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Marina CMI [18]
3 years ago
13

A cereal company is exploring new cylinder-shaped containers. Two possible containers are shown in the diagram.x is 6 and r=4.5

y is 10.5 and r=3 If both containers have labels covering their lateral surfaces, which container will have less label area and by how much? Container Y by about 42 in.2. Container X by about 28 in.2. Container Y by about 9 in.2. Container X by about 85 in.2.

Mathematics
1 answer:
KatRina [158]3 years ago
5 0

Answer:

Container \rm X will have less label area than container \rm Y by about 9\; \rm in.

Step-by-step explanation:

A rectangular sheet of paper can be rolled into a cylinder. Conversely, the lateral surface of a cylinder can be unrolled into a rectangle- without changing the area of that surface.

Indeed, the width of that rectangle will be the same as the height of the cylinder. On the other hand, the length of that rectangle should be exactly equal to the circumference of the circles on the top and the bottom of the cylinder. In other words, if a cylinder has a height of h and a radius of r at the top and the bottom, then its lateral surface can be unrolled into a rectangle of width h and length 2\,\pi\, r.

Apply this reasoning to both cylinder \mathrm{X} and \rm Y:

For cylinder \mathrm{X}, h = 6\; \rm in while r = 4.5\; \rm in. Therefore, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be 6\; \rm in, while
  • The length of the rectangle will be 2 \, \pi \times 4.5\; \rm in = 9\, \pi\; \rm in.

That corresponds to a lateral surface area of 54\, \pi\; \rm in^2.

For cylinder \rm Y, h = 10.5\; \rm in while r = 3\; \rm in. Similarly, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be 10.5\; \rm in, while
  • The length of the rectangle will be 2\pi\times 3\; \rm in = 6\,\pi \; \rm in.

That corresponds to a lateral surface area of 63\,\pi \; \rm in^2.

Therefore, the lateral surface area of cylinder \rm X is smaller than that of cylinder \rm Y by 9\,\pi\; \rm in^2.

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Answer:

Step-by-step explanation:

From the question given in the picture,

a). Since, NR bisects a straight angle ∠MNP,

  ∠MNR ≅ ∠PNR

  m∠MNR + m∠PNR = 180°

  2(m∠MNR) = 180°

  m∠MNR = 90°

  Therefore, ∠MNR and ∠PNR are the right angles.

  Since, QN divides ∠MNR in two parts,

  Therefore, ∠QNR will be an acute angle (less than 90°).

  ∠MNR + ∠SNR = ∠MNS

  90° + ∠SNR = ∠MNS

  Therefore, m∠MNS will be more than 90°.

    ∠MNS will be an obtuse angle (greater than 90°).

(b). Since, NR divides ∠MNP and ∠QNS,

    ∠MNR ≅ ∠PNR

     ∠QNR ≅ ∠SNR

     ∠MNQ ≅ ∠PNS

(c). m∠MNR = 90°

     Since, NR bisects ∠QNS,

     ∠QNR ≅ ∠RNS

     m∠QNR = m∠RNS = 30°

     m∠QNR + m∠RNS = 30° + 30°

     m∠QNR + m∠RNS = 60°

     m∠QNS = 60° [Since, m∠QNS = m∠QNR + m∠RNS]

     m∠QNP = m∠QNS + m∠SNP

     m∠QNP = m∠QNS + (m∠PNR - m∠SNR)

     m∠QNP = 60° + (90° - 30°) = 120°        

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