SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Explain how to prove a square
For a square, the lengths of all the sides are always equal and therefore the distance between the vertices will be the same.
STEP 2: Write the vertices of the figure
[tex]\begin{gathered} A(0,4) \\ B(3,0) \\ C(7,3) \\ D(4,7) \end{gathered}[/tex]STEP 2: Write the distance formula
[tex]\mathrm{\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}[/tex]STEP 3: Find the distance AB
[tex]\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(0,\:4\right)\mathrm{\:and\:}\left(3,\:0\right)\mathrm{\:is\:}: \\ =\sqrt{\left(3-0\right)^2+\left(0-4\right)^2} \\ =5 \end{gathered}[/tex]STEP 4: Find the distance BC
[tex]\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(3,\:0\right)\mathrm{\:and\:}\left(7,\:3\right)\mathrm{\:is\:}: \\ =\sqrt{\left(7-3\right)^2+\left(3-0\right)^2} \\ =5 \end{gathered}[/tex]STEP 5: Find the distance CD
[tex]\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(7,\:3\right)\mathrm{\:and\:}\left(4,\:7\right)\mathrm{\:is\:}: \\ =\sqrt{\left(4-7\right)^2+\left(7-3\right)^2} \\ =5 \end{gathered}[/tex]STEP 6: Find the distance AD
[tex]\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(0,\:4\right)\mathrm{\:and\:}\left(4,\:7\right)\mathrm{\:is\:}: \\ =\sqrt{\left(4-0\right)^2+\left(7-4\right)^2} \\ =5 \end{gathered}[/tex]Since,
[tex]AB=BC=CD=AD=5[/tex]Therefore, following the side property of a square explained in step 1, the figure given is a SQUARE.
STEP 7: Find the perimeter of the shape
Perimeter is the distance around the shape and is calculated as:
[tex]4\times5=20[/tex]Perimeter = 20
STEP 8: Find the area of the figure
The area of a square is given as:
[tex](length\text{ of the square\rparen}^2[/tex]Area is calculated as:
[tex]5^2=5\times5=25[/tex]The area of the shape is 25