If a rectangular region has perimeter P inches and area A sq

Let the rectangular region have sides \(x\) and \(y\).
--> \(Perimeter = P = 2(x+y)\)
--> \(Area = A = xy\)

Question: Is \(x = y\) ?

(1) \(P = \frac{4}{3}*A\)

\(2(x+y) = (\frac{4}{3})xy\)

\(\frac{x+y}{xy} = \frac{4}{6} = \frac{2}{3}\)

\(\frac{1}{x} + \frac{1}{y} = \frac{4}{6} = \frac{2}{3}\)

So we have that the sum of two values equals \(\frac{2}{3}\)

If \(\frac{1}{x} = \frac{1}{y} = \frac{1}{3}\), then the answer is YES

If \(\frac{1}{x} = \frac{1}{6}\) and \(\frac{1}{y} =\frac{3}{6}\), then the answer is NO

Two different answers --> INSUFFICIENT


(2) \(P = 4\sqrt{A}\)

\(2(x + y) = 4\sqrt{xy}\)

\(x + y = 2\sqrt{xy}\)

\(x + y - 2\sqrt{xy} = 0\)

\((\sqrt{x} - \sqrt{y})^2 = 0\)

\(\sqrt{x} - \sqrt{y} = 0\)

\(\sqrt{x} = \sqrt{y}\)

\(x = y\)

--> SUFFICIENT

Lets try finding the relation between area and perimeter of a square.
We know the area of a square with length x (A) = x^2
The perimeter of the square (P) = 4x

Since P = 4x --> x = P/4

So, A = (P/4)^2
A =(P^2)/16
Therefore, P^2 = 16A
and P = 4*sqrtA

Therefore, B is suff to answer this question

Let me show you what I tried to do with this one

Is the figure a square?

Well if its a square then perimeter is 4x and area is x^2, pretty obvious up to here.

Now, statement 1

P=4/3A

So if the figure is in fact a square then it must follow that 4x=4/3 (x^2)

Is 12x = 4x^2?
Is 4x (x-3) = 0?
Is x = 3?

Well we can't solve this so insufficient

Statement 2 says that P= 4 (sqrt) A

So is 4x = 4 (sqrt (x^2))?

Well is 4x = 4 abs (x)?

If x is the side of the square then x>0, so 4x = 4x, so yes this is in fact a square

Answer is B

Hope this helps
Cheers
J

for a rectangle Perimeter P=2(l+b)
Area A=l*b

From 1) 2(l+b)=4/3*(l*b)
6(l+b)=4(l*b)
3(l+b)=2(l*b)
this equation satisfies for l=3 and b=3, dimensions of a square and at l=6,b=2 dimensions of a rectangle..hence not sufficient

From 2) 2(l+b)=4√(l*b)
(l+b)=2√(l*b) squaring both sides
(l+b)^2=4l*b
(l-b)^2=0
l=b,which is a square, Sufficient

Given: A rectangular region has perimeter P inches and area A square inches

Target question: Is the region square?
This is a good candidate for rephrasing the target question.

So, what kind of relationship between P and A do we need in order for the rectangular region to be a SQUARE?
Well, if the region is a square, then all 4 sides will be the same length
So, if P = the perimeter (sum of all 4 sides), then P/4 = the length of each side
If P/4 = the length of each side, then the AREA = (P/4)(P/4)
In other words: A = (P/4)(P/4)
Simplify: A = P²/16
Multiply both sides by 16 to get: 16A = P²
Take square root of both sides to get: 4√A = P
So, if 4√A = P, then we can be certain that the region is a square.
REPHRASED target question: Does 4√A = P?

Below, you'll find a video with tips on rephrasing the target question

Statement 1: P = (4/3)(A)
Let's TEST some values.
There are several values of P and A that satisfy statement 1. Here are two:
Case a: P = 12 and A = 9. In this case, the answer to the REPHRASED target question is YES, 4√A DOES EQUAL P
Case b: P = 4 and A = 3. In this case, the answer to the REPHRASED target question is NO, 4√A does NOT equal P
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: P = 4√A
Perfect!!
The answer to the REPHRASED target question is YES, 4√A DOES EQUAL P
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer: B

RELATED VIDEO

My approach here :

From 1: 3(l+b)= 2lb

Put in
b= 10L (or anything that makes it a non square)

33L=20L^2
33=20L

There definitely is some value of L that will satisfy this..So a rectangle does exist..

b=l

6L = 2L^2
6=2L
L=3

So a square too exists...Done A is insuff

For B

the equation comes to: 2(L+B)=4\sqrt{LB}
Square and simplify
L^2 + B^2 - 2 LB = 0
which is (L-B)^2

you can also try the above approach here:
B=10 L
101L^2- 20L^2 = 0

Which aint true..So B is suff to say that it is indeed a square

I would solve this question by simply plugging in numbers.

1. Let's assume the rectangular is indeed a square.
2. Let's assume the side of the square is 4.

If the side of this square is 4:
The perimeter would be 16 and the area would be 16 as well.

