Question

a. find the area of the square. show your work.
s = \boxed{4}, so a = \boxed{4}^2 = \boxed{16} units²
b. find the area of the triangle. show your work.
b = \boxed{ }, h = sqrt{3}, so a = \frac{1}{2}(\boxed{ })(sqrt{3}) = sqrt{3} units²
(image shows coordinate grid with points d(-2, -1), g(-1, 0), f(1, 0), c(2, -1), vertical lines at x=-2, x=-1, x=1, x=2, horizontal line at y=-1, and two triangles above the x - axis between x=-1 to 0 and x=0 to 1)

Explanation:

Part a: Area of the Square

Step1: Find the side length \( s \)

The square has vertices \( D(-2, -1) \) and \( C(2, -1) \). The distance between these two points (which is the side length \( s \)) is calculated by the difference in the \( x \)-coordinates (since the \( y \)-coordinates are the same). So \( s = 2 - (-2) = 4 \).

Step2: Calculate the area \( A \) of the square

The formula for the area of a square is \( A = s^2 \). Substituting \( s = 4 \), we get \( A = 4^2 = 16 \) square units.

Step1: Find the base \( b \) of the triangle

The triangle has vertices \( G(-1, 0) \) and \( F(1, 0) \). The distance between these two points (the base \( b \)) is the difference in the \( x \)-coordinates (since the \( y \)-coordinates are the same). So \( b = 1 - (-1) = 2 \).

Step2: Calculate the area \( A \) of the triangle

The formula for the area of a triangle is \( A = \frac{1}{2} \times b \times h \), where \( b \) is the base and \( h \) is the height. We know \( h = \sqrt{3} \) and \( b = 2 \), so substituting these values, we get \( A = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3} \) square units.

Answer:

For part a, \( s = \boldsymbol{4} \), \( A = \boldsymbol{4}^2 = \boldsymbol{16} \) units².

Part b: Area of the Triangle