Chapter 9.5, Problem 26PSC

a.

To determine

To calculate:The coordinates of the points C and D

Answer to Problem 26PSC

The coordinates of the points C and D are $\left(2P-2a,2h\right)$ and $\left(2a-2P,2h\right)$ respectively.

Explanation of Solution

Given:

ABCD is an isosceles trapezoid where $A=\left(-2a,0\right)$ and $B=\left(2a,0\right)$

The altitude of the trapezoid is $2h$ and the upper base

  $\overline{CD}$ has a length of $4p$

  

Calculation:

Here, we know that

  $x=2a-2P$ and $y=2h$

So, Coordinate of C = $\left(2P-2a,2h\right)$

And, Coordinate of D = $\left(2a-2P,2h\right)$

Conclusion: Hence, the coordinates of the points C and D are $\left(2P-2a,2h\right)$ and $\left(2a-2P,2h\right)$ respectively.

b.

To determine

To calculate:The length of the lower base.

Answer to Problem 26PSC

The length of the lower base is $4a$

Explanation of Solution

Given: ABCD is an isosceles trapezoid where $A=\left(-2a,0\right)$ and $B=\left(2a,0\right)$

The altitude of the trapezoid is $2h$ and the upper base

  $\overline{CD}$ has a length of $4p$

Concept used:

Here, we are using distance formula.

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Here, the formula is

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Here, $A=\left(-2a,0\right)$ and $B=\left(2a,0\right)$

So, distance of AB = $\sqrt{{\left(2a+2a\right)}^2}$

  →$\sqrt{{\left(4a\right)}^2}$

  $AB=4a$

Conclusion: Hence, the length of the lower base is $4a$

c.

To determine

To calculate:The length of the segment joining the mid-points of $\overline{AD}$ and $\overline{BC}$

Answer to Problem 26PSC

The length of the segment joining the mid-points of $\overline{AD}$ and $\overline{BC}$ is $2P$

Explanation of Solution

Given: ABCD is an isosceles trapezoid where $A=\left(-2a,0\right)$ and $B=\left(2a,0\right)$

The altitude of the trapezoid is $2h$ and the upper base

  $\overline{CD}$ has a length of $4p$

  

Concept used:

Here, we are using distance formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Here, $G=\left(-2a+\frac{2a-2P}{2},\frac{2h}{2}\right)$ and $H=\left(\frac{2P}{2},\frac{2h}{2}\right)$

So, $G\left(-P,h\right)$ and $H\left(P,h\right)$

Now, by using the formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

  $GH=\sqrt{{\left(2P\right)}^2}$

  $GH=2P$

Conclusion: Hence, the length of the segment joining the mid-points of $\overline{AD}$ and $\overline{BC}$ is $2P$

d.

To determine

To calculate:The length of the segment joining the mid-points of the diagonals of the trapezoid.

Answer to Problem 26PSC

The length of the segment joining the mid-points of the diagonals of the trapezoid is $4a-2P$

Explanation of Solution

Given: ABCD is an isosceles trapezoid where $A=\left(-2a,0\right)$ and $B=\left(2a,0\right)$

The altitude of the trapezoid is $2h$ and the upper base

  $\overline{CD}$ has a length of $4p$

  

Concept used:

Here, we are using distance formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Here, mid-point of AC = $\left(\frac{-2a+2P-2a}{2},\frac{2h}{2}\right)$

  →$\left(P-2a,h\right)$

Now, mid-point of BD = $\left(\frac{4a-2P}{2},\frac{2h}{2}\right)$

  →$\left(2a-P,h\right)$

Now, by using the formula

  $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

So, the required length = $\sqrt{{\left(2a-P+2a-P\right)}^2}$

  →$4a-2P$

Conclusion: Hence, the length of the segment joining the mid-points of the diagonals of the trapezoid is $4a-2P$