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=b2h2=62422 5 cm4.4721 cm a=c+2x 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 k1=a/c5/3=22.3607/13.41645/3=0 S=(a+c) h/2=(22.3607+13.4164) 4/232 571.554271.554 cm2a:c=5:3 \ \\ 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=c + 2x \ \\ 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} \ \\ 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 \doteq 32 \ \sqrt{ 5 } \doteq 71.5542 \doteq 71.554 \ \text{cm}^2



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