MGMAT Geometry Triangle

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

by nadontheway » Tue Jun 24, 2008 4:02 am

Why is A not sufficient??

From A we know that ACD is isoceles right triangle with hypothenuse AD.
Angle CAD is 45° and angle ACD is 90° so CDA must be 45°.
CD should be 5/Sqroot2.
And BD should be 2*5/Sqroot2
Why am i wrong?

Although the statement A doesn't say isoceles, you can deduce it thanks to the statement itself.
The perpendicular cuts the right triangle into two triangles creating a right angle (ACD) and splitting the right triangle ABD in two (45°). So you can deduce that angle ADC is 45°.
So A should be sufficient. Am I clear?

Thx

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Source: Beat The GMAT — Data Sufficiency | Tue Jun 24, 2008 4:02 am

codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

Re: MGMAT Geometry Triangle

by codesnooker » Tue Jun 24, 2008 4:24 am

nadontheway wrote:Why is A not sufficient??

From A we know that ACD is isoceles right triangle with hypothenuse AD.
Angle CAD is 45° and angle ACD is 90° so CDA must be 45°.
CD should be 5/Sqroot2.
And BD should be 2*5/Sqroot2
Why am i wrong?

Although the statement A doesn't say isoceles, you can deduce it thanks to the statement itself.
The perpendicular cuts the right triangle into two triangles creating a right angle (ACD) and splitting the right triangle ABD in two (45°). So you can deduce that angle ADC is 45°.
So A should be sufficient. Am I clear?

Thx

There is no such theorem. You are trying to predict the angle according to the given figure. See I can read your mind.

The figure is totally wrong. Get a rule and a D, try to make a figure having angles B (30 degree), D (60 degree) and A (90 degree).

Then try to draw the perpendicular from A t on BD. You will get your answer.

PS: You are trying to apply bisect theorem which is not fit over here.

Let me know if you still have any confusion.

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

MGMAT Geometry Triangle

by nadontheway » Wed Jun 25, 2008 11:47 pm

Thanks a lot codesnooker. I really appreciate.

My confusion is based on the definition of Perpendicular lines itself: "Perpendicular lines form right angles". Based on that information, I concluded that angle C must be 90°.

What do you think?

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

Re: MGMAT Geometry Triangle

by codesnooker » Thu Jun 26, 2008 12:00 am

nadontheway wrote:Thanks a lot codesnooker. I really appreciate.

My confusion is based on the definition of Perpendicular lines itself: "Perpendicular lines form right angles". Based on that information, I concluded that angle C must be 90°.

What do you think?

Did you try to sketch the figure that I mentioned in my earlier post on this topic.

The angle C will be equal to 90 degree only when A and D are equal to 45 degree. However that is not necessary at all. The relation between A and D is

A + D = 90 degree

So A and D could be anything, however their sum should be always equal to 90 degree.

Sketch the figure and you can derive it yourself.

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

by nadontheway » Thu Jun 26, 2008 12:17 am

I sketch the figure you mentioned in your earlier post only conceptually (I don't have a D;)).
I see that A+D= 90°.
With the perpendicular AC, A can be 45° or 30° and D can be 45° or 60° or anything else as long as A+D=90°. So at this stage we don't know if ACD is isoceles triangles or equilateral triangle or someting else. I can see that now. So A is not sufficient to determine BD.

But because perpendicular lines form right angles C must be 90° whatever angle A and D measure. Is that correct?

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Thu Jun 26, 2008 1:09 am

nadontheway wrote: But because perpendicular lines form right angles C must be 90° whatever angle A and D measure. Is that correct?

Of course, that's what PERPENDICULAR termed for.

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

by nadontheway » Thu Jun 26, 2008 1:12 am

THANKS I see where my mistake is now.

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

by nadontheway » Thu Jun 26, 2008 2:20 am

I can't believe how easy this problem is. I really need to focus and handle geometry concepts...

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anjaligeorge1: Senior | Next Rank: 100 Posts; Posts: 66; Joined: Wed Feb 20, 2008 6:23 pm; Thanked: 2 times

by anjaligeorge1 » Thu Jun 26, 2008 6:08 am

Is the OA E ?

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nadontheway: Senior | Next Rank: 100 Posts; Posts: 88; Joined: Fri Nov 16, 2007 2:05 am

by nadontheway » Thu Jun 26, 2008 6:16 am

The correct answer is C

Using statements 1 and 2, we know that AC is the perpendicular bisector of BD. This means that triangle BAD is an isosceles triangle so side AB must have a length of 5 (the same length as side AD).
We also know that angle BAD is a right angle, so side BD is the hypotenuse of right isosceles triangle BAD.
If each leg of the triangle is 5, the hypotenuse (using the Pythagorean theorem) must be 5sqrt2.

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