from a rectangular solid of metal 42cm by 30 cm by 20 CM a conical cavity of diameter 14 cm and depth 24 cm is drilled out find the surface area of the remaining solid second the volume of remaining solid third the weight of material drilled out if the weight is 7 gram per centimetre cube
Answers
Answer:
TSA - 5796 cm^2
volume - 23968 cm^3
weight - 167.776 kg
Step-by-step explanation:
(i)Original TSA of cuboid = 2(lb+ bh+hl)
= 2(42x30 + 30x20 + 20x42)
= 2(1260+600+840)
= 2(2700)
=5400cm^2
l = √24^2 + 7^2
l = √576 +49
l = √625
l = 25 cm
new TSA = 5400 - πr^2 + πrl
= 5400 - 22/7 x 7 x 7 + 22/7 x 7 x 25
= 5400 - 154 + 550
= 5400 + 396
= 5796 cm^2
(ii) volume of new solid = old volume - volume of cone
= lbh - 1/3πr^2h
= 42 x 30 x 20 - 1/3 x 22/7 x 7 x 7 x 24
= 25200 - 1232
= 23968 cm^2
(iii) weight = 23968 x 7
= 167776 g
= 167.776 kg