In quadrilateral ABCD, sides AB and BC each have length √2,

if we know four sides of the quadrilateral we can find its area...

s=a+b+c+d/4

area =sqrt of s*(s-a)*(s-b)*(s-c)*(s-d) ....

so answer A ...

Am i missing something.....

i guess with both, statements 1 and 2 together, you are able to determine the area of the given quadrilateral.

please refer to the image attached..



you can see that once you have the 2nd condition, you can find the diagonal AC of the quad.
and when you calculate the length of the diagonal, you end up getting an equilateral triangle ADC on the other side.
So, 1 you have a right triangle and 2 you have an equilateral triangle, the areas of both of which can be calculated.

Thus, I would go with option C.

To the best of my knowledge, on the GMAT only convex polygons are considered, except maybe the cases when a drawing is supplied.
This is just another example when Manhattan is going overboard.

I'm going with C here. Please provide OA ok?

Here's my reasoning

Statement 1 we have that AD which is the diagonal of the right triangle formed in the polygon is 2 therefore knowing two sides and one angle is sufficient to know the measures of the right triangle but thing is we don't know about BC is this side perpendicular to AB or not? Insufficient

Statement 2 only tells us that the angle in ABC is 90 degrees but still no way to find the area of the other right triangle.

Now with both statements together we know the area of the right triangle and we know that AB is perpendicular to BC hence the area of the given square is 2 which added to the area of the right triangle can provide the total area of the figure

Answer is thus C

Please let me know if this is sound reasoning
Cheers
J

Hi Bunnel,

When I use st1 and st2 I get one 45-45-90 triangle. and another one an equilateral triangle with site 2. I can get area of quadrilateral by adding both the areas. I choose C but official ans is E. Please clarify.

Thanks

Not sure how to clarify better than the diagram:


As you can see both those figure satisfy all the conditions and have different areas. Thus the answer must be E.

Hope it helps.

Thanks for clarification. But i think second figure is not a convex polygon(because interior ABC angle is 270 and not less than 180). And in GMAT I guess we are concerned about convex polygon only?
And if we agree that second figure is valid, then can we say ans is C? Also i feel is it really possible to draw such diagram with precision in exam ? Any alternative way to build the thought process for exam?

Thanks.
Manoj Parashar

Yes, OG defines a quadrilateral as a polygon with four sides. Next, it says that the term "polygon" will be used to mean a convex polygon, that is, a polygon in which each interior angle has a measure of less than 180°.

Hence this question violates GMAT definition of a quadrilateral, which makes the question/solution presented by MGMAT flawed.

I wouldn't worry about this question at all.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

In quadrilateral ABCD, sides AB and BC each have length √2, while side CD has length 2. What is the area of quadrilateral ABCD?

(1) The length of side AD is 2.
(2) The angle between side AB and side BC is 90°.

In the original condition, there are 5 variables (4 sides and 1 diagonal line) and 3 equations(sides AB and BC each have length √2, while side CD has length 2) and thus we need 2 more equations to match the number of variables and equations. Since there is 1 equation each in 1) and 2), the best answer is C by our DS definition. However, this type of question always shows the diagram in actual exam. If diagram is not shown, E is the answer because we don't know the alphabetical orders of vertices (ABCD? ACBD? BDAC?)

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

For these types of questions which involve ambiguity regarding the shape, does it ultimately matter what shape you draw for the different possibilities? e.g. I was able to conclude E, but didn't draw the arrow shape that you have, but somehing else.

As long as the figure you draw agrees with info given in the question then it does not matter what shape you draw.