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?

Result

S =  47 cm2

Solution:

a=16 cm b=4 cm S0=a⋅ b=16⋅ 4=64 cm2  x+y=a x2=y2+b2  x+y=16 y2=x2+42=7.52+42=18  x=15/2=152=7.5 cm y=17/2=172=8.5 cm  S1=b⋅ x2=4⋅ 7.52=15 cm2 S2=x⋅ b=7.5⋅ 4=30 cm2 S3=(y−x)⋅ b/2=(8.5−7.5)⋅ 4/2=2 cm2  S=S1+S2+S3=15+30+2=47 cm2a=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=18 \ \\ \ \\ x=15/2=\dfrac{ 15 }{ 2 }=7.5 \ \text{cm} \ \\ y=17/2=\dfrac{ 17 }{ 2 }=8.5 \ \text{cm} \ \\ \ \\ S_{1}=\dfrac{ b \cdot \ x }{ 2 }=\dfrac{ 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/2=(8.5-7.5) \cdot \ 4/2=2 \ \text{cm}^2 \ \\ \ \\ S=S_{1}+S_{2}+S_{3}=15+30+2=47 \ \text{cm}^2

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