Respuesta :
we know that
[surface area]=2*[area of the base]+[perimeter of the base]*height
area of the base=30*70-----> 2100 mm²
perimeter of the base=2*(30+70)-----> 200 mm
height=4 1/2 mm-----> (4*2+1)/2-----> 9/2 mm
[surface area]=2*[2100]+[200]*9/2-----> 5100 mm²
the answer is
5100 mm²
[surface area]=2*[area of the base]+[perimeter of the base]*height
area of the base=30*70-----> 2100 mm²
perimeter of the base=2*(30+70)-----> 200 mm
height=4 1/2 mm-----> (4*2+1)/2-----> 9/2 mm
[surface area]=2*[2100]+[200]*9/2-----> 5100 mm²
the answer is
5100 mm²
Answer:
[tex]5100mm^{2}[/tex]
Step-by-step explanation:
Given : A chocolate bar measures 30 mm wide, 70 mm long, and four and one over two mm high
To Find : how much paper will be needed to make the wrapper.
Solution :
Length of chocolate bar = 70 mm
Width of chocolate bar = 30 mm
Height of chocolate bar = [tex]4\frac{1}{2} =\frac{9}{2} =4.5 mm[/tex]
Since chocolate bar in in the shape of cuboid
Total surface area of cuboid : 2(length*width+width*height+height*length)
Thus the wrapper required to wrap chocolate = 2(70*30+30*4.5+4.5*70)
= 2100+135+315
= 2(2550)
= 5100 square mm
Hence[tex]5100mm^{2}[/tex] paper will be needed to make the wrapper.