Question

mr. koomson
row 1, 1

1. given the coordinates of abcd as follows: a(1, 1), b(9, 1), c(9, 6), and d(1, 6), find the coordinates of the mid - point of each side of abcd and label them wxyz.

(a) find the area of wxyz.
(b) find the perimeter of wxyz.

Explanation:

Step1: Find mid - points

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.
For side AB: $A(1,1)$ and $B(9,1)$, mid - point $W=(\frac{1 + 9}{2},\frac{1+1}{2})=(5,1)$.
For side BC: $B(9,1)$ and $C(9,6)$, mid - point $X=(\frac{9 + 9}{2},\frac{1 + 6}{2})=(9,\frac{7}{2})$.
For side CD: $C(9,6)$ and $D(1,6)$, mid - point $Y=(\frac{9+1}{2},\frac{6 + 6}{2})=(5,6)$.
For side DA: $D(1,6)$ and $A(1,1)$, mid - point $Z=(\frac{1+1}{2},\frac{6 + 1}{2})=(1,\frac{7}{2})$.

Step2: Analyze the shape of WXYZ

We can observe that WXYZ is a rectangle. The length $l$ between $W(5,1)$ and $X(9,\frac{7}{2})$ and $Y(5,6)$ and $Z(1,\frac{7}{2})$ can be found using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
$d_{WX}=\sqrt{(9 - 5)^2+(\frac{7}{2}-1)^2}=\sqrt{16+\frac{25}{4}}=\sqrt{\frac{64 + 25}{4}}=\sqrt{\frac{89}{4}}=\frac{\sqrt{89}}{2}$.
The width $w$ between $W(5,1)$ and $Z(1,\frac{7}{2})$ and $X(9,\frac{7}{2})$ and $Y(5,6)$:
$d_{WZ}=\sqrt{(1 - 5)^2+(\frac{7}{2}-1)^2}=\sqrt{16+\frac{25}{4}}=\frac{\sqrt{89}}{2}$.

Step3: Calculate the area of WXYZ

The area formula of a rectangle is $A=l\times w$. Here, $A = \sqrt{(9 - 5)^2+(\frac{7}{2}-1)^2}\times\sqrt{(1 - 5)^2+(\frac{7}{2}-1)^2}$.
$A = \sqrt{16+\frac{25}{4}}\times\sqrt{16+\frac{25}{4}}=\frac{89}{4}=22.25$.

Step4: Calculate the perimeter of WXYZ

The perimeter formula of a rectangle is $P = 2(l + w)$. Here, $l=\sqrt{(9 - 5)^2+(\frac{7}{2}-1)^2}$ and $w=\sqrt{(1 - 5)^2+(\frac{7}{2}-1)^2}$.
$P=2(\sqrt{(9 - 5)^2+(\frac{7}{2}-1)^2}+\sqrt{(1 - 5)^2+(\frac{7}{2}-1)^2})=2\times2\sqrt{16+\frac{25}{4}}=4\sqrt{\frac{64 + 25}{4}}=4\times\frac{\sqrt{89}}{2}=2\sqrt{89}\approx18.87$.

Answer:

(a) $22.25$
(b) $2\sqrt{89}\approx18.87$