The rectangle is:
To find the area of this rectangle, we need to find the distances of the two different sides of the rectangle. We know that the area of a rectangle is given by:
[tex]A_{\text{rectangle}}=w\cdot l_{}[/tex]Then, we need to find the distance between two points for the width, that is, it could be the distance between points C and D (segment CD) or segment AB.
To find the length, we need to find the distance of the segment AD or the distance of the segment BC.
After finding them, we need to multiply the result for w and l, and this product will be the area of the rectangle.
Finding WWe need to apply the formula for the distance between two points:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]A(-1, 4) ---> x1 = -1, y1 = 4
B(3, 3) ---> x2 = 3, y2 = 3
Then, we have:
[tex]d_{AB}=\sqrt[]{(3-(-1))^2+(3-4)^2}=\sqrt[]{(3+1)^2+(-1)^2}=\sqrt[]{(4)^2+(-1)^2}=\sqrt[]{16+1}[/tex]Therefore, this distance, which is w, is equal to:
[tex]d_{AB}=w=\sqrt[]{17}[/tex]Finding LWe can apply the same procedure to find l. We have that:
B(3,3) ---> x1 = 3, y1 = 3
C(1, -5) ---> x2 = 1, y2 = -5
Then, this distance, which is also l is:
[tex]d_{BC}=l=\sqrt[]{(1-3)^2+(-5-3)^2}=\sqrt[]{(-2)^2+(-8)^2}=\sqrt[]{4+64}=\sqrt[]{68}[/tex]Area of the Rectangle ABCDThe area is given by the product of w and l. Then, we have:
[tex]A_{\text{rectangle}}=w\cdot l=\sqrt[]{17}\cdot\sqrt[]{68}[/tex]We know that the factors of 68 are:
[tex]68=2^2\cdot17[/tex]Then, we can rewrite the area as follows:
[tex]A_{\text{rectangle}}=\sqrt[]{17}\cdot\sqrt[]{2^2\cdot17}=\sqrt[]{17}\cdot2\cdot\sqrt[]{17}=2\cdot\sqrt[]{17}\cdot\sqrt[]{17}=2\cdot(17)^{\frac{1}{2}}_{}\cdot(17)^{\frac{1}{2}}[/tex]And, finally, we have:
[tex]A_{\text{rectangle}}=2\cdot(17)^{\frac{1}{2}+\frac{1}{2}}=2\cdot17^1=34\Rightarrow A_{rec\tan gle}=34u^2[/tex]In summary, the area of the rectangle ABCD is equal to 34 square units (last option).
