Answer:
The first (left) trapezoid's area is [tex]10\sqrt{55}[/tex]m or ≈ 74.2m²
The second (right) trapezoid's area is [tex]54 +18\sqrt{3}[/tex] or ≈ 85.2 in²
Step-by-step explanation:
First trapezoid (left):
Because the first trapezoid is a normal trapezoid, we can use the equation [tex]A = \frac{a+ b}{2} * h[/tex] Where a is equal to one base length and b is equal to the other base length and h is the height of the trapezoid.
a = 7
b = 13
h = [tex]\sqrt{55}[/tex] ([tex]3^{2}+h^{2} = 8^{2}[/tex])
Plug into the equation:
[tex]A = \frac{7+13}{2} *\sqrt{55}[/tex]
A = [tex]10\sqrt{55}[/tex] or ≈74.2m²
Second trapezoid (right):
Not a normal trapezoid (split into a triangle and a square)
Let's solve for the triangle first:
using [tex]sin(30) = \frac{x}{12}[/tex] to find the right-hand side of the triangle we get x = 6
because this is a 30 60 triangle, the last side has to be [tex]6\sqrt{3}[/tex]
Now we can calculate the area of the figure:
Triangle is [tex]\frac{1}{2} * 6 * 6\sqrt{3} = 18\sqrt{3}[/tex]
Rectangle is 6 * 9 = 54
Area = [tex]54 +18\sqrt{3}[/tex] or ≈ 85.2 in²
