# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 8  and the triangle has an area of 40 . What are the lengths of sides A and B?

Apr 26, 2018

Side A = Side B = 10.77 units

#### Explanation:

Let side A=side B= $x$

Then drawing a right-angled triangle, with $x$ as your hypotenuse and $4$ as the length of your bottom side and the final side be $h$

Using pythagoras theorem,

${x}^{2}$ = ${h}^{2} + {4}^{2}$

${x}^{2} = {h}^{2} + 16$

${h}^{2} = {x}^{2} - 16$

$h = \pm \sqrt{{x}^{2} - 16}$

But since h is a length, it can only be positive

$h = \sqrt{{x}^{2} - 16}$

Area of triangle =

$\frac{1}{2} \times 8 \times \sqrt{{x}^{2} - 16} = 40$

$4 \sqrt{{x}^{2} - 16} = 40$

$\sqrt{{x}^{2} - 16} = 10$

${x}^{2} - 16 = 100$

${x}^{2} = 116$

$x = \pm 10.77$

But since x is a side, then it can only be positive

$x = 10.77$

Apr 26, 2018

$a = b = 2 \sqrt{29}$

#### Explanation:

Hey, small letters for triangle sides.

Isosceles triangle $a = b$, $c = 8$, Area $m a t h c a l \left\{A\right\} = 40.$

Let's see if we can do this different ways. First, the way the teacher probably expects:

The foot $F$ of the altitude $h$ from vertex $C$ to side $c$ is the midpoint of $A B$. So

$m a t h c a l \left\{A\right\} = \frac{1}{2} c h$

$40 = \frac{1}{2} \left(8\right) h$

$h = 10$

$A F = \frac{c}{2} = 4$ and we have a right triangle with the altitude:

${\left(\frac{c}{2}\right)}^{2} + {h}^{2} = {a}^{2}$

${4}^{2} + {10}^{2} = {a}^{2}$

$a = b = \sqrt{116} = 2 \sqrt{29}$

Let's just do it directly from Archimedes' Theorem:

$16 m a t h c a l {\left\{A\right\}}^{2} = 4 {a}^{2} {c}^{2} - {\left({b}^{2} - {c}^{2} - {a}^{2}\right)}^{2}$

The sides can be arbitrarily interchanged; I picked the form that gets some cancelling when $a = b$:

$16 m a t h c a l {\left\{A\right\}}^{2} = 4 {a}^{2} {c}^{2} - {c}^{4}$

${a}^{2} = \frac{{c}^{4} + 16 m a t h c a l {\left\{A\right\}}^{2}}{4 {c}^{2}}$

 a = \sqrt{ {8^4 + 16(40)^2}/{4(8^2)}} =2 sqrt{29} quad sqrt

That was easier.