the area of a rectangle is 100 square inches.the perimeter of the rectangle is 40 inches.a second rectangle has the same area but a different perimeter. is the second rectangle a square?
Use calculus to prove that the minimum perimeter of a rectangle of area 100 square inches is 40 inches, and that happens when all the sides are equal, namely the rectangle is a square.
Therefore any other rectangle having a different perimeter cannot be a square.
Separate a rectangle into 6 equal parts
Dhycrttfdhhcftcfgjhgfcjcyghgfccyjgjyfgjfytjfgyjgfyjygfjfgyjhfgjyfgfyjtyjtr
GfousfgydilsFDSGILUVDDIGHLLZVLIGHDC,JVZBHCCBKNJXXVC,JHBVDDJGHLDFFSLHBJVZB,JHZV,,JHBVZXCGKYIVFZDLGUIDRGB,VIJDSF,IUBZDFVIGULVFDA,HJ VDGLUIVSDAHBILVDFA. KNDADVIHG ADDBHLIDFV NM.FDVBLUIBHCDVLJVA,D. FMDADVBH.K C,MN FVD.BAKJVZFDULHIVFD.BKJAVFBHLKLIVUGZDFKHLZBDFVZHDKLVJFLBFVZIHDLBJKVFDZFGDBSLIJVBFHLSDILNJKFVDSLKUGFSHD.HIJFGZDGDLFHJZKLFDVKZJBZLBJKFZDGLBKJDFGZLBJKZLJBKDZGFLHJKFGDS
KHJFGDZJKLHFGZHJKDVZJGHKZD.JB.KZDVDVZFHBJK.FDVFDZBKJZSFBFNjldbfoihDVFJNlsdgh
Fsojuhkfdag
His.gdfanjk.lkuhrfdabui
Hirfdaukbljfvdzbkjdvzfbkhlzdvfbhjlvdfbllvdfsvfdbhkgsrdsrdjbklgreslugidrvs.jbdkfsdfalgdvfihlbjgfdlkbhfvdlgiyfvblhjfrbhlildgiyvhblidfljhbvddlbjhdfslbihbsfdhljjlhbvs,.kiuglsdfb.unjust.defygilfrebhlireflihuvddugillbkhddvilygddgb.knhulisdv, Malibu
Kbhilfsrilbdrv
Knjdcdjk
Ddjsvd
Iosgr
Ergbjo
Vndjsk
Kbjk
Dvs jkvfskjb
Bsfdvbvdsb
Sag khaki
Sjvbhhidsbfbhifsdbhkhbksdfdsfsddsbfhibihsfdbhsdjvkjytcjcyfcfjyhmcgfgummhgchmgvj,ghubilli.kbjcxx.I
Ubibulivfdjib
Dog
Jonah
Offbeat
Ink
Jkbdfjkb
Did
Dfv
Kbjdvdlhibddvbhlisdvhblidsvbilhdvsilhbddvulbi
Iubfvdu
Hi Debbie
Hidden
Iubdfv
Iubfvdi
Budfvbui
Greuiher
Uhvdskjbs
Dvkjbvdsbjk
Jkbsdf
Bjskfdk
Jsbdf
Kbjsdfb hi.vds.kbjk.jbcbjlkkhljdxvlihgvsdbjil
Jodhfghilgfdg
Vghjjvhkbkhjilihlvvdskjvhegeabrihlsefrgyukbhjldfvzj,bhVDSljhbvdsajbhlz,jcvhbxbhjvsdzlyguszdvblhksagf
Lishb
IaOHJb
Jo
But
Jobs so
s
Job
UVSd
Buoying
UIGyugogykf=‘ug&67’kuygiludfeulbidfskugvfdskvvhusefb
Hi”6*8”8*6”+87”*$”#/=8($#”/$=(8”’fhbsrlghixdvighlfdsgliybchlsidfliguss,djvhferfklbhsdilygflhreuif
Igusdflhbksefbhlidfsgiysdf.bhksdf.dsflhigsdflhkbfdsbhlicsdlhjbldihbsclgilylygiluyvyvluvyullihb🐊🐎🐊🐈🐊🐊🐊🦓🐊🦤🦛🐎🦞🕷🪰🕷🐊🦛🦎🦐🦤🦧🦎🦞🦛🦍🦎🐈🦛🐊🐈🦏🐊🦌🐊🦏🦬🦎🦧🦏🐖🪰🦏🐎🐬🦎🐎🦧🦘🦓🦑🐎🦏🐍🐎🦛🐜🐎🦤🦧🦞🪰🦛🐎🦗🦐🐎🦧🪰🕷🦧🦤🐎🦛🕷🐎🐈🦎🦞🦘🐈🪲🦞🐈🦧🪲🦞🐊🐈🕷🪰🐎🐈🦬🦚🕷🕷🐫🦌k
Needful
Housfdubillihbdfsbdsf
Bijfvdskuhhuo
Add
Oinasdlbjssdnasdb
Ouvfdzliubzdbfpihubgddhiusfppdfbiguvdfhupluhbfdvspiuhvdd
Ouhfvdaab
Hi
Huiasvblihabou
Blihvabhilhuipfebiluvdaibuvadbiuadfgouhdgafnou
Sfgdio
Hi
Joining
Throne
Ifgfsnoir
