We often come across people wondering whether they should learn up the many formulas in permutation/combination, co-ordinate geometry etc. Our take on the question is a flat ‘No’. Formulas won’t take you far in GMAT, perhaps up to 600 but certainly not further. In fact, until and unless you have an eidetic memory or a Math PHD, chances are that knowing too many formulas will be a disadvantage. Let me show you why:

Say you see this question: What is the area of a triangle with vertices at (1, 4), (7, 1) and (4, 7)?

What is the first thing that comes to your mind? I am fairly certain that my fellow engineering grads will think of either Matrices or Heron’s formula. With Matrices, the confusion will be: was it (y2 – y1) or (y1 – y2)? Or was it –x2 or +x2? With Heron’s formula, the problem will be too many calculations. So we need to get the length of the sides first, then find s, then plug it all in the formula which was ummm… ?s(s-a)… or s?(s-a)… Hope you get my point. Until and unless you spend a fair bit of time everyday with all the formulas you intend to remember and the exact cases in which they can be used, it’s a waste of time and effort. In fact, if you are too attuned to the use of formulas, you will find it hard to think of a non-formula method to solve the problem. It will be hard for you to think beyond the formula and you will spend the two minutes you get in recalling the exact formula to be used in this particular situation, that is assuming there are formulas for most situations.

Now, try to think of a non-formula method and in this, I am sure the non Math background people will do better because they are used to figuring out innovative methods of solving problems. They do not come to the floor with pre-conceived notions on ‘methods to be used while solving particular question types’ and hence can keep their minds open. I can think of and have come across at least 3 different methods of solving this problem without using any exotic formulas. We just have to think in terms of right angles since we know how to find the area of figures with right angles. Let’s discuss each of those methods:

*Question: What is the area of a triangle with vertices at (1, 4), (7, 1) and (4, 7)?*

*(A) 9/2*

*(B) 9*

*(C) 27/2*

*(D) 18*

*(E) 27*

Solution: First of all, follow the golden rule of coordinate geometry – draw the triangle.

We don’t see any right triangles so let’s make some. We know how to find the area when we have right angles around.

**Method 1: Use Trapezoids**

Area of PQR = Area of APQB + Area of QBCR – Area of APRC

Area of PQR = (1/2)*3*(4 + 7) + (1/2)*3*(7 + 1) – (1/2)*6*(4+1) = 27/2

**Method 2: Use a Rectangle**

Area of PQR = Area of ABRC – Area of AQP – Area of BQR – Area of PCR

Area of PQR = 6*6 – (1/2)*3*3 – (1/2)*3*6 – (1/2)*3*6 = 36 – (1/2)*45 = 27/2

**Method 3: Use right triangles**

Area of PQR = Area of PAQ + Area of QAC + Area of PRC

There is one complication here – we don’t know the coordinates of point C. It is easy to figure out. Just find the equation of line QR and find the value of x when y = 4.

Equation of a line is given by: y – y1 = (y2 – y1)/(x2 – x1) * (x – x1)

Equation of QR: y – 1 = (7 – 1)/(4 – 7) * (x – 7)

2x + y = 15

When y = 4, x = 11/2. So C (11/2, 4)

Area of PQR = Area of PAQ + Area of QAC + Area of PRC

Area of PQR = (1/2)*3*3 + (1/2) *(11/2 – 4)*3 + (1/2) *(11/2 -1)*3 = 27/2

Answer (C)

I am sure you can come up with some other methods if you try!

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