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Drawbridge - Geometry

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by shankar.ashwin » Mon Nov 21, 2011 9:03 am
The figure represents the design of a drawbridge that spans a channel 54 meters in width. If the two raisable sections are equal in length and overlap each other by a measure of 2 meters, what will be the clearance width of the bridge in meters, when the sections are raised to an angle of 60 degree ?

A) 25
B) 26
C) 27
D) 28
E) 54
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by vishal.pathak » Mon Nov 21, 2011 9:15 am
shankar.ashwin wrote:The figure represents the design of a drawbridge that spans a channel 54 meters in width. If the two raisable sections are equal in length and overlap each other by a measure of 2 meters, what will be the clearance width of the bridge in meters, when the sections are raised to an angle of 60 degree ?

A) 25
B) 26
C) 27
D) 28
E) 54
The overlap is 2m. Since the distance is 54 and the over lap is 2, we conclude that each raisable section = (54+2)/2 = 28m

Now the section is raised 60 degree
so the length under this section will be 28*cos(60) = 28*0.5 = 14. The save will be the length under 2nd section. So total length under them is 14*2 = 28
Clearence = 54-28 = 26 IMO
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by neelgandham » Mon Nov 21, 2011 9:23 am
From the diagram,
Let x + a be the length of each the raised sections, where a is the length of the overlapping section
Then 2x + a = 54 meters and a = 2 meters,
Implies x = 26 and x+a =28.
Cos 60 = c/(x+a)
0.5 = c /28
c = 14
Clearance width of the bridge,when the sections are raised to an angle of 60 degree = 54 - 2c = 54 -28 = 26

IMO B
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by rijul007 » Mon Nov 21, 2011 9:24 am
Length of each section = 54/2 + 1 = 28

check out the image



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26/54 = x/54
x = 26/54 * 54 = 26

Option B
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by shrey3192 » Wed Nov 12, 2014 9:31 pm
How come the non parallel sides 28?
How have we come to this..x+a(for hypotenuse)..

Also, why have we subtracted "2" once but not 2 times since its a part of both the sides as in..
why not..x+2+x+2=54(since 2 is a part of both the sides as can be seen from diagram)
Kindly explain..thanx.
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by [email protected] » Wed Nov 12, 2014 11:33 pm
Hi shrey3192,

You might find it helpful to draw the two "lines", then lay them on top of one another (as shown in the original drawing).

Here's the basic idea though….

If you had two pieces that were each 28m long and you laid them side by side, you'd have 56m total (28m + 28m = 56m)

However, the two pieces "overlap" 2m….
Lay down the first piece --> that's 28m long
Lay down the second piece, BUT the first 2m of the 2nd piece overlaps with the first piece, so only the remaining 26m creates additional length. 28m + 26m = 54m total.

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by Mathsbuddy » Fri Nov 14, 2014 6:42 am
If we consider 2 equilateral triangles (because all angles = 60 deg) overlapping by 2 metres, with their bases aligned, it is clear that the length of side is 54/2 + 1 = 28

If we pull these apart by exactly 2 metres, an inverted equilateral triangle will appear between them. All triangles in this case are congruent (all sideas and angles equal). Therefore it has side length 28 too.

When pushed back in a distance = 2, then the new gap will be 28 - 2 = 26

No need for trigonometry.
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by GMATinsight » Fri Nov 28, 2014 11:16 pm
shankar.ashwin wrote:The figure represents the design of a drawbridge that spans a channel 54 meters in width. If the two raisable sections are equal in length and overlap each other by a measure of 2 meters, what will be the clearance width of the bridge in meters, when the sections are raised to an angle of 60 degree ?

A) 25
B) 26
C) 27
D) 28
E) 54
Another Perception:

If we displace the two sections such that the overlap becomes zero then the total bottom separation of 54 wll become 54+2 = 56

Now join t he top clearance and the figure will be a composition of three Equilateral triangles with sides 56/2 = 28

Now the top clearance at this point will be 28 which has been increased by 2 as the separation has increased by 2 therefore actual top clearance with overlap = 28-2 = 26

Answer: Option B
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