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Re: Area of triangular region [#permalink]
15 Dec 2013, 22:21

Expert's post

AccipiterQ wrote:

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

To figure out whether it holds, why don't you try drawing some extreme figures, say, something like this:

Attachment:

Ques3.jpg [ 3.67 KiB | Viewed 553 times ]

Will this be true in this case? When will it be true? When the triangle is equilateral, sure. Also when the triangle is isosceles if the equal sides form the angle from which the altitude is dropped.

Don't put your faith in the figure given. It may be just one of the many possibilities or may be somewhat misleading. _________________

Re: Area of triangular region [#permalink]
24 Jan 2014, 21:10

Hi Bunuel,

I have a question.

Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why.

Re: Area of triangular region [#permalink]
25 Jan 2014, 02:39

Expert's post

dasikasuneel wrote:

Hi Bunuel,

I have a question.

Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why.

Please help. Thanks Suneel

What do you mean by "imagine"? How does the length of one side define the angles? _________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
07 Feb 2014, 04:27

Expert's post

virendrasd wrote:

Bunuel wrote:

manimgoindowndown wrote:

Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected?

They are not equal: \(BD=\frac{6}{\sqrt{3}}\) (from the first statement) and \(AD=6*\sqrt{3}\) (from the second statement).

I think answer is E, since it is not given that BDC points are collinear.

That's not correct.

OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.

OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
18 Mar 2014, 07:51

Expert's post

sanjoo wrote:

what if question wud have been like this..

If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC?

Then statement 1 would have been irrelevant and statement 2 would have been incorrect. If side of an equilateral triangle is 12, the altitude would be \((\sqrt{3}/2)*12 = 6*\sqrt{3}\) But AD is given to be 6. _________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
12 Apr 2015, 14:31

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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
12 Apr 2015, 16:23

Main take away from this question is: Figures are not drawn to scale.

rest explanations are already given. _________________

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Re: If AD is 6 and ADC is a right angle, what is the area of
[#permalink]
12 Apr 2015, 16:23

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