Question 884457: The length of a rectangle is 6 ft. longer than twice the width. If the perimeter is 90 ft. find the length and width of the rectangle.
Found 2 solutions by josgarithmetic, MathTherapy: Answer by josgarithmetic(39617) (Show Source): Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 6 ft. longer than twice the width. If the perimeter is 90 ft. find the length and width of the rectangle.
Let width be W, and length, L
Then L = 2W + 6 ------- eq (i)
Since perimeter = 2L + 2W, then 2L + 2W = 90
2(L + W) = 2(45) ------ Factoring out GCF, 2
L + W = 45 ------------ eq (ii)
2W + 6 + W = 45 ------- Substituting 2W + 6 for L in eq (ii)
3W = 39
W, or width = , or ft
Use this info to determine the value of L, or the length.
Then do a check!!
If you need a complete and detailed solution, let me know!!
Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.
Further help is available, online or in-person, for a fee, obviously.
For FREE info and answers to questions about the ASVAB exam, the NYS 3 – 8 city/state wide exams,GENERAL
MATH and HOMEWORK QUESTIONS, or MATH QUESTIONS related to the Regents Integrated Algebra,
Regents Geometry, Regents Algebra 2/Trigonometry, SHSAT, COOP/HSPT/TACHS, PSAT, SAT, ACT, SSAT/ISEE,
GRE, CLEP, and the TASC/GED, you can visit: http://asvabstudyzone.freeforums.net/.
|
|
|