Find area of a triangle divided into 6 parts and areas of 3 parts are known.

Question

An arbitrary triangle is divided into 6 arbitrary parts and the areas of 3 of them are known.

As you can see in the diagram, a triangle is divided into 6 parts by the three lines passing through the vertexes and an arbitrary point P inside the triangle You are given the area of the three regions, namely, area(△UPZ), area(△ZPW) and area(△WPY).

Find the area of the triangle.

Answer 1

Note that $$ UZ:ZW = \Delta UPZ:\Delta ZPW$$ and $$ UP:PY = (\Delta UPZ+\Delta ZPW):\Delta WPY$$ This allows you to reconstruct everything up to shearing

Answer 2

HINT: Knowing $A(\Delta UPW)$ and $A(\Delta PWY)$, you can find the ratio $\frac{|UP|}{|PY|}$. Also, knowing $A(\Delta UPZ)$ and $A(\Delta PZY)$, you can find the ratio $\frac{|UZ|}{|ZW|}$. Using these two ratios, you can find the whole area.

Answer 3

Denote the area of the triangle $A$ and the known areas I, II and III. The ratio $x/y$ can be expressed as,

$$ \frac{x}{y} = \frac{I+II}{III}\tag{1}$$

On the other hand

$$ \frac{x}{y} = \frac{Area(UVZ)}{Area(VYZ)} = \frac{I}{II} \cdot \frac{a+b}{a}\tag{2}$$

Combine (1) and (2)

$$\frac{I+II}{III} = \frac{I}{II} \cdot \frac{a+b}{a} \tag{3}$$

Also, the following ratio holds

$$\frac{a+b}{a} = \frac{A}{A-(I+II+III)}\tag{4}$$

Plug (4) into (3) to arrive at the area of the triangle,

$$ A = \frac{I\cdot (I+II) \cdot (I+II+III)}{II^2 + I \cdot(II-III)} $$