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Geometry - Drawing Extreme Diagrams

Let’s continue with geometry today. We would like to discuss how drawing extreme diagrams can help you solve questions. Most GMAT questions are quite intuitive and hence our non-traditional methods are perfect for them. They are not typical MATH problems per se; instead, they are logical puzzles. If you can prove why some things will not work, it means whatever is left will work.


Let me explain with the help of an official Data Sufficiency question.



In the figure above, is the area of the triangle ABC equal to the area of the triangle ADB?

Statement 1: (AC)^2=2(AD)^2

Statement 2: Triangle ABC is isosceles.



When presented with this question, people see right triangles and jump to Pythagorean theorem, isosceles triangles and then wage a war on AC, AB, CB and AD relations. Well, that is our traditional approach. But what do we do if making equations and solving for relations isn’t our style?


We make diagrams and figure out the relations! One thing that is apparent the moment we read statement 1 is that the figure is not to scale. From the figure it looks as if AD is greater than or at best, equal to AC. That itself is an indication that if you draw the figure on your own, you could see something that will make this question very simple. The question setter doesn’t want to show you that and hence he made the distorted figure.

Anyway, let’s first analyze the question. Then we will look at the statements.

We need to compare areas of ABC and ADB. Both are right angled triangles.

Area of ABC = (1/2) * AC * BC

Area of ADB = (1/2) * AD * AB

We need to figure out whether these two are the same.


Think about it this way – we are given a triangle ABC with a particular area. So the length of AD must be defined. If AD is very small, (shown by the dotted lines in the diagram given below) the area of ADB will be very close to 0. If AD is very large, the area will be much larger than the area of ABC. So for only one value of AD, the area of DAB will be equal to the area of ABC.

We need to figure out whether for the given relations, the triangles have equal area.


Statement 1: (AC)^2=2(AD)^2
This gives us AD = AC/Sqrt(2). Let’s draw AC and AD such that AD is somewhat shorter than AC. Now can we say that the areas of the two triangles are the same? No. The area of ABC is decided by AC and BC both not just AC. We can vary the length of BC to see that the relation between AC and AD is not enough to say whether the areas will be the same (see the diagrams given below).

So this statement alone is not sufficient.


(2) Triangle ABC is isosceles.
This means that AC = BC. Notice that the triangle is right angled so the hypotenuse must be the largest side. If ABC is isosceles, it means that the two legs of the triangle must be equal. Hence sides of ABC must be in the ratio 1:1:Sqrt(2) = AC:BC:AB. Since we only need to consider relative length of the sides, let’s say that AC = 1, BC = 1 and AB = Sqrt(2) or some multiple thereof.

We have no idea about the length of AD so this statement alone is also not sufficient.

Let’s consider both statements together now:

AD = AC/sqrt(2) = 1/sqrt(2) (Since AC = 1)

Area of ABC = (1/2) * AC * BC = (1/2) * 1 * 1 = 1/2

Area of ADB = (1/2) * AD * AB = (1/2) *  (1/sqrt(2) ) * sqrt(2) = 1/2

Both triangles have the same area. Sufficient!


Answer (C)


Now compare this approach with your Pythagorean approach. Is this simpler?


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