for the explanation, I now understand where I went wrong...
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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