Prove: NPRS is a rectangle.

Explanation of Solution

Given information:

NPQRST is a regular hexagon.

The figure is shown below:

  

A regular hexagon has all sides equal and all angles equal.

Sum of the interior angles of a regular hexagon is given by:

  $\begin{array}{l}\theta =(n-2){180}^{\circ }\\ =(6-2){180}^{\circ }\\ =4\times {180}^{\circ }\\ ={720}^{\circ }\end{array}$

So, an interior angle of regular hexagon can be calculated by:

  $\begin{array}{l}\frac{{720}^{\circ }}{n}=\frac{{720}^{\circ }}{6}\\ ={120}^{\circ }\end{array}$

In ΔNTS and ΔPQR

  $\begin{array}{l}NT=PQ\hfill \\ TS=QR\hfill \\ \angle T=\angle Q\hfill \end{array}$

So by SAS congruency, $\text{Δ}NTS\cong \text{Δ}PQR$ and both triangle’s are isosceles triangle also.

Hence,

  $PR=NS(\text{Congruent}\text{parts}\text{of}\text{Congruent}\text{Triangles})$

Also,

  $\begin{array}{l}\angle T=\angle Q={120}^{\circ }\\ \angle TNS+\angle TSN+\angle T={180}^{\circ }\\ \angle TNS+\angle TSN+{120}^{\circ }={180}^{\circ }\\ \angle TNS+\angle TSN={60}^{\circ }\\ 2\angle TNS={60}^{\circ }\\ \angle TNS={30}^{\circ }\\ \angle TSN={30}^{\circ }\end{array}$

Now considering hexagon at point S:

  $\angle S={120}^{\circ }$

But,

  $\begin{array}{l}\angle S=\angle TSN+\angle NSR\\ \angle TSN+\angle NSR={120}^{\circ }\\ {30}^{\circ }+\angle NSR={120}^{\circ }\\ \angle NSR={90}^{\circ }\end{array}$

Similarly at points P , R and N also the angles:

  $\begin{array}{l}\angle NPR={90}^{\circ }\\ \angle PRS={90}^{\circ }\\ \angle SNP={90}^{\circ }\end{array}$

Also,

  $PR=NS$ (equal sides of regular hexagon).

Therefore, NPRS is rectangle.