Triangle

We see trigonmetry might be a possible solution to the problem, but here is a non-trigonmetry solution.

Instead, we focus on the fact that this triangle is an isosceles triangle, so we can draw a perpendicular to solve this problem.

Consider the diagram below.

We can now use a system of equations to solve this problem. $\sqrt{{x}^{2}+{y}^{2}}=7, \frac{2xy}{2}=14$

Simplifying, we get ${x}^{2}+{y}^{2}=49, xy=14$. This is not an easy system to solve, but there is a small trick I often use for a problem like this. We recognize the structure of $x+y=\sqrt{{x}^{2}+2xy+{y}^{2}}=\sqrt{49+14*2}=\sqrt{77}$. Now we can use a little trick with the Vieta's formulas. We recognize that x and y are the two roots of the equation ${a}^{2}-\sqrt{77}a+14$. (because the sum of the roots is sqrt(77) and the product of the roots is 14). We recoginze here, that x and y are interchangable, because we can assign any of the roots to x or y.

But we need to read the problem. It is not asking the roots, but the product of the perimeters. Because we know x and y are the two roots, or the two solutions to the base, we calculate $(14+2x)(14+2y)=196+28(x+y)+4xy=196+28\sqrt{77}+56=252+28\sqrt{77}$. So our answer is 252+28sqrt(77).

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 $252 + 28\sqrt{77}$.