What's AB?

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What's AB?

by Juggernaut_86 » Mon Sep 12, 2011 5:27 pm
In a triangle ACB, if
a) angle ACB is a right angle
b) CE intersects AB
c) and CB = 5

what is the length of side AB?

(1) CE is perpendicular to AB

(2) AE = EB

OA is C But I am getting A

Could anyone please explain?

Thanks!

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by n@resh » Mon Sep 12, 2011 7:44 pm
Juggernaut_86 wrote:In a triangle ACB, if
a) angle ACB is a right angle
b) CE intersects AB
c) and CB = 5

what is the length of side AB?

(1) CE is perpendicular to AB

(2) AE = EB

OA is C But I am getting A

Could anyone please explain?

Thanks!
It's Answer is C!

Only by using two statements, you'll be able to calculate the Side AB for the triangle.

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by sl750 » Tue Sep 13, 2011 8:01 am
Statement 1 tells us that CE bisects the Hypotenuse, therefore BE=AE, we have two congruent triangles
As BC=5, we can use the 45:45:90 rule and find out the length of AB, which is, 5*sqrt(2). Sufficient

Statement 2 tells us the same thing. If BE=AE, that means CE is perpendicular to AB. Sufficient

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by gmatboost » Thu Sep 15, 2011 8:24 am
This is incorrect.

First, note that there is nothing that tells us that E is on the line segment AB. We just know that CE intersects AB. This doesn't actually affect the answer though.
(1) CE is perpendicular to AB
Even if E is on AB, this does NOT mean that AE = BE. This just means what it says, which is that CE and AB intersect at a right angle. If you draw a 5-12-13 right triangle, and draw a line CE that is perpendicular to the hypotenuse, you will see that this does not split AB into equal parts. Statement 1 is insufficient.
(2) AE = EB
Without known anything about CE, this fact alone tells us nothing about AB. It just tells us that E happens to be its midpoint.

Only when we combine the statements can we conclude that ACB is a 45-45-90 triangle, which allows us to find AB.
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by thestartupguy » Thu Sep 15, 2011 11:56 am
Yes, Answer: C

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by 1947 » Fri Sep 16, 2011 9:38 am
gmatboost wrote:This is incorrect.

First, note that there is nothing that tells us that E is on the line segment AB. We just know that CE intersects AB. This doesn't actually affect the answer though.
(1) CE is perpendicular to AB
Even if E is on AB, this does NOT mean that AE = BE. This just means what it says, which is that CE and AB intersect at a right angle. If you draw a 5-12-13 right triangle, and draw a line CE that is perpendicular to the hypotenuse, you will see that this does not split AB into equal parts. Statement 1 is insufficient.
(2) AE = EB
Without known anything about CE, this fact alone tells us nothing about AB. It just tells us that E happens to be its midpoint.

Only when we combine the statements can we conclude that ACB is a 45-45-90 triangle, which allows us to find AB.
Only when we combine the statements can we conclude that ACB is a 45-45-90 triangle, which allows us to find AB.
HOW CAN WE SAY that ACB IS 45-90-45 triangle ?? bcos we don know if EC=EB

please help
If my post helped you- let me know by pushing the thanks button. Thanks

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by gmatboost » Fri Sep 16, 2011 10:02 am
Assuming E is the midpoint of AB, consider triangles AEC and BEC:

AE = EB
Both triangles share side CE
Angles AEC and BEC are both right angles
That means that the hypotenuses of the triangles are equal: AC = CB

That is enough for us to know that ACB is 45-45-90

We could go a step further to answer your specific question:
Since ACB = 90
and angle ACE = angle BCE it must be the case that angle ACE = angle BCE = 45

This makes triangle ACE and BCE both 45-45-90 triangles.
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by saketk » Sun Sep 18, 2011 4:19 am
1947 wrote:
HOW CAN WE SAY that ACB IS 45-90-45 triangle ?? bcos we don know if EC=EB

please help
Hi- you can see the attached doc to clear your doubt.
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45-90-45.doc
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by 1947 » Sun Sep 18, 2011 4:59 am
Hi sakett,

You are correct in proving that but bcos of limited time available we need to understand that

since CE is common side and AEC=CEB=90 there can be no other way but AC=CB

Thanks again !!
If my post helped you- let me know by pushing the thanks button. Thanks

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by saketk » Sun Sep 18, 2011 5:58 am
1947 wrote:Hi sakett,

You are correct in proving that but bcos of limited time available we need to understand that

since CE is common side and AEC=CEB=90 there can be no other way but AC=CB

Thanks again !!
Well I was pressing the quote button and pressed Thank :)

Anyways, Dude I know this from my school days-- :).. I was trying to clear your doubt because you asked the question in your previous post..