Question

Find the perimeter and area of the shape shown. need help ong

Answer 1

Check the picture below. We know that the rectangle has a length of AB and a width of AD, so simply let's find those distances to get the perimeter and area, recall that the perimeter is simply two lengths plus two widths, and the area is just length times width.

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{4}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AB=\sqrt{[4 - 2]^2 + [5 - 3]^2}\implies AB=\sqrt{2^2+2^2}\implies \boxed{AB=2\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad D(\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill AD=\sqrt{[3 - 2]^2 + [2 - 3]^2} \\\\\\ AD = \sqrt{1^2+(-1)^2}\implies \boxed{AD=\sqrt{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large Perimeter}}{2\sqrt{2}+2\sqrt{2}+\sqrt{2}+\sqrt{2}}\implies 6\sqrt{2} \\\\\\ \stackrel{\textit{\large Area}}{2\sqrt{2}\cdot \sqrt{2}\implies 2\sqrt{2^2}}\implies 4$