SOLUTION: Find the dimensions of a rectangle whose length is 5 feet more than the width. The area of the rectangle is 36 sq. ft. Find the dimensions of the rectangle.
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Question 763786: Find the dimensions of a rectangle whose length is 5 feet more than the width. The area of the rectangle is 36 sq. ft. Find the dimensions of the rectangle.
Found 2 solutions by Alan3354, 2897696:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the dimensions of a rectangle whose length is 5 feet more than the width. The area of the rectangle is 36 sq. ft. Find the dimensions of the rectangle.
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4 by 9
Answer by 2897696(96) (Show Source): You can put this solution on YOUR website!
set variables for the length and width
width: x
length: x+5
now make and equation to solve for x
(since finding are is l*w, so you multiply the variables together)
x(x+5)=36
x^2+5x=36 ==>x^2+5x-36=0 <== ax^2+bx+c=0
use quadratic formula which is (-(b)ħsqrt(b^2-4ac))/2a
a=1 b=5 c=-36 (plug into the quadratic formula)
(-(5)ħsqrt(5^2-4(1)(-36))/2(1)
solve
(-5ħsqrt(25+144))/2
(-5ħsqrt(169))/2
(-5ħ13)/2
x= (-5+13)/2 ==>4 or x= (-5+-13)/2==>-9 * the width cannot be -9 because no width can be negative, so the only length is 6)
now to find the length, plug 6 back into the variable for solving length which is 4+5 =9
so the width is 4
the length is 9
Answer: width: 4 feet
length:9 feet
check by 4*9=36
36=36 check
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