1. What is the area of this figure? One Question

1. What is the area of this figure?

To check the area you can use the Pick's theorem:

Given a simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon's vertices are grid points, Pick's theorem provides a simple formula for calculating the area A of this polygon in terms of the number i of lattice points in the interior located in the polygon and the number b of lattice points on the boundary placed on the polygon's perimeter:

\({\displaystyle A=i+{\frac {b}{2}}-1.}\)

Example:

\(\begin{array}{|rcll|} \hline i &=& 45 \\ b &=& 20 \\ A &=& 45 + \frac{20}{2} -1 \\ \mathbf{A} &=& \mathbf{54} \\ \hline \end{array}\)

Let's split this big shape into four easier shapes: two triangles on top, one triangle on the left side, and a rectangle connecting them. Then all we have to do is find the areas of each shape and add them together.

You can just find the base and height by counting the squares on the grid. Remember that in a triangle the height is the length of a line drawn at a right angle from the base to the angle opposite the base.

Starting with the top left triangle.

A = (1/2)(B)(H)

A=(1/2)(3)(2)

A= 3 square units

Next let's do the top right triangle:

A = (1/2)(6)(1)

A = 3 square units

Next let's do the triangle on the left side:

A = (1/2)(6)(2)

A = 6 square units

Lastly let's find the area of the rectangle:

A = BH

A = (7)(6)

A = 42 square units.

Now to add them all together: 3 + 3 + 6 + 42 = 54 square units