Square inscribed inside triangle, solve for area around square

A square is inscribed into and isosceles triangle with base length 10 and legs 13.

I don't know where to start. Maybe use Pythagorean theorem to find height? And then what?

• Consider the similar triangle which lies on top of the square (i.e. the base is the top of the square). It has the same angles as the original triangle (which can be determined using the side lengths given), so you can find the base, which corresponds to the length of the square. – Dave Nov 24 '16 at 1:55
• @Dave How do I find the angles based on side lengths? – Gerard L. Nov 24 '16 at 1:59
• Well, a way that always works is the cosine law. If the sides of a triangle (of any type) are $a,b,c$, then $a^2=b^2+c^2-2bc\cos (A)$ where $A$ is the angle opposite of side $a$. This can be repeated for the other angles, and since you know the side lengths you can rearrange to solve for $A$ as such: $$\cos (A)=\frac{b^2+c^2-a^2}{2bc}$$ – Dave Nov 24 '16 at 2:02
• There are $3$ squares inscribed in a triangle (two of them equal for an isosceles triangle) so you'll have to decide which one to work on. See Maximum area of a square in a triangle for some possible ideas. – dxiv Nov 24 '16 at 2:10
• Can anyone post a straightforward solution? Haven't learned Trig yet. – Gerard L. Nov 24 '16 at 2:14

If you consider a square with a side upon the base ($10$) of the triangle, you can call the side of the square as $2x$; the height of the triangle is $12$ as you can check by Pythagorean theorem. You can draw the height so that the isosceles triangle is divided into equal parts. Now if you consider similar right triangles, one on the top of the square (it will have base $x$ and height $12-2x$) and the other one on the right side of the square (it will have base $5-x$ and height $2x$), you can write the following proportion between the heights and the bases of the two triangle:
The area of the isosceles triangle is $\frac{10*12}{2}=60$, for the square we have $\left(\frac{60}{11}\right)^2$...now it is easy.
• How is it $2x^2$, shouldnt it be $4x^2$ – Gerard L. Nov 24 '16 at 2:25
• @GerardL. No because it comes from $2x*x$ in the proportion.. – MattG88 Nov 24 '16 at 2:31