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Re: Area of triangular region [#permalink]
17 Sep 2010, 04:22

I am not 100% sure of my answer but I think it's correct:

(1) it tells us that we have a triangle 90-60-30 so now, we are able to place the points on the triangle (where is A, where is D and where is C). So we know that A is the right angle, D the 60 degree angle and C the 30 degree angle.

That is not sufficient though...

(2) AD = 6, AC = 12 but which side is the hypothenuse ?????? we can't compute the area ! Insufficient !

(1) and (2)

We know that AC is the hypothenuse, AD the height and AC the basis.

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

If AD is 6 and ADC is a right angle, what is the area of [#permalink]
15 Apr 2012, 14:11

with statement 1 we can only figure out that the left triangle is a 30-60-90. We cannot just assume that the left triangle is also a 30-60-90. The sides could be longer or shorter than shown. the same idea as statement 1 for 2. no assumptions can be made for the left triangle given that the side is 12. if we combine both questions together then we can find out vital information for both triangles. _________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
29 Jun 2012, 03:44

but isn't triangle ABD similar to triangle ADC since AD is the perpendicular. so using the ratio of 2:1:√3 we can find all sides of triangle ABD and since AD is common using proportions we can find DC ? and thus the total area ?

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
29 Jun 2012, 04:05

Expert's post

shradhagrover18 wrote:

but isn't triangle ABD similar to triangle ADC since AD is the perpendicular. so using the ratio of 2:1:√3 we can find all sides of triangle ABD and since AD is common using proportions we can find DC ? and thus the total area ?

No, for (1) we cannot conclude that ABD and ADC are similar. We only know that ABD and ADC are both right triangles (one angle) and share the common side AD (one side), which is not enough to conclude that ABD and ADC are similar.

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
04 Apr 2013, 10:00

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? _________________

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Answer: C.

I thought if you dropped something that was perpindicular to the base of a triangle then it was automatically a bisector; so if you found BD using (1) BD MUST equal dc, no?

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Answer: C.

I thought if you dropped something that was perpindicular to the base of a triangle then it was automatically a bisector; so if you found BD using (1) BD MUST equal dc, no?

Of course not. Are you saying that the height and the median in any triangle coincide? That's not true. _________________

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
28 Oct 2013, 14:05

The triangles are similar if you take both statements together. (1) gives you all angles of the left triangle. (2) gives you the length of two sides of a triangle, and it's given that it's a right triangle From the above you know that the triangles are similar and equal to each other (because they share the same side that's relative to the same angle)...

Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]
11 Dec 2013, 08:36

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

(1) Angle ABD = 60°

If ABD is 60 then we know triangle ABD = 30:60:90 triangle. Coupled with the one measurement we are given we can find the length of the other two sides of this triangle. However, we know nothing about AC or DC, only that triangle ADC is right - we don't know the measurements of the angles so we cannot find their leg lengths. For example, DC could be six inches long or 20. We don't know and therefore cannot find the area of this triangle. Insufficient.

(2) AC = 12 We have a leg and a hypotenuse of a right triangle so we can find the length of the second leg but we know nothing about triangle ABD, except for one leg length. Insufficient.

1+2) 1 Gives us the length of all 3 sides of ABD. 2 gives us the length of all three sides of ADC. Sufficient.

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio \(1:\sqrt{3}:2\) --> as AD=6 (larger leg opposite 60 degrees angle) then \(BD=\frac{6}{\sqrt{3}}\) (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Answer: C.

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?

gmatclubot

Re: Area of triangular region
[#permalink]
15 Dec 2013, 15:05

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

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