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OpenStudy (anonymous):
The perimeter of a rectangle is 53cm and its width is 11cm . what is the area of the rectangle ? PLZ HELP SOMEONE !!!
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OpenStudy (anonymous):
Wait I'm confused @mathway
OpenStudy (anonymous):
What is confusing?
OpenStudy (anonymous):
Where did u get 35 from ?
OpenStudy (anonymous):
Kk
OpenStudy (anonymous):
Lol its okay
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OpenStudy (anonymous):
\[\huge 53=2l+2(11)\]
OpenStudy (anonymous):
Where did u get 21 from ?
OpenStudy (anonymous):
Do you know how to solve the perimeter of a rectangle?
OpenStudy (anonymous):
Hnmm I forgot
OpenStudy (anonymous):
\(P=2(l+w)\) or \(P=2l + 2w\)
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OpenStudy (anonymous):
Just plug in the numbers. Perimeter= 53 and width=11. Then find the length (\(l\)).
OpenStudy (anonymous):
Once you've found the length, use the formula in getting the area of a rectangle. \[\huge A=l \times w\]
OpenStudy (anonymous):
Okay I'll do that
OpenStudy (anonymous):
Sure! Let me know your answer afterwards. :)
OpenStudy (anonymous):
Hold on I'm a little confused again
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OpenStudy (anonymous):
What's confusing this time?
OpenStudy (anonymous):
What should I plug in ?
OpenStudy (anonymous):
Where?
OpenStudy (anonymous):
P = 2( L+w) or p = 2L+2w
OpenStudy (anonymous):
Either way.
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OpenStudy (anonymous):
Now I have 22
OpenStudy (anonymous):
And the perimeter is 53.
OpenStudy (anonymous):
I did 53= 2L + 2(11)
OpenStudy (anonymous):
So it's now...\[53=2l+22\]
OpenStudy (mathstudent55):
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