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In the figure above, ABCD is a square, and the two diagonal [#permalink]

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03 Mar 2013, 01:36

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In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

a. 3√2 – 2√3 b. 3√2 – √6 c. √2 d. 3√2 / 2 e. 2√3 – √6

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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03 Mar 2013, 02:33

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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

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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03 Mar 2013, 03:01

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greatps24 wrote:

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

a. 3√2 – 2√3 b. 3√2 – √6 c. √2 d. 3√2 / 2 e. 2√3 – √6

Can you solve this in 2 min?

Area of 45-45-90 triangle = \(\frac{1}{2}leg^2\) = 3 (Area of square divided by 3)

So, leg = \(\sqrt{6}\)

So leg of the small triangle (the one formed in the upper left corner) = \(3-\sqrt{6}\)

Sides of a 45-45-90 triangle are in the ratio \(1:1:\sqrt{2}\)

So, hypotenuses of this small triangle = w = \(\sqrt{2}(3-\sqrt{6})\) = \(3\sqrt{2} - 2\sqrt{3}\)

Answer is A
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Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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03 Mar 2013, 03:04

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Impenetrable wrote:

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 = \(1:\sqrt{3}:2\)
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Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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05 Mar 2013, 21:21

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QUICK SOLUTION (UNDER 2 MIN):

If AB = 3, then the area of each region is equal to 3.

Consider the diagonal BC: \(BC = w + d = 3.\sqrt{2}\)

Where d is the diagonal of a smaller square, composed of the two corner regions. Now, these two regions combined have a total area of 3 + 3 = 6, and therefore: \(d = \sqrt{6}\sqrt{2} = 2.\sqrt{3}\)

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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09 Mar 2013, 22:06

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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}

caioguima wrote:

QUICK SOLUTION (UNDER 2 MIN):

If AB = 3, then the area of each region is equal to 3.

Consider the diagonal BC: \(BC = w + d = 3.\sqrt{2}\)

Where d is the diagonal of a smaller square, composed of the two corner regions. Now, these two regions combined have a total area of 3 + 3 = 6, and therefore: \(d = \sqrt{6}\sqrt{2} = 2.\sqrt{3}\)

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.

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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26 Mar 2013, 01:49

MacFauz wrote:

greatps24 wrote:

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

a. 3√2 – 2√3 b. 3√2 – √6 c. √2 d. 3√2 / 2 e. 2√3 – √6

Can you solve this in 2 min?

Area of 45-45-90 triangle = \(\frac{1}{2}leg^2\) = 3 (Area of square divided by 3)

So, leg = \(\sqrt{6}\)

So leg of the small triangle (the one formed in the upper left corner) = \(3-\sqrt{6}\)

Sides of a 45-45-90 triangle are in the ratio \(1:1:\sqrt{2}\)

So, hypotenuses of this small triangle = w = \(\sqrt{2}(3-\sqrt{6})\) = \(3\sqrt{2} - 2\sqrt{3}\)

Answer is A

Hi I have found your explanation easy for me.thanks for nice explanation. 1 kudos for you

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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04 Sep 2013, 01:49

i had the same approach as Impenetrable but couldn't come up with the right answer Could you help me please: what exactly i was doing wrong?

x-a side of a big triangle y -a side of a small triangle based on an area i can find x which is √6 y=3-x --> y=3-√6--> 2y^2=w^2 --> 2(3-√6)^2=w^2 --> 2(15-6√6)=w^2 --> 30-12√6=w^2 --> √(30-12√6)=w

and further i cannot see the proper solution which will bring me to the right answer

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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04 Sep 2013, 01:55

Galiya wrote:

i had the same approach as Impenetrable but couldn't come up with the right answer Could you help me please: what exactly i was doing wrong?

x-a side of a big triangle y -a side of a small triangle based on an area i can find x which is √6 y=3-x --> y=3-√6--> 2y^2=w^2 --> 2(3-√6)^2=w^2 --> 2(15-6√6)=w^2--> 30-12√6=w^2 --> √(30-12√6)=w

and further i cannot see the proper solution which will bring me to the right answer

Can you tell me why you do the highlighted step? It is unneccessarily time consuming to square the value and then find its root again..
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Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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04 Sep 2013, 02:22

Galiya wrote:

2(3-√6)^2=w^2

2(3^2-3*√6*2-√6^2)=2(9-6√6+6)=2(15-6√6)=30-12√6

Sorry for the typo... I meant to say "why"... As you can see the highlighted step would take at least 10-15 seconds to work out... You could directly take the root of the entire equation without expanding.. For what its worth.. The expression you have arrived at is the same as the right answer.. i.e You'll get the same value if you calculate both on a calculator...
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Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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06 Oct 2013, 09:27

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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!

Attachments

Solution.JPG [ 14.95 KiB | Viewed 27205 times ]

MGMATCAT.JPG [ 29.98 KiB | Viewed 27204 times ]

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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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12 Oct 2013, 14:49

MacFauz wrote:

greatps24 wrote:

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

a. 3√2 – 2√3 b. 3√2 – √6 c. √2 d. 3√2 / 2 e. 2√3 – √6

Can you solve this in 2 min?

Area of 45-45-90 triangle = \(\frac{1}{2}leg^2\) = 3 (Area of square divided by 3)

So, leg = \(\sqrt{6}\)

So leg of the small triangle (the one formed in the upper left corner) = \(3-\sqrt{6}\)

Sides of a 45-45-90 triangle are in the ratio \(1:1:\sqrt{2}\)

So, hypotenuses of this small triangle = w = \(\sqrt{2}(3-\sqrt{6})\) = \(3\sqrt{2} - 2\sqrt{3}\)

Answer is A

Hi- how do you know that the three regions of equal size?

Since the square is divided into 3 equal regions, the area of triangle DWY = 3. Thus: (1/2)(DY)(WD) = 3. Since triangle DWY is isosceles, DY=WD. Thus: (1/2)(WD)(WD) = 3 WD² = 6 WD = √6.

Since AD = 3 and WD = √6, AW = 3 - √6. Triangle AXW is a 45-45-90 triangle. (Don't worry about a proof. It should be clear from the figure that AXW is an isosceles right triangle.) Since the sides in triangle AXW are proportioned 1:1:√2, we get: WX = (3 - √6)√2 = 3√2 - √12 = 3√2 - 2√3.

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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09 Dec 2013, 05:28

Area of each region = 3 Diagonal of the triangle = 2H(say)+W = 3√2 Let the side of each of the small triangles be 'a'. Since it is an isosceles RAT, a^2=3; Therefore a=√6 = > Hypotenuse of each triangle = √12(1:1:√2) Now H = half of the Hypotenuse(property of a RAT) = √3

Re: In the figure above, ABCD is a square, and the two diagonal [#permalink]

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11 Dec 2013, 13:00

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

If the length of AB = 3 and ABCD is a square then the area of the square is 9. The area of each of the three divided regions is 3 which means the area of the two triangles is 3. These are 45:45:90 triangles, though we don't know the base or the height we know that they are the same so the area of this 45:45:90 triangle is A=1/2(b*b) 3 = 1/2 (b*b) 6 = b^2 √6=b

So the length of the sides of these triangles are √6. The length of the point from the end of the triangle down to C is the total length (3) - the length of the leg of the triangle (3-√6) We can find w by finding the hypotenuse of a smaller 45:45:90 triangle formed opposite angle C. we know the legs of this triangle are 3-√6 and with all 45:45:90 triangles, the lengths are in proportion x:x:x√2. W = 3√2-2√3

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