To determine if the second rectangle is a square, we first need to find the dimensions of both rectangles. We are given that the area of the first rectangle is 100 square inches and the perimeter is 40 inches.
Let's denote the length of the rectangle as L and the width as W. The area of the rectangle is given by the formula A = L * W, and the perimeter is given by the formula P = 2L + 2W.
For the first rectangle:
Area = 100 square inches
Perimeter = 40 inches
Using these formulas, we can set up the following equations:
100 = L * W (equation 1)
40 = 2L + 2W (equation 2)
To simplify equation 2, we divide both sides by 2:
20 = L + W (equation 3)
Now, we can solve these equations simultaneously to find the dimensions of the first rectangle.
From equation 3, we can express W in terms of L:
W = 20 - L (equation 4)
Substituting equation 4 into equation 1:
100 = L * (20 - L)
Expanding the equation:
100 = 20L - L^2
Rearranging to form a quadratic equation:
L^2 - 20L + 100 = 0
Now we can solve the quadratic equation to find the value(s) of L. By factoring or using the quadratic formula, we find that L = 10.
Substituting the value of L back into equation 4:
W = 20 - 10 = 10
So, the dimensions of the first rectangle are L = 10 inches and W = 10 inches, making it a square.
Since the second rectangle has the same area (100 square inches) but a different perimeter, we need to determine its dimensions.
The second rectangle's dimensions can be denoted as L2 and W2.
The area of the second rectangle is again given by A = L2 * W2, and the perimeter is given by P2 = 2L2 + 2W2.
Since the area is the same as the first rectangle (A = 100), we have:
100 = L2 * W2 (equation 5)
Now, let's say the perimeter of the second rectangle is P2, which is different from the first rectangle's perimeter (40). So:
P2 ≠ 40
To determine if the second rectangle is a square, we need to compare the dimensions. If the dimensions are equal, then it is a square.
Let's analyze the potential values of L2 and W2:
- If both L2 and W2 are equal to 10 (same as the first rectangle), then the second rectangle is a square.
- If L2 = 5 and W2 = 20, the perimeter would be 50, which does not match the given perimeter of 40.
- If L2 = 20 and W2 = 5, the perimeter would be 50, which also does not match the given perimeter of 40.
- If L2 = 2 and W2 = 50, the perimeter would be 104, which does not match the given perimeter of 40.
- If L2 = 50 and W2 = 2, the perimeter would be 104, which also does not match the given perimeter of 40.
Thus, we can conclude that the second rectangle is not a square since it does not have the same dimensions as the first rectangle.