Question

A polygon has vertices A (5, -1), B (21, 11), C (26, -1), and D (2, -8).

Part A. What is the perimeter of ABCD to the nearest tenth of a unit?
Part B. What is the area of ABCD to the nearest tenth of a square unit?
Enter the correct answers in the boxes.
A. Perimeter: units
B. Area: square units

Answer 1

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad B(\stackrel{x_2}{21}~,~\stackrel{y_2}{11}) ~\hfill AB=√((~~ 21- 5~~)^2 + (~~ 11- (-1)~~)^2) \\\\\\ ~\hfill AB=√(( 16 )^2 + ( 12)^2)\implies \boxed{AB=20}`

`B(\stackrel{x_1}{21}~,~\stackrel{y_1}{11})\qquad C(\stackrel{x_2}{26}~,~\stackrel{y_2}{-1}) ~\hfill BC=√((~~ 26- 21~~)^2 + (~~ -1- 11 ~~)^2) \\\\\\ ~\hfill BC=√(( 5)^2 + ( -12)^2)\implies \boxed{BC=13} \\\\\\ C(\stackrel{x_1}{26}~,~\stackrel{y_1}{-1})\qquad D(\stackrel{x_2}{2}~,~\stackrel{y_2}{-8}) ~\hfill CD=√((~~ 2- 26~~)^2 + (~~ -8- (-1)~~)^2) \\\\\\ ~\hfill CD=√(( -24)^2 + ( -7)^2)\implies \boxed{CD=25}`

`D(\stackrel{x_1}{2}~,~\stackrel{y_1}{-8})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill DA=√((~~ 5- 2~~)^2 + (~~ -1- (-8)~~)^2) \\\\\\ ~\hfill DA=√(( 3)^2 + ( 7)^2)\implies \boxed{DA=√(58)} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\LARGE Perimeter}}{20~~ + ~~13~~ + ~~25~~ + ~~√(58) ~~ \approx ~~ \text{\LARGE 65.6}}`

now, as far as the area goes, let's check the picture below, hmmm, there are two triangles, both have a base of 21, and you can see their heights there, well, let's simply get the area of each triangle and sum them up.

`\cfrac{1}{2}(21)(12)~~ + ~~\cfrac{1}{2}(21)(7)\implies 126~~ + ~~73.5\implies \text{\LARGE 199.5}`