Chapter 7.4, Problem 34WE

a.

To determine

To find: The distance from $H$ to each side of the square.

Explanation of Solution

Given:

  $ABCD$ is a square.

  

Calculation:

Drawing a perpendicular line from point G to side AB gives a rectangle APGD where AG are diagonals.

  

In a rectangle the diagonals bisect each other at the centre of rectangle, hence point H is the centre of rectangle APGD. Now a line drawn from H to side AD perpendicularly bisects the side AD.

  

Hence in $\Delta AHQ$ using Pythagoras` theorem,

  $\begin{array}{c}AQ{)}^2+{(}^H={(}^A\\ {(}^8+{(}^H={(}^{10}\\ {(}^H={(}^{10}-{(}^8\\ {(}^H=100-64\\ {(}^H=36\\ HQ=6\end{array}$

Thus distance of H from side AD is 6 units.

Now as the segment QH was perpendicular to side of the square, the distance of point H from side BC will be equal to $16-6=10$ units.

For distance of H from side AB, a small segment HR can be drawn perpendicular to side AB.

  

As all the angles formed in rectangle ARHQ are right angles the given quadrilateral is a rectangle.

Thus by definition, as opposite sides are equal $AQ=HR=8$ units.

And as HR is perpendicular to the side of square ABCD, the distance of H from side CD will be $16-8=8$ units.

Conclusion:

The distance of H from side AB, BC, CD and DA are 8 units, 10 units, 8 units, 6 units respectively.

Also, Point $H$ is not in the center of square $ABCD$ .

b.

To determine

To find: $BF,FC,CG,DE,EA,EH$ and $HF.$

Answer to Problem 34WE

  $\begin{array}{l}BF=0.5\\ FC=5.5\\ CG=4\\ ED=3.5\\ EA=12.5\\ EH=7.5\\ FH=12.5\end{array}$

Explanation of Solution

Given: $ABCD$ is a square.

Calculation:

In $\Delta AGD$ , $\tan (\angle AGD)=\frac{16}{12}=\frac{4}{3}$

Thus, $\angle AGD={53}^{\circ }$ .

Now, draw a perpendicular from H to side BC of square.

  

  $\angle DGH=\angle GHZ$ as they are alternate angles.

And $\angle HFZ={53}^{\circ }$ , by sum of angles in a triangle.

Now in $\Delta HFZ$ ,

  $\begin{array}{c}\tan (\angle HFZ)=\tan ({53}^{\circ })=\frac{10}{x}\\ \frac{4}{3}=\frac{10}{x}\\ x=\frac{10\times 3}{4}\\ x=7.5=FZ\end{array}$

As, earlier found out that BZ will be equal to 8, hence BF will be equal to $8-7.5=\mathrm{0.5.}$

Thus, $BF=0.5$

Now for FC,

  $FC=BC-BF$

  $=16-0.5$

  $=15.5$

Thus $FC=15.5$ .

For CG,

  $CG=DC-DG$

  $=16-12=4$

Thus, $CG=4$

For ED, consider $\Delta ADG$

Using sum of all angles in a triangle, $\angle DAG={37}^{\circ }$

Now, consider the smaller triangle AHE

Using trigonometric function,

  $\begin{array}{c}\cos (\angle HAE)=\cos ({37}^{\circ })=\frac{10}{AE}\\ \frac{4}{5}=\frac{10}{AE}\\ AE=\frac{10*5}{4}\\ AE=12.5\end{array}$

Thus $AE=12.5$

  $\begin{array}{c}AD=AE+ED=16\\ ED=16-12.5\\ ED=3.5\end{array}$

Thus, $ED=3.5$

For EH, use Pythagoras` theorem in $\Delta AHE$

  $\begin{array}{c}$ {(AE)}^2={(AH)}^2+{(EH)}^2$$ {(12.5)}^2={(10)}^2+{(EH)}^2$$ {(EH)}^2=156.25-100$$ {(EH)}^2=56.25$EH=7.5\end{array}$

Thus, $EH=7.5$

For FH, simply refer to the figure drawn before

  

Here, in $\Delta HFZ$ , apply trigonometric functions.

  $\begin{array}{c}\sin (\angle HFZ)=\sin ({53}^{\circ })=\frac{10}{HF}\\ \frac{4}{5}=\frac{10}{HF}\\ HF=\frac{10\times 5}{4}\\ HF=12.5\end{array}$

Thus, $HF=12.5$

Conclusion:

Hence the values are:

  $\begin{array}{l}BF=0.5\\ FC=15.5\\ CG=4\\ ED=3.5\\ EA=12.5\\ EH=7.5\\ FH=12.5\end{array}$