Answer:
[tex] A_t = A_1 +A_2 +A_3 = 1723.414+ 2287.462+4787.75= 8798.626 ft^2[/tex]
Step-by-step explanation:
We assume that the plot attaced is the illustration for the problem.
For this case we need to begin finding the values for x and y using the cosine law given by:
[tex] a^2 = b^2 + c^2 - a bc cos (a)[/tex]
We can begin finding the value of x like this:
[tex] x^2 = 50^2 + 70^2 -2 (50)(70) cos 100 =8615.537[/tex]
[tex] x= 92.820 ft[/tex]
And similar in order to find the value of y we got this:
[tex] y^ 2 = 125^2 + 100^2 - 2(125) (100) cos 50=9555.310[/tex]
[tex] y=97.751 ft[/tex]
We can find the area for the 3 triangls on this way, beggining from the left triangle:
[tex] A_1 = \frac{bh}{2}= \frac{50 * 70 sin (100)}{2}=1723.414 ft^2[/tex]
For the traingel in the most left part we have this:
[tex] A_3 = \frac{bh}{2}= \frac{125 * 100 sin (50)}{2}=4787.75 ft^2[/tex]
We obtain this value because if we find the angle a we got:
[tex] sin a = \frac{100}{97.751} sin (50) = 0.784[/tex]
And then the angle a = 51.60 and for b = 180-51.60-50=78.402 degrees
Then we can create two equations in terms of h like this:
[tex] (97.75)^2 = h^2 + (100-x)^2 [/tex]
[tex] (125)^2 = h^2 +x^2[/tex]
If we subtract both equation we got this:
[tex] 125^2 -97.75^2 = x^2 -(100-x)^2 [/tex]
[tex] 200 x = 16069.742[/tex]
[tex] x = 95.755[/tex]
And then the area is just [tex] A_3 = \frac{100*95.755}{2}= 4787.75 ft[/tex]
For the middle area we can begin finding the midsegment like this:
[tex] r= \frac{1}{2} (50+92.820+97.75) = 120.285 ft[/tex]
And now we can find the area 2 like this:
[tex] A_2 = 2287.462[/tex]
And then we just need to add the areas and we got:
[tex] A_t = A_1 +A_2 +A_3 = 1723.414+ 2287.462+4787.75= 8798.626 ft^2[/tex]
