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Manager  B
Joined: 19 Aug 2016
Posts: 80
If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(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.

Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 57074
If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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zanaik89 wrote:
Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(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.

Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks

No. The height and the median coincide only in isosceles/equilateral triangle, when the height is dropped from the vertex formed by equal sides to the base.
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Manager  B
Joined: 19 Aug 2016
Posts: 80
If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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zanaik89 wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(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.

Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks[/quote]

So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?
Math Expert V
Joined: 02 Sep 2009
Posts: 57074
Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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zanaik89 wrote:

So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?

Yes, we know that AD is perpendicular to BC but we don't know whether it bisects BC.
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Intern  B
Joined: 23 Jun 2018
Posts: 25
Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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

Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(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 the area. Sufficient.

Answer: C.

Bunuel - My doubt is if a triangle like this which has a 30-60-90 angles, shouldn't the the triangle ADC should also be 30-60-90 making triangle ABC an equilateral triangle? because an equilateral triangle can be formed by joining two identical triangles, each with 30-60-90 angles.
Manager  B
Joined: 18 Feb 2018
Posts: 122
Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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Bunuel VeritasKarishma The stem says that ADC is a right angle. That means AD is perpendicular to DC. Why are we assuming that B, D and C are in a straight line?
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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GittinGud wrote:
Bunuel VeritasKarishma The stem says that ADC is a right angle. That means AD is perpendicular to DC. Why are we assuming that B, D and C are in a straight line?

BC is a straight line and D lies on it. Note sure why you think right triangle ADC messes that up.
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Manager  B
Joined: 18 Feb 2018
Posts: 122
Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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VeritasKarishma We have no information in the stem to decide whether BC is a straight line. Why are we assuming it is?
Manager  B
Joined: 23 Jul 2015
Posts: 63
Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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VeritasKarishma wrote:
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

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.

Hi! How do we know that the ADC is also a 90 30 60 triangle? Re: If AD is 6 and ADC is a right angle, what is the area of triangular re   [#permalink] 10 Jul 2019, 11:12

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