SOLUTION: The perimeter of a standard high school basketball court is 268ft. The length is 34ft longer than the width. Find the dimensions of the court.

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Question 225094: The perimeter of a standard high school basketball court is 268ft. The length is 34ft longer than the width. Find the dimensions of the court.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
The perimeter of a standard high school basketball court is 268ft. The length is 34ft longer than the width. Find the dimensions of the court.

Step 1. The perimeter P means adding up all the four sides of a rectangle.

Step 2. Let w be the width.

Step 3. Let w+34 be the length since the length is 34 ft longer than the width.

Step 4. Then, P=w+w+w+34+w+34=268.

Step 5. Solving yields the following steps

Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: w=50.
  • Graphical form: Equation w%2Bw%2Bw%2B34%2Bw%2B34=268 was fully solved.
  • Text form: w+w+w+34+w+34=268 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+w%2Bw%2Bw%2B34%2Bw%2B34=268+%29
    For tutors: simplify_cartoon( w+w+w+34+w+34=268 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at w%2Bw%2Bw%2Bhighlight_red%28+34+%29%2Bw%2Bhighlight_red%28+34+%29=268.
Added fractions or integers together
It becomes w%2Bw%2Bw%2Bhighlight_green%28+68+%29%2Bw=268.
Look at w%2Bw%2Bw%2Bhighlight_red%28+68+%29%2Bw=268.
Moved 68 to the right of expression
It becomes w%2Bw%2Bw%2Bw%2Bhighlight_green%28+68+%29=268.
Look at .
Eliminated similar terms highlight_red%28+w+%29,highlight_red%28+w+%29,highlight_red%28+w+%29,highlight_red%28+w+%29 replacing them with highlight_green%28+%281%2B1%2B1%2B1%29%2Aw+%29
It becomes highlight_green%28+%281%2B1%2B1%2B1%29%2Aw+%29%2B68=268.
Look at .
Added fractions or integers together
It becomes %28highlight_green%28+4+%29%29%2Aw%2B68=268.
Look at highlight_red%28+%28highlight_red%28+4+%29%29%2Aw+%29%2B68=268.
Remove unneeded parentheses around factor highlight_red%28+4+%29
It becomes highlight_green%28+4+%29%2Aw%2B68=268.
Look at 4%2Aw%2B68=highlight_red%28+268+%29.
Moved these terms to the left highlight_green%28+-268+%29
It becomes 4%2Aw%2B68-highlight_green%28+268+%29=0.
Look at 4%2Aw%2Bhighlight_red%28+68+%29-highlight_red%28+268+%29=0.
Added fractions or integers together
It becomes 4%2Aw%2Bhighlight_green%28+-200+%29=0.
Look at 4%2Aw%2Bhighlight_red%28+-200+%29=0.
Removed extra sign in front of -200
It becomes 4%2Aw-highlight_green%28+200+%29=0.
Look at highlight_red%28+4%2Aw-200+%29=0.
Solved linear equation highlight_red%28+4%2Aw-200=0+%29 equivalent to 4*w-200 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: w=50.

Universal Simplifier and Solver


Done!



For w=50 , then w+34=84 and P=2(50+84)=268 which is a true statement.

Step 6. ANSWER: The width is 50 feet and he length is 84 feet.

I hope the above steps were helpful.

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J

drjctu@gmail.com

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