Chapter 5.1, Problem 6PSA

a.

To determine

To Find: The coordinates of points E and F.

Answer to Problem 6PSA

Coordinate of point E is $\left(2a,2a\right)$ and coordinate of point F is $\left(0,2a\right)$ .

Explanation of Solution

Given:

  

ODEF is a square with side length 2a.

Concept Used:

All sides and angles of square are equal in measure.

Calculation:

Since, all sides of square are equal in measure.

  $\Rightarrow l\left(\overline{OD}\right)=l\left(\overline{DE}\right)=l\left(\overline{EF}\right)=l\left(\overline{FO}\right)=2a$ .

Here, point E is in X-Y plane just above point D having the coordinate $\left(2a,0\right)$ . Thus, the coordinate of point E will be $\left(2a,2a\right)$ . Also, the point F is on Y-axis at same distance as the point E lying above D. Thus, the coordinate of point F will be $\left(0,2a\right)$ .

Conclusion:

Coordinate of point E is $\left(2a,2a\right)$ and coordinate of point F is $\left(0,2a\right)$ .

b.

To determine

To Calculate: The area of the square ODEF.

Answer to Problem 6PSA

  $\text{Area of Square = }4a^2\text{ sq}\text{.units}$

Explanation of Solution

Given:

  

ODEF is a square with side length 2a.

Concept Used:

All sides and angles of square are equal in measure.

  $\text{Area of Square =}{\left(Side\right)}^2$

Calculation:

Since, all sides of square are equal in measure.

  $\Rightarrow l\left(\overline{OD}\right)=l\left(\overline{DE}\right)=l\left(\overline{EF}\right)=l\left(\overline{FO}\right)=2a$ .

  $\begin{array}{l}\because \text{Area of Square =}{\left(Side\right)}^2\\ \Rightarrow \text{Area of Square =}{\left(2a\right)}^2=4a^2\text{ sq}\text{.units}\end{array}$

Conclusion:

  $\text{Area of Square = }4a^2\text{ sq}\text{.units}$

c.

To determine

To Calculate: The mid-point of $\overline{FD}$

Answer to Problem 6PSA

  $\text{Coordinate of Mid-point }Q=\left(a,a\sqrt{2}\right)$

Explanation of Solution

Given:

  

Here, FD is the diagonal of the square ODEF with side length 2a.

Concept Used:

All sides and angles of square are equal in measure.

  $\text{Diagonal of }ODEF,FD=\sqrt{F{O}^2+F{D}^2}$

Calculation:

Since, all sides of square are equal in measure.

  $\begin{array}{l}\Rightarrow l\left(\overline{OD}\right)=l\left(\overline{DE}\right)=l\left(\overline{EF}\right)=l\left(\overline{FO}\right)=2a\\ \Rightarrow \text{Diagonal of }ODEF,FD=\sqrt{F{O}^2+F{D}^2}\\ \Rightarrow \text{Diagonal of }ODEF,FD=\sqrt{{\left(2a\right)}^2+{\left(2a\right)}^2}=2a\sqrt{2}\\ \Rightarrow \text{Disatnce of Mid-point }Q\text{from}D=\frac{FD}{2}=a\sqrt{2}\\ \therefore \text{Coordinate of Mid-point }Q\text{of}\text{FD}=\left(a,a\sqrt{2}\right)\end{array}$ .

Conclusion:

  $\text{Coordinate of Mid-point }Q=\left(a,a\sqrt{2}\right)$

d.

To determine

To Calculate: The mid-point of $\overline{OE}$

Answer to Problem 6PSA

  $\text{Coordinate of Mid-point }Q\text{of}\text{OE}=\left(a,a\sqrt{2}\right)$

Explanation of Solution

Given:

  

Here, FD is the diagonal of the square ODEF with side length 2a.

Concept Used:

All sides and angles of square are equal in measure.

  $\text{Diagonal of }ODEF,OE=\sqrt{O{D}^2+D{E}^2}$

Calculation:

Since, all sides of square are equal in measure.

  $\begin{array}{l}\Rightarrow l\left(\overline{OD}\right)=l\left(\overline{DE}\right)=l\left(\overline{EF}\right)=l\left(\overline{FO}\right)=2a\\ \Rightarrow \text{Diagonal of }ODEF,OE=\sqrt{O{D}^2+D{E}^2}\\ \Rightarrow \text{Diagonal of }ODEF,OE=\sqrt{{\left(2a\right)}^2+{\left(2a\right)}^2}=2a\sqrt{2}\\ \Rightarrow \text{Disatnce of Mid-point }Q\text{from}O=\frac{OE}{2}=a\sqrt{2}\\ \therefore \text{Coordinate of Mid-point }Q\text{of}\text{OE}=\left(a,a\sqrt{2}\right)\end{array}$ .

Conclusion:

  $\text{Coordinate of Mid-point }Q\text{of}\text{OE}=\left(a,a\sqrt{2}\right)$