Important Math Concepts for the New GRE

1:48 AM

A farmer has a 40-yard fence. The farmer wants to create a pen for his livestock. Which of the following shapes must the farmer arrange the fence into, if he wants to maximize the area of the pen?

a square
a rhombus
a circle
a rectangle
an equilateral triangle

The more advanced GRE-math questions can be tough going from the very get-go. You may scratch your head the way you would if trying to read a papyrus scroll, or you may disdainfully ask yourself, who cares? At the end of the day, neither approach, of course, will help your score (though not wanting to help the above farmer construct a pen is totally understandable). Below are some helpful tips when caught in such a quandary.

Always Move Forward

Many of my students are about to embark on the right solution to a problem but stop short and say….no, that’s probably not the right way. Remember, any approach is better than no approach, so unless an approach is ridiculously time-consuming, then you should always try to follow through on it.

Experimenting with Numbers – Look for a Pattern

One way is to play around with different numbers and scenarios, with the greater goal of looking for a pattern. Oftentimes, you may think that experimenting with numbers at random will take an eternity. Until, that is, you suddenly discover a pattern, and voila!, the answer is yours.

In the case above, we will need to work through each answer choice. So, let’s take our 40-yard fence and try to fashion it into a livestock pen. If we were to make it into a square (A), then we would get a 10×10 pen with an area of 100. Good, that’s a start.

Now, let’s try the other figures – a rhombus will always have a smaller area than a square with the same sides. And, the more the angles of the rhombus diverge, the less the area of the rhombus (think of a flattened rhombus).

Right now, the square is the largest. What about (C) a circle? To find the radius, you would take circumference and divide by

. That gives us 20/pi

as a radius. Solving for the area pi*r^2

, we get

, which equals about 130.

For (D) rectangle, you may notice a pattern – the more rectangular a shape becomes (the more the length and width diverge), the less the area. For instance a 6 x 4 rectangle is much larger, area-wise, than a 1 x 9 rectangle (note both rectangles conform to the original problem, which calls for a perimeter of 40-yards, the length of the fence). Therefore, a square must be larger. And, because a circle is larger than a square, the answer is still (C).

Next, we have an equilateral triangle. Each side is going to equal 40/3

. The formula, $(s^2*sqrt{3})/4$ gives us the area of the equilateral, where s is the length of each side. The math here is pretty foreboding, so you can ditch 40 and use your own number. Remember, the problem is only asking you to compare shapes—the actual quantity for the length is not the same, as long as it holds constant across shapes. So, I will use 12, a number that is divisible by both 3 and 4.

A square with such dimensions (3 x 3) will have an area of 9. An equilateral triangle, plugging in 4 for each side give us 4sqrt{3}

, which equals less than 9 (note that 4sqrt{4}=8

). Therefore, the square has a larger area, but still loses out to the circle. If the farmer wants to maximize the area of the pen, he should make it into a circle.

Final Takeaway

One last lesson: to maximize a fixed perimeter, always use a circle. And, the more sides you add to a figure, while keeping the perimeter fixed, the greater the area of that figure. Therefore, an octagon will have a greater area than a hexagon with an equal perimeter, which will have a greater area than a pentagon with an equal perimeter.