Find Area of A Rectangle With Shortcut

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SUMMARY

To find the area of a rectangle with a perimeter of 72 cm, one must understand that there are infinite rectangles that can satisfy this condition. The area can be expressed as A = bh, where b (base) and h (height) are related through the equation b = 36 - h, derived from the semi-perimeter of 36 cm. The maximum area achievable under these constraints is 324 cm², occurring when both dimensions are equal, specifically at 18 cm each.

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  • Understanding of basic geometry concepts, specifically perimeter and area.
  • Familiarity with algebraic manipulation of equations.
  • Knowledge of the properties of rectangles and semi-perimeters.
  • Ability to interpret mathematical expressions and inequalities.
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  • Study the relationship between perimeter and area in different geometric shapes.
  • Learn about optimization techniques in geometry to maximize area.
  • Explore algebraic methods for solving equations involving multiple variables.
  • Investigate the properties of rectangles and squares in greater detail.
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Students, educators, and anyone interested in geometry, particularly those looking to simplify calculations involving the area and perimeter of rectangles.

susanto3311
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hi all...

how do you find area of a rectangle, if its perimeter of a rectangle is = 72 cm?

i mean how to easy find it without hard work.

do you have a formula or just tricks similar like..

http://calculus-geometry.hubpages.com/hub/How-to-Find-the-Area-Perimeter-and-Diagonal-of-a-Rectangle

thanks in advance...

susanto
 
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For a given perimeter, there are an infinite number of rectangles that will have that perimeter. We need more information about the rectangle.
 
MarkFL said:
For a given perimeter, there are an infinite number of rectangles that will have that perimeter. We need more information about the rectangle.

if perimeter of a rectangle is = 72 cm, counting area of a rectangle =...

i need simple formula to calculate it.
possible?
 
Well, suppose we let the height $0<h<36$ be a free variable, then the base $b$ is $b=36-h$. Thus the area $A$ is:

$$A=bh=(36-h)h$$

Thus we find:

$$0<A\le324$$
 
The sum of the base and the height must be equal to the semi-perimeter, which is 36...half of 72. And the maximum area comes from the base and height being equal. So the upper bound is $18^2=324$.
 

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