# SAT Right triangle geometry [closed]

In the figure above, tan B = $$\frac{3}{4}$$. If BC = 15 and DA = 4, what is the length of DE?

• What have you tried? Which sides are you able to find the length of? – Arthur Aug 28 at 10:59
• We can help if you tell us where you are stuck exactly – Klangen Aug 28 at 11:07
• Sorry! I don't have any idea of where to start, and I'm completely stuck! – Felicia Natalia Rivera Aug 30 at 0:55

Let $$x$$ the lenght of $$AC$$ and $$y$$ the lenght of $$BD$$, so I have: $$\left\{\begin{matrix} x^2+(y+4)^2=15^2 \\ \frac{x}{y+4}=\frac{3}{4} \end{matrix}\right.$$

That is the same as:

$$\left\{\begin{matrix} y^2+8y-128=0 \\x=\frac{3}{4}\cdot(y+4) \end{matrix}\right.$$

This has solutions: $$x=9$$ and $$y=8$$. The two right triangle are similar, so: $$\frac{BD}{AB}=\frac{DE}{AC}$$. From this, I obtain: $$\frac8{12}=\frac{DE}{9}$$ and $$DE=6$$.

AC=3x
AB=4x
(3x)²+(4x)²=225
x=3
BD=8
AB/AC=BD/DE
12/9=8/DE
DE=6