Quadrilaterals II

In the ABCDEFGHIJKL, the two adjacent sides are perpendicular to each other, and all sides except the AL and GF sides are identical. The AL and GF sides are twice as long as the other sides. The lines BG and EL intersect at point M and divide the dodecagon into six shapes (three triangles, two quadrilaterals, and one pentagon). The EFGM square has an area of 7 cm².

Determine the area of the other five departments.

Final Answer:

S₁ =  7 cm²
S₂ =  5 cm²
S₃ =  2 cm²
S₄ =  2 cm²
S₅ =  1 cm²

Step-by-step explanation:

$$\begin{gathered} S_6=7{\text{cm}}^2 \\ {S}_6=(1/2+1/4+1)S_0 \\ {S}_0=S_6/(1+\frac{1}{2}+\frac{1}{4})=7/(1+\frac{1}{2}+\frac{1}{4})=4{\text{cm}}^2 \\ {S}_0=a\cdot a \\ a=\sqrt{S_0}=\sqrt{4}=2\text{cm} \\ {S}_1=S(ABML) \\ {S}_1=S_6=7=7{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} S_2=S(CDEMB) \\ {S}_2=(1+1/4)\cdot S_0=(1+1/4)\cdot 4=5{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} S_3=S(JKL) \\ {S}_3=1/2\cdot S_0=1/2\cdot 4=2{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} S_3=S(GIH) \\ {S}_4=S_3=2=2{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} S_5=S(MIJ) \\ {S}_5=\frac{1}{4}\cdot S_0=\frac{1}{4}\cdot 4=\frac{1\cdot 4}{4}=\frac{4}{4}=1{\text{cm}}^2 \end{gathered}$$

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