How Does Quadrupling The Side Lengths Of A Triangle Affect Its Area?

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embizze | High School Teacher | (Level 2) Educator Emeritus

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Here is a good rule of thumb: a dilation creates a similar figure. For all similar figures, the ratio of corresponding lengths is the same and is called the scale factor. In your example the scale factor is 4.

If the scale factor is k, all ratios of corresponding lengths will be k. (lengths of corresponding medians, altitudes, sides, apothems, etc...)

If the scale factor is k, the ratio of corresponding areas (including the areas of the figures) is k^2. For your example, the dilated triangle is 4^2=16 times as large.

If the figure is 3-dimensional, the ratio of corresponding volumes is k^3.

linear 1:k
area   1:k^2
volume 1:k^3

where k is the scale factor (or dilation constant.)

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on


The simplest method here is to use Hero's formula: the area of a triangle with sides' lengths a, b and c is equal to

`A = sqrt(p*(p-a)*(p-b)*(p-c)),`

where `p=(a+b+c)/2.`

When we quadruple each side, sides become a'=4a, b'=4b and c'=4c.
p' = (a'+b'+c')/2 = (4a+4b+4c)/2 = 4p. Also p'-a'=4p-4a=4*(p-a) and so on.


`A' = sqrt(p'*(p'-a')*(p'-b')*(p'-c')) = sqrt(4p*4(p-a)*4(p-b)*4(p-c)) = `

`sqrt(4*4*4*4)*sqrt(p*(p-a)*(p-b)*(p-c)) = (4*4)*A = 16A.`

So the answer is:
area of the new triangle is 16 times more than the area of initial one.

If you don't know Hero's formula, look at the picture: the new triangle can be divided by 16 triangles congruent to the initial one.

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