Percent Increase what would be the percent increase in the area of the plot?

[mathdad]

If the length and width of a rectangle garden plot were each increased by 20 percent, what would be the percent increase in the area of the plot?

My Work:

Here I am thinking geometry mixed with algebra.
The area of a rectangle is found by using A = bh.

A = (0.20)(0.20)

Correct set up?

 

[MarkFL]

No, that would be a way to begin computing the amount of decrease of the area if both length and width were decreased by 80%.

 

[mathdad]

No, that would be a way to begin computing the amount of decrease of the area if both length and width were decreased by 80%.

Can you set it up? Explain the process.

 

[MarkFL]

Based on what you initially posted, and my reply to it, can you hazard another attempt?

 

[mathdad]

I will attempt to answer this when time allows. Time to go back to my post.

 

[MarkFL]

I will attempt to answer this when time allows. Time to go back to my post.

Please wait until you have progress to post, rather than just posting to say you will post later. Such posts needlessly bump threads. ;)

 

[mathdad]

100(1.20^2 - 1)

Note: 1.00 + 0.20 = 1.20

Answer is 44 percent.

 

[MarkFL]

100(1.20^2 - 1)

Note: 1.00 + 0.20 = 1.20

Answer is 44 percent.

Yes, I would write:

$$100\frac{\Delta A}{A}\%=100\frac{1.2b\cdot1.2h-bh}{bh}\%=100(1.2^2-1)\%=44\%$$

 

[mathdad]

I like the little set up here. I now know whst to do should I come across similar questions.

 

[HOI]

If the length and width of a rectangle garden plot were each increased by 20 percent, what would be the percent increase in the area of the plot?

My Work:

Here I am thinking geometry mixed with algebra.
The area of a rectangle is found by using A = bh.

A = (0.20)(0.20)

Correct set up?

No. If the original length and width were x and y, respecively, so that the area is A= xy. Since x is increased by 20% then then new length is 100%+ 20%= 120% of x: 1.20x. Similarly the new width is 120% of y: 120y. the new area is (1.2x)(1.2y)= 1.44xy= 1.44A. So the area has increased by 44%

 

[mathdad]

No. If the original length and width were x and y, respecively, so that the area is A= xy. Since x is increased by 20% then then new length is 100%+ 20%= 120% of x: 1.20x. Similarly the new width is 120% of y: 120y. the new area is (1.2x)(1.2y)= 1.44xy= 1.44A. So the area has increased by 44%

I have already answered the question. See previous post.