Question

example 2

1. abcd is a quadrilateral with vertices a(-4, -1), b(-3, 4), c(-1, 4), and d(0, -1).

a. verify abcd is a trapezoid.
b. is abcd an isosceles trapezoid? explain.

1. qrst is a quadrilateral with vertices q(2, 4), r(4, 2), s(3, -3), and t(-2, 2).

a. verify
b. is qrst an isosceles trapezoid? explain.

Explanation:

Step1: Recall slope formula

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Find slopes of sides of $ABCD$

For side $AB$ with $A(-4,-1)$ and $B(-3,4)$:
$m_{AB}=\frac{4-(-1)}{-3 - (-4)}=\frac{4 + 1}{-3+4}=5$.
For side $BC$ with $B(-3,4)$ and $C(-1,4)$:
$m_{BC}=\frac{4 - 4}{-1-(-3)}=\frac{0}{2}=0$.
For side $CD$ with $C(-1,4)$ and $D(0,-1)$:
$m_{CD}=\frac{-1 - 4}{0-(-1)}=\frac{-5}{1}=-5$.
For side $DA$ with $D(0,-1)$ and $A(-4,-1)$:
$m_{DA}=\frac{-1-(-1)}{-4 - 0}=\frac{0}{-4}=0$.
Since $m_{BC}=m_{DA} = 0$, $BC\parallel DA$. A quadrilateral with one - pair of parallel sides is a trapezoid. So $ABCD$ is a trapezoid.

Step3: Recall distance formula

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step4: Find lengths of non - parallel sides

For side $AB$:
$d_{AB}=\sqrt{(-3+4)^2+(4 + 1)^2}=\sqrt{1 + 25}=\sqrt{26}$.
For side $CD$:
$d_{CD}=\sqrt{(0 + 1)^2+(-1 - 4)^2}=\sqrt{1+25}=\sqrt{26}$.
Since $d_{AB}=d_{CD}$, $ABCD$ is an isosceles trapezoid.

For quadrilateral $QRST$:

Step1: Find slopes of sides of $QRST$

For side $QR$ with $Q(2,4)$ and $R(4,2)$:
$m_{QR}=\frac{2 - 4}{4 - 2}=\frac{-2}{2}=-1$.
For side $RS$ with $R(4,2)$ and $S(3,-3)$:
$m_{RS}=\frac{-3 - 2}{3 - 4}=\frac{-5}{-1}=5$.
For side $ST$ with $S(3,-3)$ and $T(-2,2)$:
$m_{ST}=\frac{2+3}{-2 - 3}=\frac{5}{-5}=-1$.
For side $TQ$ with $T(-2,2)$ and $Q(2,4)$:
$m_{TQ}=\frac{4 - 2}{2+2}=\frac{2}{4}=\frac{1}{2}$.
Since $m_{QR}=m_{ST}=-1$, $QR\parallel ST$. So $QRST$ is a trapezoid.

Step2: Find lengths of non - parallel sides

For side $RS$:
$d_{RS}=\sqrt{(3 - 4)^2+(-3 - 2)^2}=\sqrt{1 + 25}=\sqrt{26}$.
For side $TQ$:
$d_{TQ}=\sqrt{(2 + 2)^2+(4 - 2)^2}=\sqrt{16 + 4}=\sqrt{20}=2\sqrt{5}$.
Since $d_{RS}
eq d_{TQ}$, $QRST$ is not an isosceles trapezoid.

Answer:

a. For $ABCD$, it is a trapezoid because $m_{BC}=m_{DA} = 0$, so $BC\parallel DA$. For $QRST$, it is a trapezoid because $m_{QR}=m_{ST}=-1$, so $QR\parallel ST$.
b. $ABCD$ is an isosceles trapezoid since $d_{AB}=d_{CD}=\sqrt{26}$. $QRST$ is not an isosceles trapezoid since $d_{RS}=\sqrt{26}$ and $d_{TQ}=2\sqrt{5}$ and $d_{RS}
eq d_{TQ}$.