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## Simple techniques to solve geometry questions on GRE- Part I

A Quantitative comparison question in Geometry involves variables that could be tricky at times. Let us look at one example-

1.

In the above figure, △ ABC is an Isosceles triangle. AB is the diameter of the circle with centre O and AB = d

Quantity A                                                                 Quantity B

Area of shaded region                                                   d / 2

A. Quantity A is greater.

B. Quantity B is greater.

C. The two quantities are equal.

D. The relationship cannot be determined from the information given.

Explanation:

In this question, we need to first find the area of the shaded region and then compare the quantities A and

B.We know that area of the triangle is:

In △ ABC, we can consider the base of the triangle AB which is same as diameter of circle.  Hence Base = d.

Now, the question states that the △ ABC is Isosceles, which implies that AC = BC. Join ‘O’ and ‘C’. Since the triangle is an Isosceles triangle, OC will be the height and the median of the triangle. Now OC is also the radius of the circle and hence OC = d/2.

Now the area of triangle becomes:

So now the question becomes –

Quantity A                                                Quantity B

A. Quantity A is greater.

B. Quantity B is greater.

C. The two quantities are equal.

D. The relationship cannot be determined from the information given.

When quantities have variables, it is always better to substitute easy numbers in the variables and compare the resultant quantities. So let us plug in a number and compare two quantities.

Let’s substitute d=2, we get both the values of quantities A and B as 1.

It is better to plug in some other number to confirm whether answer choice C is correct.

Now let us try a different number for d. Let’s take d = 1

we find that Quantity B is greater than Quantity A. This shows that the quantities are not equal. Hence answer choice C too gets eliminated and the correct answer is answer choice D!

One need not always remember the formulae for area of many figures in Geometry. Instead it is better to look for simple regular figures, which you are familiar with, to find the area.

2. In the diagram given below ABC is a right isosceles triangle. The lengths of the line segments AC and DE are x and y respectively. In terms of x and y, what is the area of shaded region?

The shaded region is a Trapezium. It is always a good idea to work with area of the triangle formula which is known to all. The area of the shaded region can be found by removing the smaller unshaded area of DBE from the bigger area of ABC.

Area of shaded region = area of ABC – area of DBE

Knowing this, one can easily eliminate options B and E as there is no subtraction happening there. Now we are left with answer choices A, C and D. Since ABC is isosceles, ∠ABC = ∠ ACB = 45 degrees. As DE is parallel to AC, ∠BED = ∠BCA = 45 degrees. This shows that DBE is an Isosceles triangle with height and base equal to y. We already know that ABC is an Isosceles triangle with height and base equal to x.

Now, using area of triangle formula: we can say that area of ABC is: and area of DBE is :,Therefore, area of the shaded region is

ALTERNATE WAY
This question can also be solved in another way. Let us construct a square ABGC of side x and a smaller square DBFE of side y as shown below.

Then the area of the shaded region = (area of the square ABGC – area of square DBFE)/2

(The total area has to be divided by 2 to avoid redundancy which can be seen in grey shaded region)

Now the side length of the square ABGC is x units and that of the square DBFE is y units. We also know that the area of square is:

square units. Therefore, the area of the shaded region =For geometry, we need to use concepts, formulae and techniques, such as process of elimination, to quickly arrive at the answer.

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