Math CentralQuandaries & Queries


Question from Gilbert, a student:

I'm having trouble figuring out how to reduce the over all size of an irregular pentagon.
The pentagon I'm trying to reduce has 2 right angles adjacent to each other.
Basically the shape of a house. I have width, length of the legs and the height of the peak.
But cannot figure out how to get the size of a reduced version that keeps the same proportion
and angles.


Hi Gilbert,

I am not sure what you mean by reducing the over all size. If you want the width of the reduced pentagon to be r times the width of the original pentagon (r some fraction smaller than 1) then you accomplish this by multiplying the width, length and height of the peak in the original pentagon by r. If however you want the area of the smaller pentagon to be some factor, say k, of the larger pentagon then yo need to reduce the lengths of the sides by √k. I can illustrate with a simpler example.

Suppose you have a rectangle that is x units by y units then its area is xy square units. If you multiply the length of each side by √k then you have a rectangle that is √k x units by √k y units and hence an area of

(√k x)(√k y) = kxy square units.

I hope this helps,

About Math Central


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS