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3 votes
the measured dimensions of a rectangle are 9 m by 8 m to the nearest whole unit. find the minimum and maximum possible areas of the rectangle

2 Answers

6 votes

Answer: 63.75 square meters

Step-by-step explanation:

3 votes

Answer:

Min = 63.75 square meters

Max = 80.75 square meters

Technically there is no max area. See the explanation below.

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Step-by-step explanation:

We have to think backwards. What value, when rounded to the nearest whole number, will get us to 9?

We could have something like 8.6 or 9.2 as potential candidates. Both round to 9. The smallest we can do is 8.5; there is no largest because we could have 9.49 or 9.499 or 9.4999 or 9.49999 and so on.

The possible lengths are in this interval
8.5 \le L < 9.5; all of those values in that interval round to 9 when rounding to the nearest whole number.

Use similar logic to determine the interval
7.5 \le W < 8.5 has all those W values rounding to 8.

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Let's pick on the smallest values from each interval. Use them to find the smallest possible area of the rectangle.

A = L*W

A = (8.5)*(7.5)

A = 63.75

The smallest area possible is 63.75 square meters.

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Since each interval technically doesn't have a largest value, this extends to the idea that there is no largest area either.

Let's picked the so called "largest" value from each interval however, and follow the same idea as the previous section.

A = L*W

A = (9.5)*(8.5)

A = 80.75

Therefore, the ceiling value of the area is 80.75 square meters. The area will get closer and closer to this upper bound without actually reaching it.

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Summary:

The area value (A) is in the interval
63.75 \le A < 80.75

In other words: The area is between 63.75 and 80.75, including the first endpoint but excluding the second endpoint.

User Kannan S
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