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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on June 26, 2009, 05:13:06 pm

Title: Q.
Post by: khalil on June 26, 2009, 05:13:06 pm: One side of a rectangle is three times the other. If the perimeter increases by 2%, what is the percentage increase in the area?
Title: Re: Q.
Post by: kamil9876 on June 26, 2009, 05:29:40 pm: P=3x+x+3x+x
=8x

let ' mean new thing, ie: P' means new perimeter etc.

$P'=\frac{102}{100}8x$
$8x'=\frac{102}{100}8x$
$x'=\frac{102}{100}x$ (1)

so x also increases by 2%.

However:
$<br />A=3x^2$

$A'=3(x')^2$

sub in (1) and expand the square:

$A'=3x^2 * (\frac{102}{100})^2$
$A'=A*(\frac{102}{100})^2$

So the area was multiplied by a ratio of $(\frac{102}{100})^2=\frac{102^2}{100^2}$

which is $\frac{102^2}{100}$ percent. However converting it into "how much percent it increased by" We have ot subtract 100%.(ie. just like increasing by 2% means multiplying by 102/100, we're going backwards now)

So in the end its like 4.04%

Note: I have assumed that the 3:1 ratio of length to width is preserved even after this increase of Perimeter. Which you havn't meant clear but the question probably said though, otherwise there would be no answer.
Title: Re: Q.
Post by: khalil on June 26, 2009, 07:16:54 pm: thanks Kamil, you've taken a Further maths approach to it, but never mind it yields the correct answer!