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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  A quadrilateral has a perimeter of 16. What is the area of the quadril

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A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

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Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

quadrilaterl can be square , rhombus or a rectangle or ∆
#1

insufficient ; but we dont know whether figure is ∆ or a rectangle or any other shape quadrilateral
#2
all angles are same ; so we cannot determine anything in specific insufficient
from 1 & 2
sides are equal measure and given info about l & w ; we can say that its a rectangle
so
let l = x so w = 0.6x
given x+x+0.6x+0.6x=16
x= 5 ; l and w = 3
area = 15
IMO C
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Originally posted by Archit3110 on 23 May 2019, 10:04.
Last edited by Archit3110 on 24 May 2019, 09:22, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
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Archit3110 wrote:
Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

quadrilaterla can be square , rhombus or a rectangle
#1
let l = x so w = 0.6x
given x+x+0.6x+0.6x=16
x= 5 ; l and w = 3
area = 15
sufficient
#2
all angles are same ; so we cannot determine anything in specific insufficient
IMO A
Bunuel & chetan2u ; hello I am not able to understand why answer given as C , could you please explain the same..

Quadrilateral can be of any shape..
You are confusing the two terms quadrilateral and parallelogram.
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Archit3110 wrote:
Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

quadrilaterla can be square , rhombus or a rectangle
#1
let l = x so w = 0.6x
given x+x+0.6x+0.6x=16
x= 5 ; l and w = 3
area = 15
sufficient
#2
all angles are same ; so we cannot determine anything in specific insufficient
IMO A
Bunuel & chetan2u ; hello I am not able to understand why answer given as C , could you please explain the same..

+1 to what Chetan said *and* I'll add - if you don't know statement 2, that you're dealing with all 90-degree angles and thus a rectangle, then your multiplying the length times the width isn't necessarily how you'd find the area. If you don't have 90-degree angles (a square or rectangle) then you have to multiply base times height, where the height is perpendicular to the base. That distance in other parallelograms is different from the side (you generally have to form a right triangle to calculate that perpendicular height). So statement 1 isn't sufficient: 3 * 5 = 15 is one possible area, but if the shape is a non-rectangle then it would be 5 * h and h would not be 3.

A strategic note here - this is a great example of a strategy we teach at Veritas Prep called "Why Are You Here?" Statement 2 is pretty clearly insufficient on its own, but if you liked Statement 1 by assuming you had a rectangle and then see statement 2 specifically telling you that you have a rectangle (4 equal angles), you should consider "hey do I need them to tell me that or do I already know that?" To me that's an invitation to redraw the shape with something *other* than 90-degree angles and that should show you that a parallelogram is possible.
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thanks a lot VeritasPrepBrian & chetan2u for the explanation, I now understand where I went wrong...

VeritasPrepBrian wrote:
Archit3110 wrote:
Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

quadrilaterla can be square , rhombus or a rectangle
#1
let l = x so w = 0.6x
given x+x+0.6x+0.6x=16
x= 5 ; l and w = 3
area = 15
sufficient
#2
all angles are same ; so we cannot determine anything in specific insufficient
IMO A
Bunuel & chetan2u ; hello I am not able to understand why answer given as C , could you please explain the same..

+1 to what Chetan said *and* I'll add - if you don't know statement 2, that you're dealing with all 90-degree angles and thus a rectangle, then your multiplying the length times the width isn't necessarily how you'd find the area. If you don't have 90-degree angles (a square or rectangle) then you have to multiply base times height, where the height is perpendicular to the base. That distance in other parallelograms is different from the side (you generally have to form a right triangle to calculate that perpendicular height). So statement 1 isn't sufficient: 3 * 5 = 15 is one possible area, but if the shape is a non-rectangle then it would be 5 * h and h would not be 3.

A strategic note here - this is a great example of a strategy we teach at Veritas Prep called "Why Are You Here?" Statement 2 is pretty clearly insufficient on its own, but if you liked Statement 1 by assuming you had a rectangle and then see statement 2 specifically telling you that you have a rectangle (4 equal angles), you should consider "hey do I need them to tell me that or do I already know that?" To me that's an invitation to redraw the shape with something *other* than 90-degree angles and that should show you that a parallelogram is possible.

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Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

I don't under stand.

statement 1: it says the quad has length and width. so no matter what the shape of quad is, the perimeter will 2( L + W). right?
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Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

Given Quadrilateral- A closed figure with four sides- can be square, rectangle or parallelogram, rhombus etc.

Equal measure of angles can be a square/rectangle we don't know. So insufficient.

Statement 1.
L=100x, w=40x, but again it can be any rectangle, kite etc. we don't know.

1+2: Definitely a rectangle-> Find x and then find the area. (C)
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Originally posted by LeoN88 on 24 May 2019, 20:34.
Last edited by LeoN88 on 24 May 2019, 20:45, edited 1 time in total.
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pulak1988 wrote:
Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

I don't under stand.

statement 1: it says the quad has length and width. so no matter what the shape of quad is, the perimeter will 2( L + W). right?

pulak1988

True, if it's a parallelogram then the area would be= h*base
For rectangle, area= width*height; in different cases we are getting different areas.

Check GMAT Book page 69
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pulak1988 wrote:
Bunuel wrote:
A quadrilateral has a perimeter of 16. What is the area of the quadrilateral?

(1) The width is 40% of the length
(2) All angles are of equal measure

I don't under stand.

statement 1: it says the quad has length and width. so no matter what the shape of quad is, the perimeter will 2( L + W). right?

I guess you are assuming the shape as a rectangle and so assuming the formula as 2( L + W). From statement 1 its clearly not clear what is the shape
of the quadrilateral .(It can be rectangle/square/rhombus/trapezoid/or any other quad with all four sides different.)

Press Kudos if it helps!! Re: A quadrilateral has a perimeter of 16. What is the area of the quadril   [#permalink] 24 May 2019, 20:56
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