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I am currently stuck on a problem that should be easy enough... In an isosceles triangle the two equal sides are $5m$ and the area is $12m^2$. How can I find out the length of the third side?

My approach so far is this:

$24 = b·h$

$5^2 = (\frac{b}{2})^2 · h^2$

When you do the substitution you always get the nasty $x^4$ type equation. Any one that can help me here?

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$5^2 = 3^2 + 4^2$, and $12=3*4$. This doesn't work in general, but in this case there's a nice right triangle that gives you the two possible answers. –  Jonathan Christensen Dec 20 '12 at 21:19
    
Thank you for the answer! How did you arrive at the numbers $3$ and $4$? –  Lukas Arvidsson Dec 20 '12 at 21:25
    
You can use Heron's formula. $A = \sqrt{s(s-a)(s-b)(s-c)},$ where $s=\frac{a+b+c}2$ half perimeter. In your triange, let's length of the sides are $a, a,$ and $x$. Substitute values, square both sides and solve the resulting quadratic equation in $x$. There will be two solutions: one positive on negative. –  karakfa Dec 20 '12 at 21:39
    
@LukasArvidsson by recognizing that 5 being the hypotenuse and 12 being the product of the legs meant that each half of isoceles triangle must be a 3-4-5 right triangle. –  Jonathan Christensen Dec 20 '12 at 21:50
    
@LukasArvidsson unfortunately, there isn't a unique solution because you don't know whether 3 is the height or half of the base. So the solution might be $6m$ or $8m$. –  Jonathan Christensen Dec 20 '12 at 21:52
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3 Answers 3

up vote 6 down vote accepted

I think you want to use $\displaystyle 5^2 = \left(\frac{b}{2}\right)^2 + h^2\tag{1},$

assuming you mean to be using the Pythagorean Theorem - you need to sum the terms on the right of $(1)$, in contrast to what you've written in your question: $ 5^2 = \left(\frac{b}{2}\right)^2 \cdot h^2$.

$24 = b\cdot h \implies h = \dfrac{24}{b} \tag{2}$

So from $(1)$ we get $\quad 25 = \dfrac{b^2}{4} + h^2\tag{3}.$

Substitute your value for $h$ (found from $(2)$) into $(3)$, and solve for $b$.

$$25 = \dfrac{b^2}{4} + \left(\dfrac{24}{b}\right)^2 \implies 100b^2 = b^4 + 4\cdot24^2\tag{3}\implies b^4 - 100 b^2 + 2304 = 0$$ $$\iff (b^2 - 36)(b^2 - 64) = 0\tag{4}$$


Both factors in $(4)$ are a "difference of squares": so there will be four solutions to $(4)$, two of which are negative, so you need to throw those out (can't have negative length!).

That leaves you with two possible solutions for the base: $b_1 = x_1,$ or $b_2=x_2$, and respectively, when $b_1 = x_1, \;h_1=\dfrac{x_2}{2}$, or when $b_2 = x_2, \; h_2 = \dfrac{ x_1}{2}$.

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Thank you for the answer, my problem is that there are two unknown variables, both the base $b$ and the height $h$. To get rid of one of them I thought I would use substitution from the known fact that $b·h = 24$ How do you move on from there? –  Lukas Arvidsson Dec 20 '12 at 21:23
    
You will end up with a 4th degree polynomial, but you can solve it like you would a quadratic. –  amWhy Dec 20 '12 at 21:29
    
Yes, so far I am with you ;) My problem arrives when substituting $h$ with $\frac{24}{b}$. Can you please explain how you take the next step? Thank you very much for you help! –  Lukas Arvidsson Dec 20 '12 at 21:30
    
Ahh ok, that is where my problem lies :) Have to try again! –  Lukas Arvidsson Dec 20 '12 at 21:30
    
Thank you for an excellent answer! –  Lukas Arvidsson Dec 20 '12 at 21:42
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It is convenient to let the unknown third side be $2b$. Let the height be $h$. The area condition yields $bh=12$. By the Pythagorean Theorem, $b^2+h^2=25$.

Now instead of using a quadratic equation, we use a technique that goes back to Neo-Babylonian times. Note that $$(b+h)^2=b^2+h^2+2bh=49.$$ Similarly, $(b-h)^2=b^2+h^2-2bh=1$.

Thus $b+h=7$ and $b-h=\pm 1$. Add: We get $2b=8$ or $2b=6$.

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Thank you for an excellent answer! This was very interesting and I think my math book might actually have had this solution in mind. –  Lukas Arvidsson Dec 23 '12 at 2:10
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After the substitution, you'll have a biquadratic equation of the form $ax^4+bx^2+c=0$, which is solved simply by taking $y=x^2$ and you'll have a usual grade 2 equation, solve for y and later $x=+\sqrt{y}$ (only the positive solution is valid here, since it's a length).

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