# Isosceles trapezoid

Calculate the content of an isosceles trapezoid whose bases are at ratio 5:3, the arm is 6cm long and it is 4cm high.

Result

S =  71.554 cm2

#### Solution:

$a:c = 5:3 \ \\ b = 6 \ cm \ \\ h = 4 \ cm \ \\ x = \sqrt{ b^2-h^2 } = \sqrt{ 6^2-4^2 } = 2 \ \sqrt{ 5 } \ cm \doteq 4.4721 \ cm \ \\ a = c + 2x \ \\ 3a = 5c \ \\ 3 \cdot \ (c+2x) = 5c \ \\ 3c + 6x = 5c \ \\ 2c = 6x \ \\ c = 3 \cdot \ x = 3 \cdot \ 4.4721 = 6 \ \sqrt{ 5 } \ cm \doteq 13.4164 \ cm \ \\ a = 5/3 \cdot \ c = 5/3 \cdot \ 13.4164 = 10 \ \sqrt{ 5 } \ cm \doteq 22.3607 \ cm \ \\ k_{ 1 } = a/c - 5/3 = 22.3607/13.4164 - 5/3 = -0 \ \\ S = (a+c) \cdot \ h/2 = (22.3607+13.4164) \cdot \ 4/2 = 32 \ \sqrt{ 5 } \doteq 71.5542 = 71.554 \ cm^2$

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