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Geometry: In isosceles triangle ABC

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Source: — Data Sufficiency |
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by vinay1983 » Thu Sep 12, 2013 9:49 pm
Yaj wrote:In isosceles triangle ABC, what is the length of AB?

(1) AC < BC/2

(2) BC = 7

[spoiler]OA: C[/spoiler]

Thank You!
Statement 2 Insufficient

Statement 1 We do not know the value of any 1 side of the triangle. Atleast One side value is needed. Insufficient

C or E

BC = 3.5

So AB is less than 3.5 so AB can either be 3.5 or be the value equal to that of AC

The point here is the inequality triangle law. the third side should be greater than difference of the remaining 2 sides and lesser than the sum of the remaining 2 sides.

So Suppose AC = 3.4, then Ab can be 3.4 or 3.5 itself, since in a scalene triangle any 2 sides are equal. So these value will follow the above mentioned rule. Hence Sufficient.

I hope I am right.

I have a question though, do we need to find the exact value of AD?Can somebody generalize this?
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by [email protected] » Thu Sep 12, 2013 10:04 pm
Hi Yaj,

This DS questions asks us for side AB of an ISOSCELES triangle?

Fact 1: AC < BC/2
There's no numeric information here, so there's no way to answer the question.
Fact 1 is INSUFFICIENT

Fact 2: BC = 7
This doesn't tell us AB, but it does tell us something about the triangle. Since it's ISOCELES....

It's either 7 7 -
Or it's 7 - - in which the two blanks are the same
Fact 2 is INSUFFICIENT

Combined, we know that...
It's an ISOSCELES triangle
BC = 7
AC < 3.5
AB = ? but either matches AC or BC

Using the Triangle Inequality Theorem, we know that AB + AC MUST BE > 7

If AB = AC, then that WOULDN'T HAPPEN.

So, AB MUST = BC

Thus AB = 7

Final AnswerC

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by Java_85 » Fri Sep 13, 2013 9:57 am
Yaj wrote:In isosceles triangle ABC, what is the length of AB?

(1) AC < BC/2

(2) BC = 7

[spoiler]OA: C[/spoiler]

Thank You!
IMO it's E and there is no possible value for AB. Here is why:

(1) not sufficient
(2) not sufficient
(1)(2) AC=AB <3.5 and at the same time we have the constraint AB+AC<7 (Please notice it is not AB+AC<=7) --> putting this two constraint together there is no possible value for AB. --> Just E can be the answer. Please correct me if I'm wrong.
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by [email protected] » Fri Sep 13, 2013 12:48 pm
Hi Java_85,

Take a look at your work again.

Have you factored in:

1) That the triangle is ISOSCELES (we're told that in the prompt)
2) Every real triangle is based on the Triangle Inequality Theorem (so any 2 sides of the triangle that are added > 3rd side).

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by Java_85 » Fri Sep 13, 2013 5:01 pm
Thanks Rich.C, I got it now.
I was making a mistake and for no reason was assuming the two ISOSCELES are AB and AC.
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by lunarpower » Fri Sep 13, 2013 11:27 pm
When you see statement 1, make sure you think about it for a bit. It turns out to be insufficient here, but it's still significant, because it determines exactly which sides are the congruent ones.

If the problem says this instead:
Yaj wrote:In isosceles triangle ABC, is AB = BC?

(1) AC < BC/2

(2) BC = 7

Thank You!
... then the answer is (a). Make sure you know why.
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by shinys » Sat Sep 14, 2013 7:32 pm
Ron, should this be how we approach the question in the updated case?

Statement 1:
If AC = x, then BC is greater than 2x.
By Triangle property, third side, AB, has to be between
>2x - x = >x
and
>2x + x = >3x

Since triangle is isosceles, AB = BC because it is greater than AC.

SUFF

Statement 2:
It doesnt tell us anything about the other two sides. INSUFF
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by lunarpower » Sun Sep 15, 2013 4:02 am
That's one way to do it.

If you haven't memorized that rule for triangles, you can also just try drawing the triangle both ways.

I.e., try drawing a triangle with sides x, x, and something that's less than half of x -- this will work just fine (you'll get a long skinny triangle, where the 2 congruent sides are the long ones).
Now try drawing a triangle with sides x, something that's less than half of x, and the same something that's less than half of x (or, equivalently, sides of y, y, and more than 2y). It won't work -- you won't be able to get the two short sides to join each other. They're too short to bridge the long side.

I'm not really capable of memorizing abstract rules like the one above, so this is how I'd go about doing it.
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