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## anonymous one year ago 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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1. anonymous

Wait I'm confused @mathway

2. anonymous

What is confusing?

3. anonymous

Where did u get 35 from ?

4. anonymous

Kk

5. anonymous

Lol its okay

6. anonymous

$\huge 53=2l+2(11)$

7. anonymous

Where did u get 21 from ?

8. anonymous

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

9. anonymous

Hnmm I forgot

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

Okay I'll do that

14. anonymous

Sure! Let me know your answer afterwards. :)

15. anonymous

Hold on I'm a little confused again

16. anonymous

What's confusing this time?

17. anonymous

What should I plug in ?

18. anonymous

Where?

19. anonymous

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

20. anonymous

Either way.

21. anonymous

Now I have 22

22. anonymous

And the perimeter is 53.

23. anonymous

I did 53= 2L + 2(11)

24. anonymous

So it's now...$53=2l+22$

25. mathstudent55

|dw:1439244874901:dw|

26. anonymous

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

27. mathstudent55

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

28. anonymous

I'm doing that now @mathway

29. mathstudent55

|dw:1439245071530:dw|

30. anonymous

|dw:1439245092875:dw|

31. anonymous

Awesome! Take your time. :) @Halia

32. anonymous

I have 31 = 2L

33. anonymous

|dw:1439245240622:dw|

34. mathstudent55

|dw:1439245252424:dw|

35. 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

36. mathstudent55

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

37. anonymous

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

38. anonymous

Is the area 341cm²

39. anonymous

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

40. anonymous

I got 15 for the length @mathway

41. anonymous

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

42. anonymous

Oh yeah your right

43. anonymous

So what's the length now? :)

44. anonymous

15.0 I believe dont I have to divide 31÷2

45. anonymous

Use your calculator. :)

46. anonymous

Kk I did it & I got 15.5

47. anonymous

Great!

48. anonymous

Now what should I do?

49. anonymous

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

50. anonymous

$w=11;~ l=15.5$

51. anonymous

Kk I'll do that right now :)

52. anonymous

$A=l \times w$$A=15.5 \times 11$

53. anonymous

Mkay I got 170.5cm²!!!

54. anonymous

Yes! Great job!

55. anonymous

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

56. anonymous

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

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