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The width of a rectangle is twice as long as the length. if the length is increased by 50% and the width is decreased by 20%, the perimeter becomes 248. find the width and length of the original rectangle.

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Answer:

Explanation:

The percents here make this more tricky than it originally seems to be. We'll make a table and see where it takes us:

original new

length

width

And we'll fill it in according to our rules given. Starting with the original, we are told that the width is twice as long as the length. We don't know the length, so we'll call that L, and if the width is twice that, the width is 2L:

original new

length L

width 2L

Now here's the tricky part. What I'm going to do is fill in the "new" column with the expressions and then we'll simplify them in the next step.

The length is increased by 50%. So we have 100% of the original length and we are adding another 50% to that:

original new

length L 100%L + 50%L

width 2L

The width is decreased by 20%, so we have 100% of 2L and we are subtracting 20% of 2L from that:

original new

length L 100%L + 50%L

width 2L 100%(2L) - 20%(2L)

And now we'll simplify that "new" column:

original new

length L 150%L = 1.5L

width 2L 80%(2L) = 160%L = 1.6L

Now we're ready for the perimeter part. The formula for the perimeter of a rectangle is P = 2L + 2w, so filling in from our "new" column, since 248 is the perimeter given for AFTER the rectangle's length and width are manipulated:

248 = 2(1.5L) + 2(1.6L) and

248 = 3L + 3.2L and

248 = 6.2L so

L = 40 and that means that w = 80 (because in the "original" column, the width is twice the length)

User Ian Chu
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