Geometry Word Problems: Complex Examples (page 4 of 6) Sections: Introduction, Basic examples, Triangle formulas, Complex examples, The Box Problem & the Goat Problem, Max / Min problems
The perimeter P of this rectangular area with (asyet unknown) length L and width W is given by: 2L + 2W = 148 The area A is given by: L×W = 1320 I will divide my "perimeter" equation above by 2, so I am dealing with smaller numbers. This gives me the following system (or "set") of equations: L
+ W = 74
I can solve either one
of these equations for either one of the variables, and then plug this
into the other equation. I think I'll solve the addition equation, and
plug the result into the multiplication equation: L
= 74 – W
(solving the first
equation for L)
Once again, I've come up with two valid solutions. If W = 30, then L = 74 – W = 74 – 30 = 44. If W = 44, then L = 74 – W = 74 – 44 = 30. The important point is that the shorter side (whether I refer to it as the "width" or the "length") is across the front of the lot. The garden is 44 feet by 30 feet, with the 30ft length along the front. Note that we cannot say which of the dimensions is the length or the width, since no information was provided regarding which was longer. So "this by that" is as accurate an answer as we can give.
The first statement, "three times the width exceeds twice its length by three inches", compares the length L and the width W. I'll start by doing things orderly, with clear and complete labelling: the width: W
The second statement, "four times its length is twelve more than its perimeter", compares the length L and the perimeter P. I will again be complete with my labelling: four times its length:
4L
So now I have my two equations: Copyright © Elizabeth Stapel 20002011 All Rights Reserved 3W
= 2L + 3
There are various ways of solving this; the way I do it (below) just happens to be what I thought of first. I'll take the first equation and solve for W: 3W
= 2L + 3
Now I'll simplify the second equation, and then plug in this above expression for W: 4L
= 2L + 2W + 12
Then: W
= ( ^{2}/_{3} )L + 1
The question didn't ask me to "Find the values of the variables L and W". It asked me to "Find the dimensions of the rectangle," so the actual answer is: The length is 21 inches and the width is 15 inches. << Previous Top  1  2  3  4  5  6  Return to Index Next >>



Copyright © 20002012 Elizabeth Stapel  About  Terms of Use 




