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.

Correct result:

S =  71.554 cm2

Solution:

a:c=5:3 b=6 cm d=b=6 cm h=4 cm  x=b2h2=62422 5 cm4.4721 cm a=x+c+x 3a=5c 3 (c+2x)=5c 3c+6x=5c 2c=6x  c=3 x=3 4.47216 5 cm13.4164 cm a=5/3 c=5/3 13.416410 5 cm22.3607 cm  S=(a+c) h/2=(22.3607+13.4164) 4/232 571.554271.554 cm2   Correctness test:  k1=ac=22.360713.4164531.6667a:c=5:3 \ \\ b=6 \ \text{cm} \ \\ d=b=6 \ \text{cm} \ \\ h=4 \ \text{cm} \ \\ \ \\ x=\sqrt{ b^2-h^2 }=\sqrt{ 6^2-4^2 } \doteq 2 \ \sqrt{ 5 } \ \text{cm} \doteq 4.4721 \ \text{cm} \ \\ a=x+c+x \ \\ 3a=5c \ \\ 3 \cdot \ (c+2x)=5c \ \\ 3c + 6x=5c \ \\ 2c=6x \ \\ \ \\ c=3 \cdot \ x=3 \cdot \ 4.4721 \doteq 6 \ \sqrt{ 5 } \ \text{cm} \doteq 13.4164 \ \text{cm} \ \\ a=5/3 \cdot \ c=5/3 \cdot \ 13.4164 \doteq 10 \ \sqrt{ 5 } \ \text{cm} \doteq 22.3607 \ \text{cm} \ \\ \ \\ S=(a+c) \cdot \ h/2=(22.3607+13.4164) \cdot \ 4/2 \doteq 32 \ \sqrt{ 5 } \doteq 71.5542 \doteq 71.554 \ \text{cm}^2 \ \\ \ \\ \text{ Correctness test: } \ \\ k_{1}=\dfrac{ a }{ c }=\dfrac{ 22.3607 }{ 13.4164 } \doteq \dfrac{ 5 }{ 3 } \doteq 1.6667



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