Chapter 5.1, Problem 7PSA

a.

To determine

To Find: The coordinates of point C in terms of a and b.

Answer to Problem 7PSA

Coordinate of point C is $\left(2a,2b\right)$

Explanation of Solution

Given:

  

RECT is a rectangle of length 2a units and breadth 2b units.

Concept Used:

Opposite sides of rectangle are equal in measure.

Calculation:

Since, opposite sides of rectangle are equal in measure.

  $\begin{array}{l}\Rightarrow l\left(\overline{RE}\right)=l\left(\overline{EC}\right)=2a\\ l\left(\overline{OT}\right)=l\left(\overline{FO}\right)=2b\end{array}$ .

Here, point C is in X-Y plane just above point E having the coordinate $\left(2a,0\right)$ . Thus, the coordinate of point C will be $\left(2a,2b\right)$ .

Conclusion:

Coordinate of point C is $\left(2a,2b\right)$

b.

To determine

To Find: Whether $\overline{RC}\cong \overline{ET}$ ?

Answer to Problem 7PSA

Yes, $\overline{RC}\cong \overline{ET}$

Explanation of Solution

Given:

  

RECT is a rectangle of length 2a units and breadth 2b units.

Concept Used:

Opposite sides of rectangle are equal in measure. Each angle is a right angle. RC and ET are the diagonals of the rectangle RECT.

Calculation:

Since, opposite sides of rectangle are equal in measure.

  $\Rightarrow l\left(\overline{RE}\right)=l\left(\overline{EC}\right)=2a,l\left(\overline{OT}\right)=l\left(\overline{FO}\right)=2b$

Also, since each angle is a right angle, thus the length of the diagonal of rectangle obeys Pythagoras theorem.

  $\begin{array}{l}\therefore \text{Diagonal of RECT},RC=\sqrt{R{E}^2+E{C}^2}\\ \Rightarrow \text{Diagonal of RECT},RC=\sqrt{{\left(2a\right)}^2+{\left(2b\right)}^2}\\ \Rightarrow \text{Diagonal of RECT},RC=2\sqrt{a^2+b^2}\left(1\right)\\ \text{Diagonal of RECT},ET=\sqrt{R{E}^2+E{T}^2}\\ \Rightarrow \text{Diagonal of RECT},ET=\sqrt{{\left(2a\right)}^2+{\left(2b\right)}^2}\\ \Rightarrow \text{Diagonal of RECT},ET=2\sqrt{a^2+b^2}\left(2\right)\\ \therefore l\left(RC\right)=l\left(ET\right)\\ \Rightarrow \overline{RC}\cong \overline{ET}\end{array}$

Conclusion:

  $\overline{RC}\cong \overline{ET}$