Since ∆ ABC and ∆ EFD are similar, their corresponding sides are in proportion. Let's say the ratio of the lengths of their corresponding sides is k. Then:
EF = k * AB
FD = k * BC
DE = k * AC
The area of a triangle is given by 1/2 * base * height. Let h1 and h2 be the heights of ∆ ABC and ∆ EFD, respectively.
Area of ∆ ABC = 1/2 * AB * h1 = 81
h1 = 162/AB
Using the ratios above, we can express the heights of ∆ EFD in terms of h1:
h2 = (EF * h1) / DE = (k * AB * h1) / (k * AC) = AB/AC * h1
h2 = (FD * h1) / DE = (k * BC * h1) / (k * AC) = BC/AC * h1
Now we can find the area of ∆ EFD:
Area of ∆ EFD = 1/2 * EF * h2
= 1/2 * k * AB * (AB/AC * h1)
= 1/2 * k * AB * (AB/AC * (162/AB))
= 81 * k * (AB/AC)
We know that ∆ ABC has an area of 81, so:
81 = 1/2 * AB * h1
162 = AB * h1
h1 = 162/AB
Substituting this into the equation for the area of ∆ EFD:
Area of ∆ EFD = 81 * k * (AB/AC)
= 81 * k * (AB/AC) * (h1/162)
= 81 * k * (AB/AC) * (AB/(2h1))
= 81 * k * (AB/AC) * (AB/(2 * 162/AB))
= AB^2 * k * (1/2AC)
Therefore, the area of ∆ EFD is:
Area of ∆ EFD = AB^2 * k * (1/2AC) = (AB/AC)^2 * Area of ∆ ABC
= (EF/DE)^2 * Area of ∆ ABC
So the area of ∆ EFD is (EF/DE)^2 times the area of ∆ ABC. Substituting in the expressions we found earlier for EF and DE:
Area of ∆ EFD = (k * AB / k * AC)^2 * Area of ∆ ABC
= (AB/AC)^2 * Area of ∆ ABC
Since we know that the area of ∆ ABC is 81, we can substitute this in to get:
Area of ∆ EFD = (AB/AC)^2 * 81
So the area of ∆ EFD is (AB/AC)^2 times 81. If we're given the ratio of AB to AC, we can plug that in to find the area of ∆ EFD.