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An isosceles triangle has two sides of length $7$ and an area of $14.$ What is the product of all possible values of its perimeter?

 Feb 9, 2024
 #1
avatar+394 
+3

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).

 Feb 10, 2024
edited by hairyberry  Feb 10, 2024
 #3
avatar+1624 
+2

Clever solution hairy! If you don't know trig, this is the way to go- smiley

proyaop  Feb 10, 2024
 #2
avatar+1624 
+2

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 \(252 + 28\sqrt{77}\)cool

 Feb 10, 2024

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