Question

On your paper, construct a rectangle on a coordinate plane that satisfies these criteria.

Answer 1

Answer:

$A = (5,3)$

$B = (-5,3)$

$C = (-5,-8)$

$D = (5,-8)$

Step-by-step explanation:

Required

$Perimeter = 42$

Construct a rectangle whose perimeter is 42 units and satisfies the given conditions.

First, name the rectangle ABCD.

Such that:

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

$D = (x_4,y_4)$

For the rectangle to be either horizontal or vertical, then:

$y_1 = y_2$ and $y_3 = y_4$

We have that:

$Perimeter = 42$

Replace perimeter with its formula

$2(AB + BC) = 42$

Divide both sides by 2

$AB + BC = 21$

This implies that, the distance between adjacent sides (through the edges) must be equal to 21

Having said that: a set of coordinates that satisfy the given conditions are:

$A = (5,3)$ -- First quadrant

$B = (-5,3)$ -- Second quadrant

$C = (-5,-8)$ -- Third quadrant

$D = (5,-8)$ -- Fourth quadrant

The above quadrants satisfy the condition:

$y_1 = y_2$ and $y_3 = y_4$

HOW TO KNOW THE PERIMETER IS 42

To do this, we simply calculate the distance between the edges and add them up

Distance is calculated as:

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

For AB

$A = (5,3)$

$B = (-5,3)$

$D_1 = \sqrt{(5 - (-5))^2 + (3 - 3)^2}= \sqrt{(10)^2 + (0)^2} = \sqrt{100} = 10$

For BC

$B = (-5,3)$

$C = (-5,-8)$

$D_2 = \sqrt{(-5 - (-5))^2 + (3 - (-8))^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

For CD

$C = (-5,-8)$

$D = (5,-8)$

$D_3 = \sqrt{(-5 -5)^2 + (-8 - (-8))^2}= \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10$

For DA

$D = (5,-8)$

$A = (5,3)$

$D_4 = \sqrt{(5 -5)^2 + (-8 -3)^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

So, the perimeter is:

$P = D_1 + D_2 + D_3 + D_4$

$P = 10 + 11 + 10 +11$

$P = 42$

See attachment for rectangle