Simple triangle problem.

Question

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?

Answer 1

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 $b_2 = x_2, \; h_2 = \dfrac{ x_1}{2}$.

Answer 2

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).

Answer 3

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 $2b=6$.