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Simple techniques to solve geometry questions on GRE – Part II

Geometry is an integral part of GRE. You may encounter a Geometry question that contains seemingly meagre information and makes you seriously consider “cannot be determined” as your answer choice. However, this answer choice can be a trap and you might have all the data to solve the question. Let us look at one such question involving overlapping figures.

In the figure shown below, line AB is parallel to line DE, AB=CD, BC=DE and ∠ B = ∠ D =  90 degree
Find the degree measure of ∠AEC







A. 90 degree

B. 60 degree

C. 45 degree

D. 30 degree

E. cannot be determined

We find three triangles in the diagram. In △ABC and △CDE, AB = CD and BC = DE is given. ∠ B = ∠ D =   90 degree is also given in the question. Therefore the △ABC and △CDE are congruent (by SAS property). As the triangles are congruent we can say that AC = CE (sides opposite to corresponding equal angles are equal). Hence △ACE becomes an Isosceles triangle.

Since △ABC and △CDE are congruent triangles, the angles opposite to the equal sides will also be equal ( ∠ACB = ∠CED and ∠BAC = ∠ECD) as shown in the diagram below.







From the above figure it becomes obvious that in △ABC and △CDE , x+y = 90 degree (sum of angles of a triangle = 180)

We know that angles on a straight line add up to 180 degree. Therefore x + ∠ ACE + y = 90 degree.

⇒ ∠ACE = 90 degree

As △ACE is an Isosceles  triangle the degree measure of ∠AEC is 45 degree.

To solve Geometry problems we need to use concepts, formulae and strategies like approximation and process of elimination to arrive at the answer quickly.

4. PQRS is a square. PQ is tangent to circle with radius r and OM = MQ. Then what is the ratio of the area of circle to the area of square?

Capture (1)













Well, here we see three figures: circle , triangle and a square. On observation, we can see that the base of the triangle is also the side of the square and the height of triangle is the radius of circle. Now  we have been asked to find the ratio of the area of circle to the area of square. We very well know that the area of circle is πr2 and the area of square is side x side. It means that the ratio should have a π term in numerator. Thus answer choices D and E, which do not have the π in numerator, get eliminated. After eliminating D and E we are now left with answer choices A, B and C.
Now, the question states OM = MQ which means OQ is twice the radius. Since PQ is tangent to the circle, we can say that ∠OPQ =  90 degree as radius is perpendicular to the tangent.
Now let us apply Pythagoras theorem on the △POQ.






Thus  PQ, which is also the side of square, is equal to  r√3.  The area of square becomes:

Capture5                                                                                                                                                                          Hence the ratio of the area of circle to square is:




Therefore the correct answer is B .


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