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+192=50522.4722 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) 323.3912/32422.4377=2422.438a = 12 \ \\ b = 19 \ \\ c = \sqrt{ a^2+b^2 } = \sqrt{ 12^2+19^2 } = \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 = 2422.438
S=π h (a+b)=3.1416 10.1459 (12+19)988.0996=988.1S = \pi \cdot \ h \cdot \ (a+b) = 3.1416 \cdot \ 10.1459 \cdot \ (12+19) \doteq 988.0996 = 988.1



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