Lesson FIND RECTANGLE DIMENSIONS USING GIVEN PERIMETER AND AREA

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This Lesson (FIND RECTANGLE DIMENSIONS USING GIVEN PERIMETER AND AREA) was created by by josgarithmetic(39635) About Me : View Source, Show
About josgarithmetic: Academic and job experience with beginning & intermediate Algebra. Tutorial help mostly for Basic Math and up through intermediate algebra.
FIND RECTANGLE DIMENSIONS USING GIVEN PERIMETER AND AREA

x, y, dimensions of rectangle as unknowns
A and p, area and perimeter, both known

system%28xy=A%2C2x%2B2y=p%29
-
2y=p-2x
y=%28p-2x%29%2F2=p%2F2-x
-
substitute for y in the area equation.
x%28p%2F2-x%29=A
%28p%2F2%29x-x%5E2=A
-x%5E2%2B%28p%2F2%29x=A
-x%5E2%2B%28p%2F2%29x-A=0
%28-1%29%28-x%5E2%2B%28p%2F2%29x-A%29=%28-1%29%2A0
x%5E2-%28p%2F2%29x%2BA=0
You can alternatively choose to multiply both sides by 2 so the coefficient on the x
term may be easier to use:
highlight_green%282x%5E2-px%2B2A=0%29

Continuing this way, x=%28p%2B-+sqrt%28p%5E2-4%2A2%2A2A%29%29%2F%282%2A2%29
highlight%28x=%28p%2B-+sqrt%28p%5E2-16A%29%29%2F%284%29%29

The MINUS form will be one of the dimensions and the PLUS form will be the other dimension.
Be aware that your academic exercises could be with factorable quadratic equations, and so
choosing a fully symbolic answer all the way to the general solution may not be needed;
instead, the actual A and p values can be used at any time.


SPECIFIC EXAMPLE:
find dimensions if A=60 and p=34.
Find dimensions, x and y.

x=%2834-sqrt%2834%5E2-16%2A60%29%29%2F4
x=%2834-14%29%2F4
x=20%2F5
highlight%28x=5%29
Using xy=A, the other dimension can be determined.
y=A%2Fx
y=60%2F5
highlight%28y=12%29
-
The dimensions are 5 and 12.
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