The area of a triangle is \(ab\sin{C}\over{2}\), along with b*h*1/2, but I prefer the first one for this problem. (You can look the proof up, it is pretty simple)
We know that a = b = 7, so the area of the triangle = 49sinC/2 = 14. Then solving for sinC = 4/7.
Because we want to find the third length, it is best to use the Law of Cosines for C, but how?
Using the identity \(\sin^2\theta+\cos^2\theta=1\), where \(\theta=C\), we know that \(\cos^2C=1-\sin^2C=1-({4\over7})^2=1-{16\over{49}}={33\over{49}}\). Square rooting both sides, you get \(\cos{C}=\pm{\sqrt{33}\over{7}}\). Now, we can effectively do Law of Cosines: c^2 = a^2 + b^2 - 2ab*cosC.
We will do 2 seperate calculations of c from law of Cosines based on the two possible values of cosC.
Case 1: \(\cos{C} = +{\sqrt{33}\over{7}}\), keep in mind that c is the third side length we don't know, and a = b = 7.
\({c_1}^2=7^2+7^2-2*7*7*{\sqrt{33}\over{7}}=98-14\sqrt{33}\), and since c has to be positive to be a real triangle:
\(c_1 = \sqrt{98-14\sqrt{33}}\)
Case 2: \(\cos{C}=-{\sqrt{33}\over{7}}\); a = b = 7 still.
\({c_2}^2=7^2+7^2-2*7*7*-{\sqrt{33}\over{7}} = 98 + 14\sqrt{33}\), which yields
\(c_2=\sqrt{98+14\sqrt{33}}\), (again, there is no plus/minus because c has to be positive to be a real triangle.)
Setting c1 = a + b*sqrt(c) is a great strategy for radical simplification, and then you can square both sides and solve system of equations (with substitution, and yes, I did it)! Oh and by the way, don't expect a good answer for this problem...
After simplification, you should reach \(c_1=\sqrt{77}-\sqrt{21}; c_2=\sqrt{77}+\sqrt{21}\).
Product of 2 possible perimeters = (7 + 7 + c1)(7 + 7 + c2) = [14 + sqrt(77) - sqrt(21)][14 + sqrt(77) + sqrt(21)], and you can do difference of squares: a^2 - b^2 = (a + b)(a - b), where a = 14 + sqrt(77) here, and b = sqrt(21):
= 252 + 28sqrt(77)
Hence, our final answer is \(252 + 28\sqrt{77}\). 
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