 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
Regular pentagon QRSTU has a side length of 12 centimeters and an area of 248 square centimeters. Regular pentagon VWXYZ has a perimeter of 140 centimeters. I need to find its area. |
Hi there.
When you have similar shapes and the ratio of the linear dimensions is x, then the ratio of the area dimensions is x2.
What I mean is this: The side length of the second pentagon is 28. So the ratio of the linear dimensions is 28/12 = 7/3.
So the ratio of the areas is (7/3)2. This means the area of the second pentagon is the area of the first times this ratio:
A = (248) (7/3)2.
You can also calculate it the longhand way. See the following examples:
http://mathcentral.uregina.ca/QQ/database/QQ.09.06/s/dana1/dana1.html
http://mathcentral.uregina.ca/QQ/database/QQ.09.07/h/sarah1.html
http://mathcentral.uregina.ca/QQ/database/QQ.09.04/susan1.html
Or just type pentagon into the Quick Search.
Cheers,
Stephen La Rocque
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.