Question

$\Delta ABC$ is an equilateral triangle. From the figure $AB=3AE$, what is the likelihood that any randomly selected point inside $\Delta ABC$ is inside $\Delta AED$.
(Note: The perpendicular drawn from the vertex of the equilateral triangle to the opposite side bisects it into equal halves.)

Answer 1

For a random point inside $\Delta ABC$ to be a part of $\Delta AED$, the probability in percentage will be given by

$P=\frac{\text{Area of}\left(\Delta AED\right)}{\text{Area of}\left(\Delta ABC\right)}\times 100\%$ eq (1)

$P=\frac{\text{Area of}\left(\Delta AED\right)}{\text{Area of}\left(\Delta ABD\right)}\times \frac{\text{Area of}\left(\Delta ABD\right)}{\text{Area of}\left(\Delta ABC\right)}\times 100$

To find the probability, we have to find the ratio of area of $\left(\Delta AED\right)$ to area of $\left(\Delta ABD\right)$, and also the ratio of area of $\left(\Delta ABD\right)$ to area of triangle $\left(\Delta ABC\right)$.

Area of $\left(\Delta AED\right)=\frac{1}{2}\times FD\times AE$ eq (2)

Since,
$AB=3AE$

We have,
$AE=\frac{1}{2}\times AB$

Therefore,

Area of $\left(\Delta AED\right)=\frac{1}{2}\times FD\times \frac{1}{3}\times AB$

Area of $\left(\Delta AED\right)=\frac{1}{6}\times FD\times AB$ eq(3)

Now,
Area of $\left(\Delta ABD\right)=\frac{1}{2}\times FD\times AB$ eq(4)

From eq(3) and eq(4),

$\frac{A\left(\Delta AED\right)}{A\left(\Delta ABD\right)}=\frac{\frac{1}{6}\times FD\times AB}{\frac{1}{2}\times FD\times AB}$

$=\frac{\frac{1}{6}}{\frac{1}{2}}$

$=\frac{1}{6}\times \frac{2}{1}$

$\frac{A\left(\Delta AED\right)}{A\left(\Delta ABD\right)}=\frac{1}{3}$ (5)

Again,

$A\left(\Delta ABD\right)=\frac{1}{2}\times AD\times BD$

$=\frac{1}{2}\times AD\times \frac{1}{2}\times BC$
(Since $D$ is the midpoint of $BC$)

$A\left(\Delta ABD\right)=\frac{1}{4}\times AD\times BC$ eq (6)

$A\left(\Delta ABC\right)=\frac{1}{2}\times AD\times BC$ eq (7)

From eq(6) and eq(7)

$\frac{A\left(\Delta ABD\right)}{A\left(\Delta ABC\right)}=\frac{\frac{1}{4}\times AD\times BC}{\frac{1}{2}\times AD\times BC}$

$=\frac{\frac{1}{4}}{\frac{1}{2}}$

$=\frac{1}{4}\times \frac{2}{1}$

$\frac{A\left(\Delta ABD\right)}{A\left(\Delta ABC\right)}=\frac{1}{2}$ eq(8)

From eq (5) and eq (8),

$\frac{A\left(\Delta AED\right)}{A\left(\Delta ABD\right)}\times \frac{A\left(\Delta ABD\right)}{A\left(\Delta ABC\right)}=\frac{1}{3}\times \frac{1}{2}$

$\frac{A\left(\Delta AED\right)}{A\left(\Delta ABD\right)}=\frac{1}{6}$

Using in equation (1) we get,

$P=\frac{1}{6}\times 100$

$P=16\frac{2}{3}\%$

There is $16\frac{2}{3}\%$ chance that the random point inside $\Delta ABC$ is also inside $\Delta AED$.