Surface area of the top

A cylinder is three times as high as it is wide. The length of the cylinder’s diagonal is 20 cm. Find the surface area of the top of the cylinder.

Result

S =  31.416 cm2

Solution:

u=20 cm h=3D u2=h2+D2 u2=9D2+D2 D=u9+1=209+1≐2 10 cm≐6.3246 cm r=D/2=6.3246/2≐10 cm≐3.1623 cm h=3⋅ D=3⋅ 6.3246≐6 10 cm≐18.9737 cm u2=h2+D2=18.97372+6.32462=20 cm u=u2(check−ok) S=π⋅ r2=3.1416⋅ 3.16232≐31.4159 S=10 π=31.416 cm2u=20 \ \text{cm} \ \\ h=3D \ \\ u^2=h^2 + D^2 \ \\ u^2=9D^2 + D^2 \ \\ D=\dfrac{ u }{ \sqrt{ 9+1 } }=\dfrac{ 20 }{ \sqrt{ 9+1 } } \doteq 2 \ \sqrt{ 10 } \ \text{cm} \doteq 6.3246 \ \text{cm} \ \\ r=D/2=6.3246/2 \doteq \sqrt{ 10 } \ \text{cm} \doteq 3.1623 \ \text{cm} \ \\ h=3 \cdot \ D=3 \cdot \ 6.3246 \doteq 6 \ \sqrt{ 10 } \ \text{cm} \doteq 18.9737 \ \text{cm} \ \\ u_{2}=\sqrt{ h^2 + D^2 }=\sqrt{ 18.9737^2 + 6.3246^2 }=20 \ \text{cm} \ \\ u=u_{2} (check - ok) \ \\ S=\pi \cdot \ r^2=3.1416 \cdot \ 3.1623^2 \doteq 31.4159 \ \\ S=10 \ \pi=31.416 \ \text{cm}^2

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