Find the unknown area by different methods

Question

this problem has many ways of calculating, could provide some solutions, particularly using proportions

EDIT : A it's not midpoint

Answer 1

If we view $AEB$ as having base $AE$ and triangle $FEB$ as have base $FE$, then the both have the same height--- the altitude of line $AF$ to point $B$.

Since triangles $AEB$ and $FEB$ both have the same height then their bases are in proportion to their areas. So $EF = \frac 32 AE$.

Triangle $DEF$ is similar $AEB$ and the sides are that $EF$ corresponds to $AE$ and so the sides are $\frac 32$ as long. So the area of the $DEF$ (white) area is $2cm^2 *(\frac 32)^2 = 4.5 cm^2$.

So the area of the white plus the yellow is $4.5cm^2 + 3cm^2 = 7.5 cm^2$.

As this is a rectangle triangles $DCB$ and $DFB$ are congruent and have the same area.

And so the area of the blue plus the green is $7.5 cm^2$. So the area of the blue is $7.5 cm^2 - 2cm^2 = 5.5 cm^2$

Answer 2

Area of the triangle $ABF=5$.

And $A$ is the midpoint of the side $CB$.

So the area of rectangle $BCDF$= 4 × area of triangle$ ABF$ $=4×5= 20$.

Thus area of the triangle $BDC$ is half of the area of the rectangle =$10$.

Thus area of the blue reigion= $10-2=8$.

Answer 3

Here are some hints toward different approaches:

1. The shoelace algorithm allows you to calculate the area of any polygon given that you know the coordinates of their vertices
2. The law of cosines paired with the formula for the area of any triangle could also be useful. The law of cosines states that for triangle $ABC$, with side lengths $a,b,c$ which are the sides opposite to the angles they correspond with, that $a^2=b^2+c^2-2ab(cos(A))$ and the area of any triangle is $\frac{1}{2}ab(cos(C))$

These are obviously not answers, if you are looking for new ways to break down those geometries, these are both powerful methods.

Answer 4

Slightly shorter: Let $w$ be the width of the rectangle, $h$ the height, and let $x$ be the length of segment $AB$. Argue as in @fleablood's answer (+1) that $AE=\frac23 EF$. By similar triangles, conclude $x=\frac23 w$.

The area of triangle $\triangle ABF$ is then $\frac12xh=\frac13wh$. Since this equals $5$, the area of the rectangle is $wh=15$.