A closed box with a square base is to be wrapped with a square sheet

Probably one of the first things that comes to our mind when we look at this question – ‘Whoa, really? Am I supposed to solve this question in 2 minutes?’
Well, there’s no other option, really!

In reality though, this is a question that looks difficult but, IS NOT difficult. You only have to be able to break it down into smaller geometrical entities for which you know the formulas for area calculation.

When you look at it in detail, you need to know the following formulas:

Area of a rectangle = length * breadth
Area of a right angled triangle = ½ * (product of perpendicular sides)
Length of a diagonal of a square in terms of its side = √2 * side

What is actually difficult though, is to visualize how the wrapping paper will wrap around the box. This is where your visual reasoning skills are tested and this is also where GMAT tells you that it is not testing you on your knowledge of Geometry alone. It’s actually not that hard if you follow the dotted lines.

To help you understand how the wrapping paper will get wrapped eventually, we have made some constructions in the diagram below:



We have marked some points with alphabets. We have connected point F to point D. You can now see that rectangle DFGH corresponds to one of the lateral (bounding) surfaces of the cube. The area of this rectangle is w * h.

There will be 4 such bounding surfaces, so the total area of these bounding surfaces will be 4wh.

Triangle DEF, when folded will cover the triangle HKG. HK represents half the diagonal of the square.

The diagonal of the square = w√2. Therefore, HK = ½ * w√2 = \(\frac{w}{√2}\)
.
The diagonals of a square bisect each other at right angles. Hence, KG will also be \(\frac{w}{√2}\). Also, HKG will be a right angled triangle.
So, area of triangle DEF = area of triangle HKG = ½ * \(\frac{w}{√2}\) * \(\frac{w}{√2}\). = \(\frac{w^2}{4}\).

We have 8 such triangles (4 at the top and 4 at the bottom). Therefore, the total area of these bounding surfaces will be 2*\(w^2\).

Triangle FGI is a right angled triangle, which represents the part of the wrapping paper, which will remain unused.

Area of triangle FGI = ½ * h * h = \(\frac{h^2}{2}\). There are 4 such triangles, so the total area of these triangles will be 2*\(h^2\).

Hence, the total area of the wrapping paper = 4wh + 2 * \(w^2\) + 2 * \(h^2\), which simplifies to 2\((w + h)^2\).

An ideal time to solve this question could be anywhere between 2 to 3 minutes. However, if you are stuck with this question beyond 3 minutes, it’s a good idea to eliminate answer options like B, D and E, based on the area of the rectangles (4wh), because none of these answer options give 4wh. Beyond this stage, option C can be eliminated because it does not take care of the unused portion of the wrapping paper.

So, even when you are stuck, you have a way out to zero in on the right answer, if you have practiced eliminating options based on simple mathematical logic.

I hope this helped!
Thanks