anonymous
  • anonymous
The perimeter of a rectangle is 53cm and its width is 11cm . what is the area of the rectangle ? PLZ HELP SOMEONE !!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Wait I'm confused @mathway
anonymous
  • anonymous
What is confusing?
anonymous
  • anonymous
Where did u get 35 from ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Kk
anonymous
  • anonymous
Lol its okay
anonymous
  • anonymous
\[\huge 53=2l+2(11)\]
anonymous
  • anonymous
Where did u get 21 from ?
anonymous
  • anonymous
Do you know how to solve the perimeter of a rectangle?
anonymous
  • anonymous
Hnmm I forgot
anonymous
  • anonymous
\(P=2(l+w)\) or \(P=2l + 2w\)
anonymous
  • anonymous
Just plug in the numbers. Perimeter= 53 and width=11. Then find the length (\(l\)).
anonymous
  • anonymous
Once you've found the length, use the formula in getting the area of a rectangle. \[\huge A=l \times w\]
anonymous
  • anonymous
Okay I'll do that
anonymous
  • anonymous
Sure! Let me know your answer afterwards. :)
anonymous
  • anonymous
Hold on I'm a little confused again
anonymous
  • anonymous
What's confusing this time?
anonymous
  • anonymous
What should I plug in ?
anonymous
  • anonymous
Where?
anonymous
  • anonymous
P = 2( L+w) or p = 2L+2w
anonymous
  • anonymous
Either way.
anonymous
  • anonymous
Now I have 22
anonymous
  • anonymous
And the perimeter is 53.
anonymous
  • anonymous
I did 53= 2L + 2(11)
anonymous
  • anonymous
So it's now...\[53=2l+22\]
mathstudent55
  • mathstudent55
|dw:1439244874901:dw|
anonymous
  • anonymous
You subtract 22 on both sides. What did you get?
mathstudent55
  • mathstudent55
The figure above shows the rectangle and the perimeter. 2 sides measure 11 cm each. The entire perimeter measures 53 cm.
anonymous
  • anonymous
I'm doing that now @mathway
mathstudent55
  • mathstudent55
|dw:1439245071530:dw|
anonymous
  • anonymous
|dw:1439245092875:dw|
anonymous
  • anonymous
Awesome! Take your time. :) @Halia
anonymous
  • anonymous
I have 31 = 2L
anonymous
  • anonymous
|dw:1439245240622:dw|
mathstudent55
  • mathstudent55
|dw:1439245252424:dw|
mathstudent55
  • 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
  • mathstudent55
I waited until you wrote this. This pretty much is the side. |dw:1439245350104:dw|
anonymous
  • anonymous
Now I'm just confused !!!!!!
anonymous
  • anonymous
Is the area 341cm²
anonymous
  • anonymous
@Halia Don't worry. I got you. What is the length first?
anonymous
  • anonymous
I got 15 for the length @mathway
anonymous
  • anonymous
Partly correct, but is that the exact length? The length has a decimal. :)
anonymous
  • anonymous
Oh yeah your right
anonymous
  • anonymous
So what's the length now? :)
anonymous
  • anonymous
15.0 I believe dont I have to divide 31÷2
anonymous
  • anonymous
Use your calculator. :)
anonymous
  • anonymous
Kk I did it & I got 15.5
anonymous
  • anonymous
Great!
anonymous
  • anonymous
Now what should I do?
anonymous
  • anonymous
And now, for the area (finally!!!), multiply the length and width.
anonymous
  • anonymous
\[w=11;~ l=15.5\]
anonymous
  • anonymous
Kk I'll do that right now :)
anonymous
  • anonymous
\[A=l \times w\]\[A=15.5 \times 11\]
anonymous
  • anonymous
Mkay I got 170.5cm²!!!
anonymous
  • anonymous
Yes! Great job!
anonymous
  • anonymous
Thank you so much for helping me !!! :) I really appreciate it
anonymous
  • anonymous
You're welcome! And it was YOU who did the work, not me, so you're awesome! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.