from a solid cylinder whose height is 2.4 cm and diameter is 1.4 CM a conical cavity of the same height and same diameter is hollowed out find the total surface area of the remaining solid to the nearest centimetres
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l=√r^2 + h^2
=√(0.7)^2 + (2.4)^2
=√0.49 + 5.76
=√6.25=2.5 cm
Remaining solid= CSA of cylinder + CSA of cone + area of base
= 2πrh + πrl+ πr^2
=πr(2h+l+r)
= 22/7* 0.7(2*2.4 + 2.5+0.7)
=2.2(4.8+2.5+0.7)
=17.6 cm
=app. 18 cm
=√(0.7)^2 + (2.4)^2
=√0.49 + 5.76
=√6.25=2.5 cm
Remaining solid= CSA of cylinder + CSA of cone + area of base
= 2πrh + πrl+ πr^2
=πr(2h+l+r)
= 22/7* 0.7(2*2.4 + 2.5+0.7)
=2.2(4.8+2.5+0.7)
=17.6 cm
=app. 18 cm
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