In the figure above, ABCD is a square, and the two diagonal lines divi

That's a tricky one, took me a while


AB=3, thus the area is 3*3 and therefore 9.

The three areas are of equal size, thus each area is 9/3

Now we can figure out the lengths of the sides of both triangles (right triangles):
Area of Triangle:
9/3 = (x*x)/2 =>
6 = x*x
√6 = x = each side of the triangle


Now we can calculate the distance between the point where the triangle ends and the actual corner of the square.
AB - x
3 - √6


This short piece of the side can be part of a new square triangle with w as the bottom line.
(AB - x)^2 + (AB - x)^2 = w^2
2(AB - x)^2 = w^2
√2 * (AB - x) = w
√2 * (3 - √6) = w
3√2 - √2√6 = w
3√2 - √(12) = w
3√2 - √(4*3) = w
3√2 - √4√3 = w
3√2 - 2√3 = w

Answer A


I am 99.9% sure that there is an easier way, but a stupid way is better than no way



Regards, Impenetrable

Actually, your method IS pretty quick.

It would help if you could remember that

Ratio of sides of a 45-45-90 triangle = \(1:1:\sqrt{2}\)

Ratio of sides of a 30-60-90 triangle = \(1:\sqrt{3}:2\)

diagonal of bigger sqr - diagonal of smaller sqr (joining 2 rgt angled triangles at the corners)
= 3\sqrt{2} - \sqrt{6}x\sqrt{2}
=3\sqrt{2}-2\sqrt{3}

A pretty tricky one, indeed.

I'll be posting a fairly lenghty and detailed reply so bear with me

First of all one should always bear in mind the following proportions :

1/ (45°-45°-90°) triangle => 1:1:\(\sqrt{2}\)
and
2/ (30°-60°-90°) triangle => 1:\(\sqrt{3}\):2

In our case, we'll be using proportion 1.

You should also know your square area formula. (This is a 700+ question and sometimes we take certain notions for granted, so be careful )

Now, the question states that the square is divided into 3 equal areas, so, since AB = 3 and the area of a square is x^2 (x being the side of said square), then the area of ABCD will be equal to 9. Therefore, each area will be equal to 3.

Let's find the length of the sides of the two isoceles triangles (they are isoceles since they both have two angles that have the same value, in this case 45°). Using proportion 1, both sides corresponding to the 45° angle will be equal to \(\sqrt{6}\). Why ?
Well since they are right isoceles triangles, then they're half-squares, in which case their areas will be equal to \(\frac{x^2}{2}\) . And since the area is equal to 3, then that gives us \(\frac{x^2}{2}\) = 3, so x^2 = 6, therefore x = \(\sqrt{6}\).

Substract 3 to get the length of the smaller segments to get 3 - \(\sqrt{6}\).

And if you notice on the right angle corners, the "w" we're looking for is actually the hypothenuse of the (45°-45°-90°) triangle whose sides are 3 - \(\sqrt{6}\) !

So apply the proportion 1 to get the hypothenuse length which is : \(\sqrt{2}\)*(3 - \(\sqrt{6}\)) = 3*\(\sqrt{2}\)-2*\(\sqrt{3}\), which is answer A.

Hope that helped and good luck with the rest.

I have understood MGMAT's solution for the problem. However, I am unable to reach the solution by the approach I have used.

Bunuel can advise on this?

I have attached my solution too.

Considering 45/45 we have isosceles triangle and hence the sides as mentioned.

Using Pytha theorem,

We can have width rectangle 'w' as

sqrt{ (3-x)^2 + (3-x)^2} = sqrt{2} (3-x)

And length as = sqrt {2} x

Since we know the three parts have same area then:

Area of lower triangle = 1/2* base * height = x^2/2

Area of rectangle = Length * breadth = 2x(3-x)

Equating both

x^2/2 = 2x(3-x)

we have x=12/5 and when substituting this for 'w'

w= sqrt{2} (3-x)

w=sqrt{2}3/5

And this is not an option .

Plz advise !!

Rgds,
TGC!

a = area of the portion

Just rearrange the figure and it becomes lot easier to solve this question .

Please check the figure for solution

Answer: option A

Rather than calculate, we can BALLPARK.



Diagonal of a square \(= s\sqrt{2}\)
Since square ABCD has a side of 3, diagonal BD \(= 3\sqrt{2} ≈ (3)(1.4) = 4.2\)
As shown in figure above, w ≈ 1/5 of diagonal BD:
\(\frac{1}{5}*4.2 = \frac{4.2}{5} = \frac{8.4}{10} = 0.84\)

The correct answer must yield a value close to 0.84.
Only A is viable:
\(3\sqrt{2} - 2\sqrt{3} ≈ 3(1.4) - 2(1.7) = 4.2 - 3.4 = 0.8\)

Let's first calculate the length of diagonal BD
Since BC = DC = 3, we can use the Pythagorean theorem to determine that BD = 3√2


We're told that "the two diagonal lines divide it into three regions of equal area."
The area of square ABCD = (3)(3) = 9
9/3 = 3, which means each region has area 3

So, if we let x = the length of each leg in the blue RIGHT triangle below...

...then it must be true that: (base)(height)/2 = 3
In other words, (x)(x)/2 = 3
Simplify: x² = 6
Solve: x = √6

We get the following:


Now let's examine the green RIGHT triangle below...

Since this is an isosceles triangle (with two 45 degree angles), we can let each leg have length y

Upon using the Pythagorean theorem to determine the value of y, we get: y = √3


Using the exact same logic, we can conclude that this other green triangle has the following lengths:


Since we already concluded that BD = 3√2, we can write the following equation: √3 + w + √3 = 3√2

Solve for w to get: w = 3√2 - 2√3

Answer: A

I used a less exact method to get to the solution but since the question asks us to approximate I think it is good enough, and more importantly saves a lot of time:

We can simply assume that the area in the middle is a rectangle where the hypotenuses of the triangles (above and below the area in the middle) constitute the sides of the rectangle with lengths 3√2.
Finally to approximate W we just need the formula for the area of a rectangle (which has an area of 3):

3√2 * W = 3

solving for W will already tell us that W must be below 1 and thus, A is our answer. What is also worth noting is that by ignoring the small isosceles triangles on each side of the rectangle we will err on the bigger side of W (since we will have to make up for the area we cut off by ignoring the small triangles). This gives us further assurance that A is our answer (since W is still below 1, even though we cut off the triangles).
Hope it is useful, cheers.

I answered this very quickly using no math:

It is not written that the drawing is not to scale. I measured the side AB with my GMAT white board, then divided into 3 equal parts, then compared that to length w. W is clearly slightly under a third of the side of the square, so under 1.

Answer A.

Definitely a back of the enveloppe method, but it works well with many GMAT questions to scale. Any way to get the right answer!

An alternative approach:

Area of all the three segments will be 3 units each.
An isosceles right-angled triangle, whose area is 3 gives us the following equation(1/2*base*height):
\(\frac{(x*x)}{2}=3\)
—> \( x =\sqrt{6}\)

For a right angle triangle, \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{h^2}\)
\(\frac{1}{x^2}+\frac{1}{x^2}=\frac{2}{6}=\frac{1}{3}\)
Since a=b =x (calculated above) for the triangle.

Now, \(\frac{1}{h^2}=\frac{1}{3}\)
This gives perpendicular length of the triangle as \(h=\sqrt{3}\)

Length of the height of two triangles = \(2*\sqrt{3}\)

Now, gap = diagonal - 2*height of the triangle

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.