Question

from a solid cylinder whose height is 3.6 cm and diameter 2.1 CM a conical cavity of the same height and the same diameter is hollowed out find the total surface area of the remaining solid to the nearest CM square​

Answer 1

Height of the solid cylinder, h = 3.6 cm

The diameter of the solid cylinder, d = 2.1 cm

The radius of the solid cylinder,

r = 2.1/2 = 1.05 cm

Since the height and diameter of the conical part that is hollowed out from the solid cylinder is given to be same.

The slant height of the cone,

I= √ [r^ 2 +h^ 2 ]= √[(1.05)^2 +(3.6)^2]=√[14.0.625]=3.75 cm

Since the conical part is cut out from the solid cylinder, therefore,

The Total Surface Area of the

remaining solid = (Curved surface area of the cylinder) + (Curved surface area of the cone) + (Area of the cylindrical base)

= (2πrh)+(πrl)+(πr^2)

=[2×(22/6)×1.05×3.6]+[(22/7)×[(22/7)×1.05×3.75]+[(22/7)+(1.05)^2]

= 23.76 + 12.375 + 3.465

= 39.6

= 40cm^2

Thus, the total surface area of the solid is 40cm^2