9514 1404 393
Answer:
(b) 7720 mm² of paper
Step-by-step explanation:
The central horizontal rectangle of the net is 80 mm high and ...
40 + 5.5 + 40 + 5.5 = 91
mm long. Its area 9 is ...
(80 mm)(91 mm) = 7280 mm²
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The two pieces of the net not accounted for above are the "flaps" above and below the central rectangle. Each of those two has an area of 40 mm by 5.5 mm, so their total area is ...
2(40 mm)(5.5 mm) = 440 mm²
Then the area of paper needed for the wrapper is ...
total area = 7280 mm² + 440 mm²
total area = 7720 mm²
