A square ABCD has a side length 10 and E is the midpoint of AD. The sq

Question

https://gmatclub.com/forum/download/file.php?id=58293 A square ABCD has a side length 10 and E is the midpoint of AD. The square is folded from the vertex A along line segment BE, such that A moves to the point F, as shown above. What is the distance between F and the line segment DC? A. 1 B. 2 C. 3 D. 4 E. 5 Are You Up For the Challenge: 700 Level Questions Untitled.png

chetan2u · Accepted Answer

Bunuel , great question and must be your own.

the distance between F and the line segment DC MEANS the perpendicular from base to point F.

Draw a parallel line to base passing through F, let it be GH.

Now take the  	riangle EGF  and  	riangle BFH 
 \angle GFH = 180 , and  \angle EFB = 90 , so if we take  \angle GFE = x , then  \angle BFH = 90-x 
But then  \angle GEF  is also 90-x, so both triangles  	riangle EGF  and  	riangle FHB  are similar triangles.

The ratio of the sides can be known from the hypotenuse ...EF:FB=5:10=1:2
Now, let EG=a and GF=b, so FH=2a and HB=2b.

We can draw further conclusions as.. 
GH=10=2a+b....(i)
Also our answer=GD=HC=5-a=10-2b.......2b-a=5....(ii)
 Multiply (ii) by 2 and add to (i).. 
4b-2a=10.......4b-2a+2a+b=10+10.........5b=20....b=4

Thus our answer = HC=10-2b=10-2*4=10-8=2

B

Bunuel · Answer

Anyone want to try this tough question?

IanStewart · Answer

If you know some coordinate geometry that goes way beyond the scope of the GMAT, there are a few ways to do the problem. For example, there's a formula you could know that tells you where a point ends up when you reflect it through a line. So if we put point D at the origin, the line EB has equation y = 0.5x + 5. We just want to know where the point A, which is (0, 10), ends up after we reflect it through y = 0.5x + 5. Using the formula, which is totally unnecessary for GMAT purposes and is quite complicated, so I won't reprint it here, you find the reflected point is at (4, 2), and since the y-coordinate of that point is the perpendicular distance from the x-axis (which is line DC), 2 is the answer. You can also just find the slope of the reflected line EF, which is simpler, but still nothing you'd ever need to do on the GMAT.

Another way to do the problem is by noticing that the area of triangle AEB is 1/4 of the area of the square, and the area of EFB is identically 1/4 of the area of the square. So the remaining area below those two triangles is 1/2 the area of the square, so is 50. If we draw a horizontal line through point F, we divide that half of the area into three simple shapes - a thin rectangle at the bottom, and two right triangles. It then becomes fairly easy to test answer choices, though I'd use an estimate first to rule out the implausible candidates.

Or if we say point F has coordinates (a, b) (taking D as the origin), and you draw a horizontal line through F, you can notice that you get two similar right triangles (one with hypotenuse EF, one with hypotenuse FB), and using similarity, you find 10 - b = 2a. You can then use Pythagoras on either of the small triangles (we know EF is 5 and FB is 10, so we only have one unknown) to finish the problem.

I'd imagine there's a simpler way than any of those I list above to do the problem, a way I'm not seeing at a quick glance.

Fdambro294 · Answer

I got to the answer via a method that is probably way too long for the GMAT.

The First thing I spot is that the Answer Choices are all nice Integers. There are no Roots or Fractions or whacky numbers.

2nd, I see the Line EB creates a Triangle with a Side-Length of 5. For some reason what pops into my head every time I see 5 with Triangles is the possibility of 2 Right Triangles:   ----or----

(Step 1) Draw the Square before it is Folded.

EA = 5

AB = 10

Now when we Fold the Corner over:

EF becomes the Side = 5

FB becomes the Side = 10

(Step 2) As Ian did above, draw a Line through Point F that is Parallel to the Bottom Side DC. Call this Line XY.

