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A museum exhibit, ABCD, has infrared beams around it for security, as shown below.
If the length of the beams is made three times the original length on each side, which statement is correct about the maximum area available for an exhibit to be displayed?
It becomes nine times the original area.
It becomes twentyseven times the original area.
It becomes three times the original area.
It becomes eightyone times the original area.
 one year ago
 one year ago
A museum exhibit, ABCD, has infrared beams around it for security, as shown below. If the length of the beams is made three times the original length on each side, which statement is correct about the maximum area available for an exhibit to be displayed? It becomes nine times the original area. It becomes twentyseven times the original area. It becomes three times the original area. It becomes eightyone times the original area.
 one year ago
 one year ago

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CompassionateBest ResponseYou've already chosen the best response.2
Give me best answer on your previous problem, then I will answer this.
 one year ago

izzy777Best ResponseYou've already chosen the best response.0
answer this one ill give you for both best award
 one year ago

CompassionateBest ResponseYou've already chosen the best response.2
Imagine x,y,z to be the lengths of the sides where they are labeled. I multiplied each portion by 6 to get 6x, 6y, and 6z. From there, we find the new size of each area, which is 36xy + 18xz. The old area was just xy + xz. So what is the difference in Areas? The parallelogram here is a trapezoid. Thus the area of this parallelogram is: (b1+b2)h/2 . In this case, b1 = AB and b2 = CD. and the height (h) is AD. When each length is timed by six, the new sides are just 6 times the original length of each. Plugging this in, the new area is: 1/2(6h(6b1+6b2),or(afterfactoring),36(1/2(b1+b2)h) . Since the original area is 1/2(b1+b2)h, the new area is just 36/3 times the original area . = 12
 one year ago
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