Irregular pentagon

A rectangle-shaped, 16 x 4 cm strip of paper is folded lengthwise so that the lower right corner is applied to the upper left corner. What area does the pentagon have?

Correct result:

S =  47 cm²

Solution:

$$\begin{gathered} a=16\text{ cm } \\ b=4\text{ cm } \\ {S}_0=a\cdot b=16\cdot 4=64{\text{cm}}^2 \\ x+y=a \\ {x}^2=y^2+b^2 \\ x+y=16 \\ {y}^2=x^2+4^2=7.5^2+4^2=16 \\ x=15\mathrm{/}2={\displaystyle \frac{15}{2}}=7.5\text{ cm } \\ y=17\mathrm{/}2={\displaystyle \frac{17}{2}}=8.5\text{ cm } \\ {S}_1={\displaystyle \frac{b\cdot x}{2}}={\displaystyle \frac{4\cdot 7.5}{2}}=15{\text{cm}}^2 \\ {S}_2=x\cdot b=7.5\cdot 4=30{\text{cm}}^2 \\ {S}_3=(y-x)\cdot b\mathrm{/}2=(8.5-7.5)\cdot 4\mathrm{/}2=2{\text{cm}}^2 \\ S=S_1+S_2+S_3=15+30+2=47{\text{cm}}^2 \end{gathered}$$

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