Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
$5^2 = 3^2 + 4^2$, 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
up vote 5 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 $b_2 = x_2, \; h_2 = \dfrac{ x_1}{2}$.

share|cite|improve this answer
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

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 $2b=6$.

share|cite|improve this answer
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

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.