Question

finding the perimeter of a trapezoid in the coordinate plane
what is the perimeter of the trapezoid with vertices q(8, 8), r(14, 16), s(20, 16), and t(22, 8)? round to the nearest hundredth, if necessary.
units

Explanation:

Step1: Recall the distance formula

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

Step2: Calculate the length of QR

For points $Q(8,8)$ and $R(14,16)$, $x_1 = 8,y_1 = 8,x_2=14,y_2 = 16$.
$QR=\sqrt{(14 - 8)^2+(16 - 8)^2}=\sqrt{6^2 + 8^2}=\sqrt{36+64}=\sqrt{100}=10$.

Step3: Calculate the length of RS

For points $R(14,16)$ and $S(20,16)$, $x_1 = 14,y_1 = 16,x_2 = 20,y_2=16$.
$RS=\sqrt{(20 - 14)^2+(16 - 16)^2}=\sqrt{6^2+0^2}=6$.

Step4: Calculate the length of ST

For points $S(20,16)$ and $T(22,8)$, $x_1 = 20,y_1 = 16,x_2 = 22,y_2 = 8$.
$ST=\sqrt{(22 - 20)^2+(8 - 16)^2}=\sqrt{2^2+( - 8)^2}=\sqrt{4 + 64}=\sqrt{68}\approx8.25$.

Step5: Calculate the length of TQ

For points $T(22,8)$ and $Q(8,8)$, $x_1 = 22,y_1 = 8,x_2 = 8,y_2 = 8$.
$TQ=\sqrt{(8 - 22)^2+(8 - 8)^2}=\sqrt{( - 14)^2+0^2}=14$.

Step6: Calculate the perimeter

The perimeter $P=QR + RS+ST+TQ$.
$P=10 + 6+8.25+14=38.25$.

Answer:

$38.25$