Determine how many acres there would be in that triangular lake, surrounded as shown by square plots of 370, 116 and 74 acres
$\displaystyle A=\sqrt{s(s-a)(s-b)(s-a)}$
$\displaystyle a=19.23538$
$\displaystyle b=10.77033$
$\displaystyle
c=8.60233$
$\displaystyle s=\frac{a+b+c}{2} = \frac{\sqrt370 + \sqrt116 +\sqrt74}{2} = 19.30402$
$\displaystyle A=\sqrt{19.30402(0.06864)(8.53368)(10.70169)}=11.0 0037=11$
$\displaystyle \frac{1}{4}$$\displaystyle (\sqrt116+\sqrt74+\sqrt370)(\sqrt116+\sqrt74+\sqrt 370)^{1/2}$
$\displaystyle \frac{1}{4}$ $\displaystyle (\sqrt116+\sqrt370-\sqrt74)(\sqrt74+\sqrt370-\sqrt116)^{1/2}$
Can you take it from here? This way is very long. Are you sure you need to long way?
Hello, sonia1!
There is another approach . . . It requires Trig . . .
Determine how many acres there would be in a triangular lake,
surrounded by square plots of 370, 116 and 74 acres.Use the Law of Cosines to find angle $\displaystyle A.$Code:* * * ___ * * __ √116 * * √74 * * * * A * * * * * * √370
. . $\displaystyle \cos A \:=\:\frac{(\sqrt{116})^2 + (\sqrt{370})^2 - (\sqrt{74})^2} {2(\sqrt{116})(\sqrt{370})} \:=\:0.994345617$
Hence: .$\displaystyle A \:\approx\:6.1^o$
Then the area of the triangle is given by:
. . $\displaystyle A \;=\;\tfrac{1}{2}(\sqrt{116})(\sqrt{370})\sin6.1^o \:=\:11.007643903 \;\approx\;11\text{ acres} $