Question

Given rectangle ABCD with vertices A(1,4), B(3,6), C(6, 3), and D(4,1), what is the area of ABCD? Leave your answer in simplified radical form.12 square units572 square units672 square units10 square units

Answer 1

The distance formula between two points is :

$d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between vertices A(1, 4) and B(3, 6) is :

$\begin{gathered} d=\sqrt[]{(3-1)^2+(6-4)^2} \\ d=\sqrt[]{2^2+2^2} \\ d=\sqrt[]{4+4} \\ d=\sqrt[]{8} \\ d=2\sqrt[]{2} \end{gathered}$

The distance between the vertices B(3, 6) and C(6, 3) is :

$\begin{gathered} d=\sqrt[]{(3-6)^2+(6-3)^2} \\ d=\sqrt[]{(-3)^2+(3)^2} \\ d=\sqrt[]{9+9} \\ d=\sqrt[]{18} \\ d=3\sqrt[]{2} \end{gathered}$

The area formula of a rectangle is length x width or AB x BC

The area will be :

$\begin{gathered} A=2\sqrt[]{2}\times3\sqrt[]{2} \\ A=6\sqrt[]{4} \\ A=6(2) \\ A=12 \end{gathered}$

The answer is A. 12 square units