We now have created a Rectangle on the Bottom of the Square that goes through Point F called: Rectangle DXYC

The Vertical Distance from Point F to Line DC will now be the Length of the Either of the OPP. Sides of the New Rectangle we created: XD -or- YC

Just to make it clear:

EX + XD = ED

BY + YC = BC

Above the Rectangle there are 2 Right Triangles: Right Triangle EXF -and- Right Triangle BYF

Triangle EXF has a Hypotenuse = 5 = Side EF

Triangle BYF has a Hypotenuse = 10 = Side BF

Once we fill in an Answer Choice for XD = YC = Opposite Sides of Rectangle we created

We can then Subtract that Distance from ED = 5 and BC = 10 and Find the Legs of Each of those Right Triangles.

Using Pythagoras, we can find the Other Leg of Each Triangle.

Finally, the 2 OTHER Legs of Each Triangle should SUM to the Opposite Side - DC - of our New Rectangle DXYC

(Step 3) Eliminate 2 Answers from our Testing Pool because they "Visually" seem too far

(E) 5
and
(D) 4

just by imagining and Visualizing the Figure, these Distances seem way to high. To be safe, start with (C) 3

(Step 4) Testing (C) 3

this means XD = YC = 3 = Perpendicular Distance from Point F to Side DC

Starting with Right Triangle EXF:

Since XD = 3:

EX would = 5 - 3 = 2

Using Pythagoras to Find the Other Leg XF

(2)^2 + (XF)^2 = (5)^2

XF = sqrt(21)

Now looking at Right Triangle BYF:

this means YC = 3

which means BY = 10 - 3 = 7

Using Pythagoras to Find the Other Leg YF:

(7)^2 + (YF)^2 = (10)^2

YF = sqrt(51)

Now it must be true, because of the Rectangle we created:

XF + YF = Side DC = 10

sqrt(21) + sqrt(51) does NOT equal 10 ---- thus (C) 3 is ELIMINATED

(Lastly)
when testing the answer choice, you can see that Both Hypotenuses of the 2 Right Triangles are Multiples of 5.

EF = 5(1) = 5

BF = 5(2) = 10

Starting with Right Triangle BYF:

if we test the 2 Right Triangles as 3-4-5 Right Triangles:

BY could equal = 4(2) = 8

or

BY could equal = 3(2) = 6

If BY = 6, this means the Distance YC = 10 - 6 = 4 ----- which as I stated above Visually seems too far of a Distance from Point F to Side DC

Let BY = 8

We would then have a 3-4-5 Right Triangle and ---

the Length of OTHER Leg YF = 6****

Now, Looking at Right Triangle EXF:

Since YC = 2 ------> Opposite Side of our Created Rectangle XD must also = 2

which means EX = 5 - 2 = 3

Since Hypotenuse EF = 5:

the OTHER Leg XF = 4*****

YF + XF = Side XY of our Created Rectangle that is Opposite SIDE DC = 10

(YF = 6) + (XF = 4) = 10 = Opposite Side DC ------- WORKS!!!!!

the Numbers work, and the Vertical Distance from Point F to Side DC must = 2

Correct Answer -B-

I Apologize if this was waaayyyy too Long. I just tried to go through my thought process step by step. If the Answer Choices were not "nice easy Integers", then I would have not been able to do this.

probably took between 3-4 minutes when quickly testing the Answers and Numbers

chetan2u · Answer

Such questions almost always require something to be drawn in a way to simplify the question. 
Here too, it becomes pretty simple, if you draw a line somewhere. Will post a solution in one hour to give time to someone who wants to try on those lines. There can be many possible answers and I may have just got one of them.

GowriPrakash · Answer

IMO B.

After drawing a line passing through F (PQ in the file attached), it gets pretty simple. We need to find the perpendicular distance between PQ and DC

Let me know if the image is clear

chetan2u · Answer

On the correct lines, but you cannot say a triangle to be 3-4-5 only because the hypotenuse is 5.

GowriPrakash · Answer

Yes, I do not have any confirmation that it's a 3-4-5 triangle. But if I assume that, then I can verify that QC = BC - BQ = 10 - 8 = 2. Also fits the 6-8-10 triangle. And since PQ is a straight line and PD = QC = 2 confirms that.

Let me know where my reasoning is going wrong here.

Is there any other way that we can indeed tell that its a 3-4-5 triangle ?

