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

- anonymous

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- anonymous

Wait I'm confused @mathway

- anonymous

What is confusing?

- anonymous

Where did u get 35 from ?

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## More answers

- anonymous

Kk

- anonymous

Lol its okay

- anonymous

\[\huge 53=2l+2(11)\]

- anonymous

Where did u get 21 from ?

- anonymous

Do you know how to solve the perimeter of a rectangle?

- anonymous

Hnmm I forgot

- anonymous

\(P=2(l+w)\) or \(P=2l + 2w\)

- anonymous

Just plug in the numbers. Perimeter= 53 and width=11. Then find the length (\(l\)).

- anonymous

Once you've found the length, use the formula in getting the area of a rectangle. \[\huge A=l \times w\]

- anonymous

Okay I'll do that

- anonymous

Sure! Let me know your answer afterwards. :)

- anonymous

Hold on I'm a little confused again

- anonymous

What's confusing this time?

- anonymous

What should I plug in ?

- anonymous

Where?

- anonymous

P = 2( L+w) or p = 2L+2w

- anonymous

Either way.

- anonymous

Now I have 22

- anonymous

And the perimeter is 53.

- anonymous

I did 53= 2L + 2(11)

- anonymous

So it's now...\[53=2l+22\]

- mathstudent55

|dw:1439244874901:dw|

- anonymous

You subtract 22 on both sides. What did you get?

- mathstudent55

The figure above shows the rectangle and the perimeter.
2 sides measure 11 cm each.
The entire perimeter measures 53 cm.

- anonymous

I'm doing that now @mathway

- mathstudent55

|dw:1439245071530:dw|

- anonymous

|dw:1439245092875:dw|

- anonymous

Awesome! Take your time. :) @Halia

- anonymous

I have 31 = 2L

- anonymous

|dw:1439245240622:dw|

- mathstudent55

|dw:1439245252424:dw|

- mathstudent55

The perimeter minus the two known sides is the same as the two unknown sides.
53 cm - 22 cm = 31 cm
Since 31 cm is the length of two sides, each side measures 31 cm / 2 = 15.5 cm

- mathstudent55

I waited until you wrote this.
This pretty much is the side.
|dw:1439245350104:dw|

- anonymous

Now I'm just confused !!!!!!

- anonymous

Is the area 341cm²

- anonymous

@Halia Don't worry. I got you. What is the length first?

- anonymous

I got 15 for the length @mathway

- anonymous

Partly correct, but is that the exact length? The length has a decimal. :)

- anonymous

Oh yeah your right

- anonymous

So what's the length now? :)

- anonymous

15.0 I believe dont I have to divide 31÷2

- anonymous

Use your calculator. :)

- anonymous

Kk I did it & I got 15.5

- anonymous

Great!

- anonymous

Now what should I do?

- anonymous

And now, for the area (finally!!!), multiply the length and width.

- anonymous

\[w=11;~ l=15.5\]

- anonymous

Kk I'll do that right now :)

- anonymous

\[A=l \times w\]\[A=15.5 \times 11\]

- anonymous

Mkay I got 170.5cm²!!!

- anonymous

Yes! Great job!

- anonymous

Thank you so much for helping me !!! :) I really appreciate it

- anonymous

You're welcome! And it was YOU who did the work, not me, so you're awesome! :)

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