Chapter 3.3, Problem 54HP

(a)

To determine

Draw the square and find other two vertices B and D.

Answer to Problem 54HP

B(2,4) and D(10,-4)

Explanation of Solution

Given:

Square ABCD has vertices at A(2,-4) and C(10,4).

Calculation:

Place the given vertices , and since the square has equal sides , so join the vertices and find other vertices:

  

So, B(2,4) and D(10,-4)

(b)

To determine

Prove that $\overline{AD}\parallel \overline{BC}$ and $\overline{AB}\parallel \overline{DC}$

Answer to Problem 54HP

B(2,4) and D(10,-4)

Explanation of Solution

Given:

From part (a) ,

Square ABCD has vertices at A(2,-4) , B(2,4) ,C(10,4), and D(10,-4).

Calculation:

Two lines are parallel if their slopes are equal.

Consider line segments $\overline{AD}\text{ and }\overline{BC}$

Slope of $\overline{AD}=\frac{-4-(-4)}{10-2}=\frac{0}{8}=0$

Slope of $\overline{BC}=\frac{4-4}{10-2}=\frac{0}{8}=0$

Since the slopes of both the segments are equal , so, $\overline{AD}\parallel \overline{BC}$

Consider line segments $\overline{AB}\text{ and }\overline{DC}$

Slope of $\overline{AB}=\frac{4-(-4)}{2-2}=\frac{8}{0}=undefined$

Slope of $\overline{DC}=\frac{4-(-4)}{10-10}=\frac{8}{0}=undefined$

Since the slopes of both the segments are equal ,so, $\overline{AB}\parallel \overline{DC}$

Hence proved.

(c)

To determine

Show that measure of each angle inside the square is $90°$

Explanation of Solution

Given:

From part (a) ,

Square ABCD has vertices at A(2,-4) , B(2,4) ,C(10,4), and D(10,-4).

Calculation:

Since the figure is a square , so all the angles are $90°$ and all the sides are equal.

Also ,

Slope of $\overline{AD}=\frac{-4-(-4)}{10-2}=\frac{0}{8}=0$

Slope of $\overline{AB}=\frac{-4-4}{2-2}=\frac{-8}{0}$

Since the product of slope of both is -1 , so , $\overline{AD}\perp \overline{AB}$ .

Hence $m\angle A=90°$ .

From part (b) , $\overline{AD}\parallel \overline{BC}$ and $\overline{AB}\parallel \overline{DC}$

So,

  $\begin{array}{l}m\angle A+m\angle B=180°\mathrm{.............}[\text{consecutive interior angles on same side of transversal]}\\ \Rightarrow m\angle B=90°\end{array}$

  $\begin{array}{l}m\angle A+m\angle C=180°\mathrm{.............}[\text{consecutive interior angles on same side of transversal]}\\ \Rightarrow m\angle C=90°\end{array}$

  $\begin{array}{l}m\angle B+m\angle D=180°\mathrm{.............}[\text{consecutive interior angles on same side of transversal]}\\ \Rightarrow m\angle D=90°\end{array}$

Hence proved.