Suppose the measures of the sides of two rectangles are proportional.a. Will their perimeters have an equivalent ratio. b. Will their areas have an equivalent ratio? Please explain your answers.

4 Answers

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

In this case, the statement in A is correct, but the statement in B is not.  This is because A only involved addition while B involves multiplication.

To see that this is correct, just do a simple example.  Imagine one rectangle whose sides are 2 and 4 units and another whose sides are 4 and 8.  The second is two times the first, right.

So now look at the perimeters.  They will be 12 and 24 units, respectively.  So the second is still two times the first.

But the second rectangle's area is four times the first (2*4 as opposed to 4*8).


w0672211's profile pic

w0672211 | Student, Undergraduate | (Level 1) eNoter

Posted on

In a supposition, no. Supposition, I am not without the lack of reality and the ultimate truth of two proportionate lines. In that a commonality is assumed (sup-positionally) between two lines. At least, of the six possible commonalities, one, commonality assumes an inevitable quality and at the most six. Thus at certain abstract, and intangible lines either on paper, printed, imagined, or otherwise falsely represented, they do still in fact lead to a sensibly logical conclusion. (Equivalency) Although their ratios may not fully encompass a perimeter or area equally. There is still a commonality (From each) to reiterate that proportionately, single elements of plural lines are equal. maintaining that plurality represents two or more.  It still remains to an extent to be an abstract question with at least 7 possible outcomes.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let the rectangle have sides a and b for the 1st and  cand cd for the 2nd.

Then the perimeters are 2(a+d) and 2(c+d) .

Since, the sides of the rectangles are proportional,

a/c=a/d each equal to k say.

Then a=kc and b=kd. So a+b = k(c+d)

Therefore, the ratio of their permetrs = 2(a+b)/[2(c+d)]

=(a+b)/(c+d) = k(c+d)/(c+d) = k. So it is  proved that their perimeters are also in the sme ratio.


The ratio of their areas = ab/(cd) = (kc*kd)/(cd) =(k^2)cd/(cd) =k^2 =(a/c)^2 = (b/d)^2

This proves that thier areas are squarely proportional to their sides.

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

The answer above is correct but not complete.

The statement (a) in the question is correct, but the statement (b) is in incorrect.


If x and  y are measure of two side of a rectangle then its perimeter and formula are given by the following formula:

Perimeter = 2*(x + y)

area = x*y

Let us say in the measure of sides of first rectangles in the question are x1 andy1, and that of the other are x2 an y2.

It is given that their two sides are proportional: Therefor:

x1/x2 = y1/y2 = constant = say "p"


x1 = p*x2, and y1 = p*y2


Ratio of perimeter of two rectangles

= [2*(x1 + y1)]/[2*(x2 + y2)]

= [2*(p*x2 + p*y2)]/[2*(x2 + y2)]

= [2*p*(x2 + y2)]/[2*(x2 + y2)]

= p

Therefore if sides of two rectangles are in proportion "p" then their perimeters will be also be in same proportion "p".

Similarly ratio of areas of two rectangles:

= (x1*y1)/(x2*y2)

= [(p*x2)*(p*y2)]/(x2*y2)

= [(p^2(x2*y2)]/(x2*y2)

= p^2

Therefore if sides of two rectangles are in proportion "p" then their areas will be in proportion "p^2".