Three cubes of silver with edges 3cm, 4cm and 5cm are melted and recast into a single cube. Find the cost of coating the surface of the new cube with gold at the rate of rs 3.50per square cm
RAJVEER41:
First of all you find the total surface area of all the cubes and then edit you will get 3900 CM square and then multiply it from rupees 3.50 you then you will get your answer 136500 rupees
Answers
Answered by
119
S1 = 3 cm
S2 = 4 cm
S3 = 5 cm
v of cube = s³
V1 = 3³ = 27 cm³
V2 = 4³ = 64 cm³
V3 = 5³ = 125 cm³
total volume = 27+64+125 = 216 cm³
V4 = 216 cm³
S4³ = 216
S4 = ³√216 = 6 cm
TSA of cube = 6s²
TSA = 6 × 6²
= 6 × 36 = 216 cm²
cost of gold plating = 216 × 3.5 = ₹756
therefore the cost of covering the new cube with gold will be ₹756
S2 = 4 cm
S3 = 5 cm
v of cube = s³
V1 = 3³ = 27 cm³
V2 = 4³ = 64 cm³
V3 = 5³ = 125 cm³
total volume = 27+64+125 = 216 cm³
V4 = 216 cm³
S4³ = 216
S4 = ³√216 = 6 cm
TSA of cube = 6s²
TSA = 6 × 6²
= 6 × 36 = 216 cm²
cost of gold plating = 216 × 3.5 = ₹756
therefore the cost of covering the new cube with gold will be ₹756
Answered by
77
volume of new cube = volume of all three cubes
a³=3³+5³+4³
a³=27+125+64
a³=216
a³=6³
a=6cm
volume of new cube=216cm³
cost of coating=3.50*216
=rs. 756
a³=3³+5³+4³
a³=27+125+64
a³=216
a³=6³
a=6cm
volume of new cube=216cm³
cost of coating=3.50*216
=rs. 756
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