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Re: M0412 [#permalink]
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30 Jan 2015, 14:55
Bunuel wrote: Official Solution:
Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of nonhypotenuse sides AB and BC. (1) AB = 6. Clearly insufficient. (2) The product of the nonhypotenuse sides is equal to 24 \(\rightarrow\) directly gives us the value of AB*BC. Sufficient.
Answer: B How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle? Thx in advance for your help.



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31 Jan 2015, 06:13
chieffarmer wrote: Bunuel wrote: Official Solution:
Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of nonhypotenuse sides AB and BC. (1) AB = 6. Clearly insufficient. (2) The product of the nonhypotenuse sides is equal to 24 \(\rightarrow\) directly gives us the value of AB*BC. Sufficient.
Answer: B How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle? Thx in advance for your help. BD is a height means that B is a right angle and AC is a hypotenuse (so BD is a height from right angle B to the hypotenuse AC).
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Re: M0412 [#permalink]
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26 Apr 2015, 18:31
first option says AB =6 doesnt it imply that the sides are 6,8,10?



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27 Apr 2015, 01:44



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Re: M0412 [#permalink]
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08 Jul 2015, 09:42
Bunuel wrote: chieffarmer wrote: Bunuel wrote: Official Solution:
Since all points are distinct and BD is a height then B must be a right angle and AC must be a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Question thus asks about the product of nonhypotenuse sides AB and BC. (1) AB = 6. Clearly insufficient. (2) The product of the nonhypotenuse sides is equal to 24 \(\rightarrow\) directly gives us the value of AB*BC. Sufficient.
Answer: B How do you know that B is a right angle? Couldn't we also have BC or BA as a hypotenuse with BD still indicating the height of the triangle? Thx in advance for your help. BD is a height means that B is a right angle and AC is a hypotenuse (so BD is a height from right angle B to the hypotenuse AC). Bunuel  in the figure above isn't BD an altitude and AB the height of triangle ABC??



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Re: M0412 [#permalink]
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12 Jul 2015, 00:10
efforts wrote:
Bunuel  in the figure above isn't BD an altitude and AB the height of triangle ABC??
Hi, there is no difference in " altitude and height".. " altitude or height" is the shortest distance from a point to the line.. so here BD is the altitude/height, when AC is the base and AB is the alt/height when base is BC... Hope it clears the query..
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Re: M0412 [#permalink]
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21 Jul 2015, 04:32
Hi Bunuel,
Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous.



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Re: M0412 [#permalink]
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22 Jul 2015, 01:16
CountClaud wrote: Hi Bunuel,
Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous. If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle.
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Re: M0412 [#permalink]
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12 Jan 2016, 07:50
buffaloboy wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. I am surprised if A,B, C, and D are distinct points and ABC is right triangle, right angled at B, then how come A , and D are not the same point. AB is the height . I am damn confused. Each triangle can have three different bases and perpendicular to each base, three different heights. In the given description of triangle ABC, angle B being right angle, if you take BA or BC as height, then D coincides with A or C. But the question also says all four points are distinct. Hence we need to take AC as base and BD as height. Thats why this problem is in 700800 difficulty level. Hope this helps.



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Re: M0412 [#permalink]
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27 Mar 2016, 23:04
wow I can't believe this question tricked me so hard. Thank you for this question. It really makes me read a LOT more carefully.



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Re: M0412 [#permalink]
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29 Apr 2016, 13:35
Bunuel wrote: CountClaud wrote: Hi Bunuel,
Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous. If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle. How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...
>> !!!
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Re: M0412 [#permalink]
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02 May 2016, 04:10
nk18967 wrote: Bunuel wrote: CountClaud wrote: Hi Bunuel,
Can you please explain why the triangle cannot have A or C as the right angle, and the height BD drawn outside of the triangle? If you flip the triangle upside down, it looks to me that you can have a height outside of the triangle, which would make the location of the right angle ambiguous. If A is a right angle, so if BA is perpendicular to CA, then the height from B to CA will be BA, making A and D to coincide, which is not possible since we are told that A, B, C, and D are distinct points on a plane. The same if C is a right angle. How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane... The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.
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Re: M0412 [#permalink]
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02 May 2016, 04:26
How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane...[/quote]
The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.[/quote]

Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'. Thanks!



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02 May 2016, 04:30



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Re: M0412 [#permalink]
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02 May 2016, 07:04
Bunuel wrote: nk18967 wrote: How about this figure? It doesn't say 'D' is a point on the triangle, just says D is a point on the plane... The figure is NOT right. We are given that BD is a height of the triangle. The height is a perpendicular dropped from one of the vertices to the opposite side.  Hmm...I guess my concept of height wasn't clear. So, height of a triangle is ALWAYS the altitude of the triangle, inside the triangle.. and that altitude changes depending on which side is the 'base'. Thanks![/quote] It's not necessary to be inside a triangle: Attachment: alt2.gif [/quote] 'VERTICES' is the magic word!! Gotcha, Thanks!



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Re: M0412 [#permalink]
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24 May 2016, 18:57
I think this is a poorquality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum.



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Re: M0412 [#permalink]
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25 May 2016, 08:43
DavidFox wrote: I think this is a poorquality question and I don't agree with the explanation. Solution is incorrect. Answer should be (E). There is no way to prove that AB or BC are both legs, or alternatively that one is a leg and the other a hypotenuse. The height BD can lie outside the triangle contrary to the discussion posted in the forum. That's not correct. The height in a right triangle is either one of the legs or the perpendicular from right angle to the hypotenuse. Thus the height of a right triangle cannot lie outside the triangle. Moreover, since A, B, C, and D are distinct points then BD cannot be the height from nonright angle because in this case it would coincide with one of the legs.
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