# Find the total length of two line segments of two overlapping right triangles Find the length of $$AE+EB$$ ?

(A) $$\frac{128}{7}$$

(B) $$\frac{112}{7}$$

(C) $$\frac{100}{7}$$

(D) $$\frac{96}{7}$$

(E) $$\frac{56}{7}$$

## My solution:

For $$\Delta AEB$$ :

$$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$$

$$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$$

$$\angle AEB = 180 - \angle BAC - \angle ABD = 120.52^{\circ}$$

Use Law of Sines:

$$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$$

$$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$$

$$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$$

$$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$$

$$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$$

$$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$$

$$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$$ Answer: (D)

My question:

Is there a solution without having to compute both arcsin $$\angle BAC$$ & $$\angle ABD$$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.

Note that $$\triangle AED\sim\triangle CEB$$. Then, letting $$AE = x$$ and $$BE = y$$, we have:

$$DE = \frac{9}{5}y,\ CE = \frac{5}{9}x$$

From the similar triangles. Now, we must have $$AE + CE = AC$$ and $$BE + DE = BD$$. Noting that $$AC = 13$$ and $$BD = 15$$:

$$x + \frac{5}{9}x = 13$$

$$y + \frac{9}{5}y = 15$$

Solving, we find $$\displaystyle x = \frac{117}{14}$$ and $$\displaystyle y = \frac{75}{14}$$. Thus, $$\boxed{AE + BE= x + y = \frac{96}{7}.}$$

• very nice solution Feb 16, 2021 at 3:18

Yes there is, drop a perpendicular down from $$E$$ to $$AB$$ and denote the point as $$X$$. Let $$EX = x$$ and $$AX = 12-y$$ and $$BX = y$$. Since $$\triangle EBX$$ ~ $$\triangle DBA$$, we have that $$\frac9x = \frac{12}y$$. Also, since we have $$\triangle CBA$$ ~ $$\triangle EXA$$, we have that $$\frac5x = \frac{12}{12-y}$$. Now, solving for $$x,y$$ using our equations, we get $$x=\frac{45}{14}, y = \frac{30}{7}$$. Plugging in these values, we get $$AX = \frac{54}{7}$$ and $$BX = \frac{30}{7}$$ and $$EX = \frac{45}{14}$$. Now using the pythagorean theorem, we get $$AE = \frac{117}{14}$$ and $$BE=\frac{75}{14}$$. Now, adding these lengths, we get that the sum is $$\frac{96}{7}$$ 