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Wait I'm confused @mathway
What is confusing?
Where did u get 35 from ?
Lol its okay
Where did u get 21 from ?
Do you know how to solve the perimeter of a rectangle?
Hnmm I forgot
\(P=2(l+w)\) or \(P=2l + 2w\)
Just plug in the numbers. Perimeter= 53 and width=11. Then find the length (\(l\)).
Once you've found the length, use the formula in getting the area of a rectangle. \[\huge A=l \times w\]
Okay I'll do that
Sure! Let me know your answer afterwards. :)
Hold on I'm a little confused again
What's confusing this time?
What should I plug in ?
P = 2( L+w) or p = 2L+2w
Now I have 22
And the perimeter is 53.
I did 53= 2L + 2(11)
So it's now...\[53=2l+22\]
You subtract 22 on both sides. What did you get?
The figure above shows the rectangle and the perimeter. 2 sides measure 11 cm each. The entire perimeter measures 53 cm.
I'm doing that now @mathway
Awesome! Take your time. :) @Halia
I have 31 = 2L
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
I waited until you wrote this. This pretty much is the side. |dw:1439245350104:dw|
Now I'm just confused !!!!!!
Is the area 341cm²
@Halia Don't worry. I got you. What is the length first?
I got 15 for the length @mathway
Partly correct, but is that the exact length? The length has a decimal. :)
Oh yeah your right
So what's the length now? :)
15.0 I believe dont I have to divide 31÷2
Use your calculator. :)
Kk I did it & I got 15.5
Now what should I do?
And now, for the area (finally!!!), multiply the length and width.
Kk I'll do that right now :)
\[A=l \times w\]\[A=15.5 \times 11\]
Mkay I got 170.5cm²!!!
Yes! Great job!
Thank you so much for helping me !!! :) I really appreciate it
You're welcome! And it was YOU who did the work, not me, so you're awesome! :)