# Octagonal mat

Octagonal mat formed from a square plate with a side of 40 cm so that every corner cut the isosceles triangle with leg 3.6 cm. What is the content area of one mat?

Result

S =  1574.08 cm2

#### Solution:

$A = 3.6^2/2 = \dfrac{ 162 }{ 25 } = 6.48 \ \\ S = 40^2 -4 \cdot \ A = 40^2 -4 \cdot \ 6.48 = \dfrac{ 39352 }{ 25 } = 1574.08 = 1574.08 \ cm^2$

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