a hall is in the form of a cylinder of diameter 12 m and height 4m surmounted by a come whose vertical angle is right Ange find the outer surface area
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Since OSA of cylinder = pi*d*h
OSA of cylinder = 3.142*12*4 = 50.7964m²
OSArea of cone=pi*r*L
By Pythagoras theorem to TriangleABD,
AB²=BD²+DA²
L²=6²+6²
L²=36+36=72
L=8.4853m
Therefore OSArea of cone=pi*r*L=pi*6*8.4853
OSA of cone=159.9438m²
Total outer surface area=50.7964+159.9438
=210.7402m²
OSA of cylinder = 3.142*12*4 = 50.7964m²
OSArea of cone=pi*r*L
By Pythagoras theorem to TriangleABD,
AB²=BD²+DA²
L²=6²+6²
L²=36+36=72
L=8.4853m
Therefore OSArea of cone=pi*r*L=pi*6*8.4853
OSA of cone=159.9438m²
Total outer surface area=50.7964+159.9438
=210.7402m²
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