Conical area

A right angled triangle has sides a=12 and b=19 in right angle. The hypotenuse is c. If the triangle rotates on the c side as axis, find the volume and surface area of conical area created by this rotation.

Result

V =  2422.438
S =  988.1

Solution:

a=12 b=19 c=a2+b2=122+19250522.4722 c1=a2/c=122/22.47226.4079 c2=b2/c=192/22.472216.0643 h=c1 c2=6.4079 16.064310.1459 S1=π h2=3.1416 10.14592323.3912 V=(c1+c2) S1/3=(6.4079+16.0643) 323.3912/32422.43772422.438a=12 \ \\ b=19 \ \\ c=\sqrt{ a^2+b^2 }=\sqrt{ 12^2+19^2 } \doteq \sqrt{ 505 } \doteq 22.4722 \ \\ c_{1}=a^2/c=12^2/22.4722 \doteq 6.4079 \ \\ c_{2}=b^2/c=19^2/22.4722 \doteq 16.0643 \ \\ h=\sqrt{ c_{1} \cdot \ c_{2} }=\sqrt{ 6.4079 \cdot \ 16.0643 } \doteq 10.1459 \ \\ S_{1}=\pi \cdot \ h^2=3.1416 \cdot \ 10.1459^2 \doteq 323.3912 \ \\ V=(c_{1}+c_{2}) \cdot \ S_{1}/3=(6.4079+16.0643) \cdot \ 323.3912/3 \doteq 2422.4377 \doteq 2422.438
S=π h (a+b)=3.1416 10.1459 (12+19)988.0996988.1S=\pi \cdot \ h \cdot \ (a+b)=3.1416 \cdot \ 10.1459 \cdot \ (12+19) \doteq 988.0996 \doteq 988.1



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