AntrikshR · Answer

EF =5,BF=10
Draw a perpendicular from F to DC and call it 'h' and let's say that perpendicular is touching DC at point K.
Let DK = l and CK = 10-l

=> We have a trapazium EFKD,a trapazium BFKC and two overlapping triangles EFB*2..they all together makes a square of 100 square unit.

Putting it all together,
Area of triangle is 25 but we have to multiply it with 2 as there are two overlapping triangles. So area of 2 triangles will be 50 and the areas of two quadrilaterals will be 1/2(5+h)l and 1/2(10+h)(10-l) respectively.

=> 50+1/2(5+h)l +1/2(10+h)(10-l) = 100
Solve it we will get 
 l = 2h------1)
Now, let's draw a perpendicular to ED from F. Let's say it's touching ED at G.
Now,length of GF will be nothing but l.

Using pythagoras theorem in Triangle EFG, we will get:
25 = (5-h)^2+ l^2------2)
Using 1 and 2,
h = 2, l= 4

B IMO

Posted from my mobile device

AntrikshR · Answer

How can you assume the sides of the triangles?...Why can't the side be in decimals?..Why integer only?

Posted from my mobile device

Fdambro294 · Answer

Following  IanStewart

Using Coordinate Geometry to solve this question is doable so long as you understand the reasoning behind Reflected Points and the Mirror Line.

Setting up the Square in the Coordinate Plane as Ian said to do:
A - (0 , 10)
E - (0 , 5)
D - (0 , 0)
C - (10 , 0)
B - (10 , 10)

Let Point F be ----> (X , Y)

if we can find the Y Coordinate for Point F, we will know the Perpendicular Distance to the X-Axis, which the Side CD lies upon.

By folding over the Vertex A -----> to Point F, we are effectively taking Point A (0 , 10) and reflecting this Point over the Line that Passes through Point E and Point B

(1st) Find the Equation of the Mirror Line - the line over which we are Reflecting Point A

If set up correctly in the Coord. Plane, Line EB will pass through Point E (0 , 5) and Point B (10 , 10)

m = Slope = +1/2

b = Y Intercept = 5

Formula for Mirror Line: y = (1/2)x + 5

(2nd) There are 2 Concepts that can help make the calculation easier:

-1- If we draw in a Line Segment connecting Original Point A and its Reflected Image Point F, the Line given by y = (1/2)x + 5 will BISECT this segment.

In fact, every Point on the Mirror Line will be Equidistant from Points A and F.

-2- the Slope of our created Line Segment AF will have a Slope that is the Negative Reciprocal of +(1/2). This is because the Mirror Line will be Perpendicular to our created Line Segment AF. (this works out because Point A is on the Y Axis at (0 , 10))

(3rd) We can use the above 2 Concepts and create 2 Equations to solve for Point F (X , Y)

-1- Find the Mid-Point of Line Segment AF and then Plug this Mid-Point into the Mirror Line's Equation of: y = (1/2)x + 5

Mid-Point is: 
X Coordinate = (0 + X)/2 = X/2 
and 
Y Coordinate = (10 + Y)/2

Plugging the Mid-Point's Coordinates into the Mirror Line's Formula: y = (1/2)x + 5

(10 + Y)/2 = (1/2)(X/2) + 5

Solving this Equation ------> X = 2Y

-2- (Slope of Mirror Line) * (Slope of Line AF) = -1

Slope of Mirror Line = +(1/2)

Slope of Line AF (again Letting Point F have the Coordinates (X , Y) =

(10 - Y) / (0 - X) = (10 - Y) / (-X)

Plugging the Slopes into the Equation above:

*   = -1

----multiplying both sides by +2-----

(10 - Y) / (-X) = -2

10 - Y = 2X

From -1- we found that: X = 2Y -------> Substituting in for X

10 - Y = 2*(2Y)

10 - Y = 4Y
 
10 = 5Y

Y = 2 = the Y Coordinate of Point F in the Coordinate Plane

Since X = 2Y, we can also find the X Coordinate of Point F ------> X = 4

Thus, Point F will be at (4 , 2)

and this Point will be exactly 2 Units above the X-Axis, which shares Side CD in our Graphed Square.

Answer = 2

-B-

A square ABCD has a side length 10 and E is the midpoint of AD. The sq

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A square ABCD has a side length 10 and E is the midpoint of AD. The sq

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