Try to find the area of the brown spots, which filled in as green, would make one big rectangle.
1. Take the whole area of the rectangle (use the area of the brown rectangle around the green, and the area of the brown triangle towards the top left).
The area for the brown rectangle A = b*h becomes A = 2*3 = 6 ft^2
Area for brown triangle (upper-left) is found by the length of the whole figure shown. The whole length is 12ft at the bottom; at the top, we know 6ft and 2ft. Therefore, the missing measurement there is 12ft - (6ft + 2ft) = 4ft
Area for the triangle (I will include some trigonometry; don’t be afraid to use it!:) Find the long side of the triangle: Pythagorean Theorem: a^2 + b^2 = c^2 which becomes: 3^2 + 4^2 = c^2 9 + 16 = c^2 c^2 = 25 sqrt(c^2) = sqrt(25) c = 5
Next, I know there are easier formulas and processes for the following, but they will work. S (semiperimeter)= 1/2(a+b+c) Counting up the sides of the triangle, then dividing by two will give you your semiperimeter.
Then, A= sqrt [(S)(S-a)(S-b)(S-c)] = 6 I believe that the S value and a value are the same because that brown triangle made earlier is a 3-4-5 triangle (the sides).
Finally, now that we know the area of the brown rectangle and the brown triangle, we can subtract them from the big rectangle! 1. The big rectangle: A= b*h becomes A = 12*5 = 60 ft^2 2. Take the 60 ft^2 and set up the following: 60 ft^2 - (6ft^2+6ft^2) = 60ft^2 - 12ft^2 = 48ft^2, as the final answer.
