# 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:

$a:c = 5:3 \ \\ b = 6 \ cm \ \\ h = 4 \ cm \ \\ x = \sqrt{ b^2-h^2 } = \sqrt{ 6^2-4^2 } = 2 \ \sqrt{ 5 } \ cm \doteq 4.4721 \ cm \ \\ a = c + 2x \ \\ 3a = 5c \ \\ 3 \cdot \ (c+2x) = 5c \ \\ 3c + 6x = 5c \ \\ 2c = 6x \ \\ c = 3 \cdot \ x = 3 \cdot \ 4.4721 = 6 \ \sqrt{ 5 } \ cm \doteq 13.4164 \ cm \ \\ a = 5/3 \cdot \ c = 5/3 \cdot \ 13.4164 = 10 \ \sqrt{ 5 } \ cm \doteq 22.3607 \ cm \ \\ k_{ 1 } = a/c - 5/3 = 22.3607/13.4164 - 5/3 = -0 \ \\ S = (a+c) \cdot \ h/2 = (22.3607+13.4164) \cdot \ 4/2 = 32 \ \sqrt{ 5 } \doteq 71.5542 = 71.554 \ cm^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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