We want to find the area of DECB given that the area of â–³ AFG is 3 cm^2. Let's take a closer look at these polygons.
Notice that we can represent the of DECB as a difference of the of â–³ ABC and the area of â–³ ADE.
A_(DECB) = A_(â–³ ABC) - A_(â–³ ADE)
Therefore, our goal will be to find the areas of the two triangles. Recall that FG is a midsegment of â–³ ADE. This means that triangles â–³ AFG and â–³ ADE are with the of 2. Recall that the is a square of the linear scale factor.
area scale factor
=
( length scale factor )^2
Since â–³ AFG and â–³ ADE are similar with the Length Scale Factor of 2, this means that the area of the latter one is 2^2 = 4 times larger then the first one.
A_(â–³ ADE) = 4 A_(â–³ AFG)
Since we are given that A_(â–³ AFG) is 3cm^2, we can substitute this value into equation above and solve the other area.
A_(â–³ ADE) = 4A_(â–³ AFG)
A_(â–³ ADE) = 4( 3)
A_(â–³ ADE) = 12
Thus, A_(â–³ ADE) is 12cm^2. Next we need to find A_(â–³ ABC). Similarly to how we found A_(â–³ ADE), we first recall that DE is a midsegment of â–³ ABC. Like previously, this means that A_(â–³ ABC) is 4 times larger than A_(â–³ ADE). Let's use this fact to find the area of the other triangle.
A_(â–³ ABC) = 4A_(â–³ ADE)
A_(â–³ ABC) = 4( 12)
A_(â–³ ABC) = 48
Therefore, A_(â–³ ABC) is 48cm^2. Finally, we can move to finding the area of A_(DECB). Let's recall how this area was related to the areas of the two larger triangles.
A_(DECB) = A_(â–³ ABC) - A_(â–³ ADE)
As we know that areas of these triangles, we can substitute them into the formula above and solve for the sought area.
A_(DECB) = A_(â–³ ABC) - A_(â–³ ADE)
A_(DECB) = 48 - 12
A_(DECB) = 36
Therefore, the area of DECB is 36cm^2.
