Progressive ratio

You can find the area of the quadrilateral by finding the combined area of the four triangles surrounding it and subtracting this from the area of the square, 1.

First, find the length of each line segment as a fraction. Since each side is 1, if the ratio is 1:1 then there are 2 parts, so one part would be 1/2. Use this method to find all line segments.

Side AE = 1/2 Side AH = 4/5 ${\frac{1}{2}}$ ${\times}$ ${\frac{4}{5}}$ = ${\frac{4}{10}}$ ${\frac{2}{5}}$ ÷ 2 = ${\frac{1}{5}}$ = ${\frac{24}{120}}$

Side EB = 1/2 Side BF = 1/3 ${\frac{1}{2}}$ ${\times}$ ${\frac{1}{3}}$ = ${\frac{1}{6}}$ ${\frac{1}{6}}$ ÷ 2 = ${\frac{1}{12}}$ = ${\frac{10}{120}}$

Side FC = 2/3 Side CG = 1/4 ${\frac{2}{3}}$ ${\times}$ ${\frac{1}{4}}$ = ${\frac{2}{12}}$ ${\frac{1}{6}}$ ÷ 2 = ${\frac{1}{12}}$ = ${\frac{10}{120}}$

Side GD = 3/4 Side DH = 1/5 ${\frac{3}{4}}$ ${\times}$ ${\frac{1}{5}}$ = ${\frac{3}{20}}$ ${\frac{3}{20}}$ ÷ 2 = ${\frac{3}{40}}$ = ${\frac{9}{120}}$

${\frac{24}{120}}$ + ${\frac{10}{120}}$ + ${\frac{10}{120}}$ + ${\frac{9}{120}}$ = ${\frac{53}{120}}$

${\frac{120}{120}}$ - ${\frac{53}{120}}$ = ${\boxed{\frac{67}{120} } }$

if we assume the coordinates of D to be (0,0) then A(0,1) B(1,1) C(1,0). Then the coordinates of the internal points are H(0,0.2), E(0.5,1), F(1,2/3) and G(0.75, 0). Then we use the formula for a quadrilateral whose 4 vertices we know that is Area,A=0.5(x1y2+x2y3+x3y4+x4y1-y1x2-y2x3-y3x4-y4x1) which gives the value 67/120.

If you complete each white triangle to a rectangle, you have paired blue area with equal white area except for a blue rectangle left in the middle with side lengths $\frac{1}{4}\times\frac{7}{15}=\frac{7}{60}$ . So the blue area is $\frac{7}{60}$ plus half of what remains: $\frac{1}{2}\left(\frac{106}{120}\right) + \frac{14}{120} = \frac{53+14}{120} = \frac{67}{120}$

(Note: the side lengths of the remaining blue rectangle come from $\frac{3}{4}-\frac{1}{2} = \frac{1}{4}$ and $\frac{4}{5}-\frac{1}{3} = \frac{7}{15}$ )

Place the unit square in Cartesian space with point D as the origin (0,0). Using the given information, it's not difficult to find the following vectors:

1. vector(EH) = (-1/2, -4/5) [To be read as the vector from E to H.]
2. vector(EF) = (1/2, -1/3)
3. vector(GH) = (-3/4, 1/5)
4. vector(GF) = (1/4, 2/3)

The area of the quadrilateral, A, is then given by the following formula: A = 1/2 |vector(EH) x vector(EF)| + 1/2 |vector(GH) x vector(GF)|. ['x' represents the cross-product.]

Calculating the cross-products, taking half their magnitudes, and adding will yield an answer of 67/120.

The usefulness of this method over the standard geometric solution is that it can be extended to any polygon so long as the locations of each vertex is known. The downside is that it requires a coordinate system.

The area of the square, 1, is equal to the sum of the areas of the triangles defined at the corners and the quadrilateral EFGH. Thus, solving for Area(EFGH):

Area(ABCD) = Area(EFGH) + Area(EBF) + Area(FCG) + Area(GDH) + Area(HAE) Area(EFGH) = Area(ABCD) - Area(EBF) - Area(FCG) - Area(GDH) - Area(HAE)

Now recall the area of a triangle is defined by (1/2)BH where B = base, H = height. Thus: Area(EBF) = (1/2) EB BF Area(FCG) = (1/2) FC CG Area(GDH) = (1/2) GD DH Area(HAE) = (1/2) HA AE

Let x denote the unit components of AB, y those of BC, z those of CD and w those of DA. It remains to find the side length components. For the side EB, note that AE/EB = 1:1 or AE/EB = 1x/1x = x/x which means AB = 1x + 1x = x + x = 2x AB/2 = x =AE = EB = 0.5AB Since ABCD is a unit square, AB = 1 and thus AE = EB = 0.5(1) = 0.5

Similarly: BF/FC = 1:2 or BF/FC = 1y/2y = y/2y which means BC = 1y + 2y = y + 2y = 3y FC/3 = y = BF and FC = 2(BF) Since ABCD is a unit square, BC = 1 and thus BF = 1(1/3) = 1/3 and FC = 2(1/3) = 2/3

CG/GD = 1:3 or CG/GD = 1z/3z = z/3z which means CD = 1z + 3z = z + 3z = 4z CD/4 = z = CG and GD = 3(CG) Since ABCD is a unit square, CD = 1 and thus CG = 1(1/4) = 1/4 and GD = 3(1/4) = 3/4

DH/HA = 1:4 or DH/HA = 1w/4w = w/4w which means DA = 1w + 4w = w + 4w = 5w DA/5 = w = DH and HA = 4(DH) Since ABCD is a unit square, DA = 1 and thus DH = 1(1/5) = 1/5 and HA = 4(1/5) = 4/5

Therefore: Area(EBF) = (1/2)(1/2)(1/3) = 1/12 = 10/120 Area(FCG) = (1/2)(2/3)(1/4) = 2/24 = 1/12 = 10/120 Area(GDH) = (1/2)(3/4)(1/5) = 3/40 = 9/120 Area(HAE) = (1/2)(4/5)(1/2) = 1/5 = 20/120

Or: Area(EFGH) = 1 - (10/120) - (10/120) - 9/120 - 24/120 = 120/120 - 2(10/120) - 24/120 = 120/120 - 24/120 - 29/120 = 120/120 - 53/120 = 67/120