Seesaw touching ground - interesting problem - The Beat The GMAT Forum

Seesaw touching ground - interesting problem

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sairamGmat: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Thu Feb 25, 2010 7:04 am; Thanked: 2 times

by sairamGmat » Fri Jul 16, 2010 7:01 am

. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C?
(1) x = 30
(2) Point B is 5 feet above the ground.

OA is B

Can any one explain why is it So? I thought it is C because of the typical pythogorous theorem calculations....

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Fri Jul 16, 2010 7:15 am

sairamGmat wrote:. In the figure above, line AC represents a seesaw that is touching level ground at point A. If B is the midpoint of AC, how far above the ground is point C?
(1) x = 30
(2) Point B is 5 feet above the ground.

OA is B

Can any one explain why is it So? I thought it is C because of the typical pythogorous theorem calculations....

When only angles are given, we cant know the length. SO A is out.

Now, Point B is 5 feet above the ground.
B is midpoint of line AC
Draw a perpendicular from B to base. let us say BD
Similarly draw one from C, call it CE
Now, BD and CE are parallel and B is the midpoint of the triangle.
By midpoint theorom, CE=2BD
hence CE = 10.

Suppose, you did not know midpoint theorem, then
You can see that Triangles ABD and ACE are similar triangles.
Hence, ratios of their sides should match.
So: AB/AC = BD/CE
we know AB/AC = 1/2
Hence we get CE =10.

Hope this helps!!

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sairamGmat: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Thu Feb 25, 2010 7:04 am; Thanked: 2 times

by sairamGmat » Fri Jul 16, 2010 7:25 am

Awesome...Thank you.....

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 17, 2010 8:09 am

I doubt the OA.
The explanation for midpoint theorem holds good when both sides are the midpoints,drawing a perpendicular does lead to similiar triangles but it would be impossible to determine that BD is half of CE as posted in the previous post.
answer should be C

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sat Jul 17, 2010 9:07 am

aloneontheedge wrote:I doubt the OA.
The explanation for midpoint theorem holds good when both sides are the midpoints,drawing a perpendicular does lead to similiar triangles but it would be impossible to determine that BD is half of CE as posted in the previous post.
answer should be C

There is a corollary to midpoint theorem:
If a line is drawn from the midpoint of the side of a triangle, parallel to its base, then it passes through the midpoint of the otherside.

Here both the perpendiculars(heights from base) are parallel.

Hope this helps!!

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 17, 2010 9:27 am

Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sat Jul 17, 2010 9:29 am

aloneontheedge wrote:Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.

Assume the triangles with both the perpendicualr lines only.

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 17, 2010 9:42 am

kvcpk wrote:
aloneontheedge wrote:Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.
Assume the triangles with both the perpendicualr lines only.

Lets take a simple triangle with base 3,hyp =5 and another side 4
according to you,If we draw a line from hyp perpendicular to side 3,it shud be equal to 2.
but that is not the case. assume degree = 30

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sat Jul 17, 2010 9:50 am

aloneontheedge wrote:
kvcpk wrote:
aloneontheedge wrote:Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.
Assume the triangles with both the perpendicualr lines only.
Lets take a simple triangle with base 3,hyp =5 and another side 4
according to you,If we draw a line from hyp perpendicular to side 3,it shud be equal to 2.
but that is not the case. assume degree = 30

according to me if we draw a line from midpoint of hypotenuse parallel to the base(4), then the length of that line will be 2 and it will pass through the midpoint of the other side (3).

I am quite sure about this rule.

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aloneontheedge: Senior | Next Rank: 100 Posts; Posts: 97; Joined: Thu Jan 28, 2010 2:07 am; Thanked: 9 times

by aloneontheedge » Sat Jul 17, 2010 9:52 am

kvcpk wrote:
aloneontheedge wrote:
kvcpk wrote:
aloneontheedge wrote:Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.
Assume the triangles with both the perpendicualr lines only.
Lets take a simple triangle with base 3,hyp =5 and another side 4
according to you,If we draw a line from hyp perpendicular to side 3,it shud be equal to 2.
but that is not the case. assume degree = 30
according to me if we draw a line from midpoint of hypotenuse parallel to the base(4), then the length of that line will be 2 and it will pass through the midpoint of the other side (3).

I am quite sure about this rule.

unfortunately, I do not know how to add the diagram else i would have shown it is not the case.

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kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sat Jul 17, 2010 9:55 am

aloneontheedge wrote:
kvcpk wrote:
aloneontheedge wrote:
kvcpk wrote:
aloneontheedge wrote:Exactly...but here we are drawing a perpendicular line and it wil not be half of the parallel side.
Assume the triangles with both the perpendicualr lines only.
Lets take a simple triangle with base 3,hyp =5 and another side 4
according to you,If we draw a line from hyp perpendicular to side 3,it shud be equal to 2.
but that is not the case. assume degree = 30
according to me if we draw a line from midpoint of hypotenuse parallel to the base(4), then the length of that line will be 2 and it will pass through the midpoint of the other side (3).

I am quite sure about this rule.
unfortunately, I do not know how to add the diagram else i would have shown it is not the case.

Read the Converse of Midpoint Theorem in this link:
https://www.tutorvista.com/content/math/ ... heorem.php

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 amFollowed by: 98 members; Thanked: 335 times

by Patrick_GMATFix » Sat Jul 17, 2010 10:44 am

aloneontheedge wrote:Lets take a simple triangle with base 3,hyp =5 and another side 4
according to you,If we draw a line from hyp perpendicular to side 3,it shud be equal to 2.
but that is not the case. assume degree = 30

If the line is drawn from the midpoint of the hypotenuse and straight down (perpendicular to the base of 3 and parallel to the height of 4), then the line drawn will be a height of 2 for sure. It will cut the base at its midpoint. kvcpk is correct on this one.

Another way to look at it (which I think kvcpk also mentioned) is as similar triangles. Similar triangles are triangles that have the same angle measures. In the example you gave, two similar triangles would be created since both triangles would have a 90 degree angle and they would also share the angle where the hypotenuse meets the base.

The 2nd property of similar triangles (besides that they have the same angle measures) is that their lengths are proportional. Because in our imaginary drawing, the hypotenuse of the newly formed triangle would be half the hypotenuse of the initial triangle, you can be certain that each side of the newly formed triangle would be half of the corresponding side in the initial triangle.

-Patrick

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