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Mathematics 16 Online
OpenStudy (anonymous):

The perimeter of a rectangle is 53cm and its width is 11cm . what is the area of the rectangle ? PLZ HELP SOMEONE !!!

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

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 \(P=2l + 2w\)

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

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.

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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