(S1) P = 4/3√A
If the rectangular is a square, all values for A area should correspond with the right value for P.
In my assumed case: if the perimeter is 16 , the area should be 16 too.
In this case : 16 is not equal to 4/3*16 , so we can't say for sure that this rectangle indeed is a square.

(S2) P = 4√A
16=4*√16
We got a match!

If you want to be sure , try any other plugin for a square.
Lets say the side of the rectangle is 2.
For it to be a square the perimeter should be 8(2+2+2+2) and the area is 4(2*2)
8=4√4
So as you can see , S2 will hold true for any plugin.A

This can be Rhombus as well right?Even Rhombus has all sides equal. So how can we confidently say that it is a square. We dont know whether the diagonals are equal

Would really appreciate if someone can clear my doubt

Thanks

Hi buddyisraelgmat,

The prompt refers to a RECTANGULAR REGION, so the four corners must be 90 degrees each. Thus, we're dealing with either a rectangle of some kind or a square.

GMAT assassins aren't born, they're made,
Rich

I would also solve by plugging in values:

Let's create two rectangular regions which are squares:

Generally: P = 4s (where s is the side of a square) and \(A = s^2\)

First square with side 2:
P = 8 and A = 4

Second square with side 3:
P = 12 and A = 9

Statement 1: Gives you a yes on the square with side 3 because P = 4/3 * A but a no for the square with side 2.
Statement 2: \(P = 4*\sqrt{A}\) - this is true for both squares we created. Fair enough, Answer B.

mh not quite satisfied with the explanations

So Ill give it a try


If a rectangular region has perimeter P inches and area A square inches, is the region square?


Perimeter of a Square = 4x
Area of a Square = x^2

(1) P = 4/3*A

4x = 4/3*x^2
3x=x^2
x=3

so in order to be a square x has to be 3, but we don't know --> insufficient



(2) P = 4√A

4x=4√x^2
x=x


--> sufficient


cheers

The goal is to determine if the shape is in fact a square.

Statement 1) P = 4/3*(A)

If true, then P=4x and A = x^2

4x = 4/3*(x^2)
x = x^2/3
x = 3

Only true if the side is length 3. We don't know if the side of the shape has length 3, so the statement is insufficient.

Statement 2) P = 4sqrt(A)

4x = 4sqrt(x^2)
4x=4x
x = x

This statement is true, so the statement is sufficient.

Asked: If a rectangular region has perimeter P inches and area A square inches, is the region square?

For square P = 4a & area A = a^2
P = 4√A

(1) P = 4/3*A
P = 4a^2/3
If a = 3
P = 12= 4*3; Area = 9; a= 3; Square
But if a=2
A = 4; P= 16/3: Not a square
NOT SUFFICIENT

(2) P = 4√A
If side = a; A = a^2 ; P = 4a = 4√A; Square
SUFFICIENT

IMO B

as a non-english native speaker.

I dont understand what the question is asking.

If a rectangular region has perimeter P inches and area A square inches, is the region square?

My interpretation : a rectangular region, since the the prompt already specified it's a rectangular, why the question is asking if that is a square?

Hi waihoe520,

By definition, a 'rectangular region' has four 90-degree angles and two 'pairs' of opposite sides that are equal in length. Thus, a square IS a rectangle (since a square fits that definition), but most rectangles are NOT squares (since a square has 4 equal sides and most rectangles do not).

GMAT assassins aren't born, they're made,
Rich

Hi

abhimahna Bunuel
Can you explain why statement 1 is not sufficient ?

As per statement 1

1) P=4/3*A

P= 2(l+b) and A l*b

hence , substituting

2(l+b)=4/3*l*b
6l+6b =4lb
3l+3b=2lb
3l+3b-2lb=0

Rearranging
3l-lb + 3b-lb=0
l(3-b)+b(3-l)=0

-> Since l,b are dimensions of a figure l,b cannot be 0 or negative
Therefore to make the above expression 0
l=b=3

Hence statement 1 is sufficient.

2(l+b)=4/3*l*b has infinitely many solutions. Foe example, if l = 10, then b = 30/17. Even if you assume that l and b are integers (which would not be correct), still 2(l+b)=4/3*l*b has more solutions. Foe example, b = 2 and l = 6.
_________________

Wondering what I did wrong here. For statement 2, we can just plug in numbers ...

P = 4√A
A = 4 ---> P = 8

With 8, you can make a square with sides = 2 all around or do 3-1-3-1

Why is this solution wrong?
_________________

Hi CEdward,

TESTing VALUES is a great way to approach this question, but you still have to follow all of the 'restrictions' that the prompt gives you (as well as the overall rules of math).

In your example, you have an AREA of 4 and a PERIMETER of 8.

With a 2x2 square, you have an area of 4 and a perimeter of 8, so this is a possible option.
With a 1x3 rectangle, you have an area of 3..... but that does NOT match the area of 4 that needs to occur, so a 1x3 rectangle is NOT possible here. The ONLY option that fits this particular combination of area and perimeter is a 2x2 square.

GMAT assassins aren't born, they're made,
Rich