Answer:
- square: 9 square units
- triangle: 24 square units
Step-by-step explanation:
Using a suitable formula the area of a polygon can be computed from the coordinates of its vertices. You want the areas of the given square and triangle.
<h3>Square</h3>
The spreadsheet in the first attachment uses a formula for the area based on the given vertices. It computes half the absolute value of the sum of products of the x-coordinate and the difference of y-coordinates of the next and previous points going around the figure.
For this figure, going to that trouble isn't needed, as a graph quickly reveals the figure to be a 3×3 square.
The area of the square is 9 square units.
<h3>Triangle</h3>
The same formula can be applied to the coordinates of the vertices of a triangle. The spreadsheet in the second attachment calculates the area of the 8×6 triangle.
The area of the triangle is 24 square units.
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<em>Additional comment</em>
We have called the triangle an "8×6 triangle." The intention here is to note that it has a base of 8 units and a height of 6 units. Its area is half that of a rectangle with the same dimensions. These dimensions are readily observed in the graph of the vertices.
Taking

and differentiating both sides with respect to

yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have

Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have

Now, when

and

, we have
It is 52 because when roundest to the whole number 52 is closest