# Math Help - lake triangle - need help

1. ## lake triangle - need help

Determine how many acres there would be in that triangular lake, surrounded as shown by square plots of 370, 116 and 74 acres

2. Originally Posted by sonia1
Determine how many acres there would be in that triangular lake, surrounded as shown by square plots of 370, 116 and 74 acres

$A=\sqrt{s(s-a)(s-b)(s-a)}$

$a=19.23538$

$b=10.77033$

$
c=8.60233$

$s=\frac{a+b+c}{2} = \frac{\sqrt370 + \sqrt116 +\sqrt74}{2} = 19.30402$

$A=\sqrt{19.30402(0.06864)(8.53368)(10.70169)}=11.0 0037=11$

3. ## how do u prove it

do u know how to work it out another way instead of using that formula..the long way

4. Originally Posted by sonia1
do u know how to work it out another way instead of using that formula..the long way
$\frac{1}{4}$ $(\sqrt116+\sqrt74+\sqrt370)(\sqrt116+\sqrt74+\sqrt 370)^{1/2}$
$\frac{1}{4}$ $(\sqrt116+\sqrt370-\sqrt74)(\sqrt74+\sqrt370-\sqrt116)^{1/2}$

Can you take it from here? This way is very long. Are you sure you need to long way?

5. 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.
Code:
                              *
*    *
___      *         *   __
√116  *              * √74
*                   *
*                        *
A *     *     *     *     *     *
√370
Use the Law of Cosines to find angle $A.$

. . $\cos A \:=\:\frac{(\sqrt{116})^2 + (\sqrt{370})^2 - (\sqrt{74})^2} {2(\sqrt{116})(\sqrt{370})} \:=\:0.994345617$

Hence: . $A \:\approx\:6.1^o$

Then the area of the triangle is given by:

. . $A \;=\;\tfrac{1}{2}(\sqrt{116})(\sqrt{370})\sin6.1^o \:=\:11.007643903 \;\approx\;11\text{ acres}$