+0

0
95
2

In triangle $ABC,$ $M$ is the midpoint of $\overline{AB}.$ Let $D$ be the point on $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC,$ and let the perpendicular bisector of $\overline{AB}$ intersect $\overline{AD}$ at $E.$ If $AB = 44,$ $AC = 30,$ and $ME = 10,$ then find the area of triangle $ACE.$

Aug 10, 2022

#1
0

Hi Guest!
First, let's draw the triangle and all of the given: Now, to find the area of triangle AEC, we already have the base (AC=30), we need the height, that is, the distance from E to AC.

Notice: AM=22 (44/2, as M is the midpoint of AB). and ME is given to be 10.

So with trigonometry: $$tan(\theta)=\frac{10}{22}=\frac{5}{11}$$

And, by pythagoras: $$AE=\sqrt{22^2+10^2}=\sqrt{584}$$

If $$tan(\theta)=\frac{5}{11} \implies sin(\theta)=\frac{5}{\sqrt{146}}$$ (Or simply using the triangle above, sin(theta) = opposite/hypotenuse which is ME/AE).

Next, in the triangle with the "Other theta" That is, the triangle having "h"

$$sin(\theta)=\frac{h}{\sqrt{584}} \iff \frac{5}{\sqrt{146}}=\frac{h}{\sqrt{584}}$$

Therefore, $$h=10$$

So, the area of triangle ACE = $$\frac{30*10}{2}=150$$

Aug 10, 2022