Question

What is the area of a rectangle with vertices at (2, 3), (7, 3), (7, 10), and (2, 10)?

Answer 1

D 35 units squared. If you would graph it on a piece of graph paper or look at the points one side would be the difference between 10 and 3 and the other side would be the difference between 7 and 2. Then taking those totals multiple them together to get the area

Answer 2

Answer:  The correct option is (d) 35 units².

Step-by-step explanation:  We are given to find the area of a rectangle with vertices at the points (2, 3), (7, 3), (7, 10), and (2, 10).

Let A(2, 3), B(7, 3), C(7, 10), and D(2, 10) represents the co-ordinates of the vertices of the given rectangle.

Then, the lengths of the sides AB, BC, CD and DA ca be calculated using distance formula as follows :

$AB=\sqrt{(7-2)^2+(3-3)^2}=\sqrt{25+0}=\sqrt{25}=5~\textup{units},\\\\BC=\sqrt{(7-7)^2+(10-3)^2}=\sqrt{0+49}=\sqrt{49}=7~\textup{units},\\\\CD=\sqrt{(2-7)^2+(10-10)^2}=\sqrt{25+0}=\sqrt{25}=5~\textup{units},\\\\DA=\sqrt{(2-2)^2+(3-10)^2}=\sqrt{0+49}=\sqrt{49}=7~\textup{units}.$

So, the area of the given rectangle will be

$Area=AB\timesBC=5\times7=35~\textup{units}^2.$

Thus, (d) is the correct option.