Answer:
\(A=42\ units^{2}\)
Step-by-step explanation:
see the attached figure with letters to better understand the problem
we know that
The figure ABCD is a rectangle
so
\(AB=CD \\AD=BC\)
The area of a rectangle is equal to
\(A=B*h\)
where
B is the base
h is the height
the base B is equal to the distance AB
the height h is equal to the distance BC
Let
\(A(2,-1),B(9,-8),C(12,-5),D(5,2)\)
Step 1
Find the distance AB
the formula to calculate the distance between two points is equal to
\(d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}\)
\(A(2,-1),B(9,-8)\)
substitute the values
\(d=\sqrt{(-8+1)^{2}+(9-2)^{2}}\)
\(d=\sqrt{(-7)^{2}+(7)^{2}}\)
\(dAB=\sqrt{98}=7\sqrt{2}\ units\)
Step 2
Find the distance BC
the formula to calculate the distance between two points is equal to
\(d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}\)
\(B(9,-8),C(12,-5)\)
substitute the values
\(d=\sqrt{(-5+8)^{2}+(12-9)^{2}}\)
\(d=\sqrt{(3)^{2}+(3)^{2}}\)
\(dBC=\sqrt{18}=3\sqrt{2}\ units\)
Step 3
Find the area
\(A=B*h\)
we have
\(B=7\sqrt{2}\ units\)
\(h=3\sqrt{2}\ units\)
substitute
\(A=7\sqrt{2}*3\sqrt{2}=42\ units^{2}\)
