SOLUTION: For a student recreation building at Northland Community College, an
architect wants to lay out a rectangular piece of land that has a perimeter
of 204 m and an area of 2565 m2
Algebra.Com
Question 1028253: For a student recreation building at Northland Community College, an
architect wants to lay out a rectangular piece of land that has a perimeter
of 204 m and an area of 2565 m2
. Find the dimensions of the land (Hint:
write a system of two equations, then solve the system)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
perimeter = 204
area = 2565
if x = length and y equal width, then the following formulas apply.
2x + 2y = 204
x * y = 2565
in the second formula of x * y = 2565, solve for y to get y = 2565 / x
in the first formula of 2x + 2y = 204, replace y with 2565 / x to get:
2x + 2 * (2565 / x) = 204x
multiply both sides of this equation by x to get:
2x^2 + 2 * 2565 = 204x
simplify to get 2x^2 + 5130 = 204x
subtract 204x from both sides of this equation to get:
2x^2 - 204x + 5130 = 0
this is a quadratic equation that can be solved by factoring or using the quadratic formula, or by graphing.
i solved using the quadratic formula and got x = 45 or x = 57.
if x * y = 2565, then:
if x = 45, y = 57
if x = 57, y = 45
you have the length is equal to 45 and the width is equal to 57, or you have the length is equal to 57 and the width is equal to 45.
perimeter = 2 * length + 2 * width = 2 * (length + width) = 2 * (45 + 57) = 2 * 102 = 204.
you have area = length * width = 45 * 57 = 2565.
numbers look good.
that's your answer.
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