Question

How do I prove this is a square? What’s the perimeter of the shape and what would be the area of the shape need help explanation for this excercise

Answer 1

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

$\begin{gathered} A(0,4) \\ B(3,0) \\ C(7,3) \\ D(4,7) \end{gathered}$

STEP 2: Write the distance formula

$\mathrm{\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

STEP 3: Find the distance AB

$\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(0,\:4\right)\mathrm{\:and\:}\left(3,\:0\right)\mathrm{\:is\:}: \\ =\sqrt{\left(3-0\right)^2+\left(0-4\right)^2} \\ =5 \end{gathered}$

STEP 4: Find the distance BC

$\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(3,\:0\right)\mathrm{\:and\:}\left(7,\:3\right)\mathrm{\:is\:}: \\ =\sqrt{\left(7-3\right)^2+\left(3-0\right)^2} \\ =5 \end{gathered}$

STEP 5: Find the distance CD

$\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(7,\:3\right)\mathrm{\:and\:}\left(4,\:7\right)\mathrm{\:is\:}: \\ =\sqrt{\left(4-7\right)^2+\left(7-3\right)^2} \\ =5 \end{gathered}$

STEP 6: Find the distance AD

$\begin{gathered} \mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\ \mathrm{The\:distance\:between\:}\left(0,\:4\right)\mathrm{\:and\:}\left(4,\:7\right)\mathrm{\:is\:}: \\ =\sqrt{\left(4-0\right)^2+\left(7-4\right)^2} \\ =5 \end{gathered}$

Since,

$AB=BC=CD=AD=5$

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:

$4\times5=20$

Perimeter = 20

STEP 8: Find the area of the figure

The area of a square is given as:

$(length\text{ of the square\rparen}^2$

Area is calculated as:

$5^2=5\times5=25$

The area of the shape is 25