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M28-40

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Bunuel
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M28-40 [#permalink] New post   16 Sep 2014, 01:31
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The length of the median BD in triangle ABC is 12 centimeters, what is the length of side AC?


(1) ABC is an isosceles triangle.

(2) \(AC^2 = AB^2 + BC^2\).
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Bunuel
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Re M28-40 [#permalink] New post   16 Sep 2014, 01:31
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Official Solution:


(1) ABC is an isosceles triangle. Clearly insufficient.

(2) \(AC^2 = AB^2 + BC^2\). This statement implies that ABC is a right triangle and AC is its hypotenuse. Important property: median from right angle is half of the hypotenuse, hence \(BD=12=\frac{AC}{2}\), from which we have that \(AC=24\). Sufficient.


Answer: B
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bigzoo
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Re: M28-40 [#permalink] New post   27 Mar 2015, 18:24
could help me please what does it mean: "The length of the median BD" ... first I thought it is the midpoint of BC, but it is not ... and since english is not my mother tounge I do not know what it means
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Bunuel
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Re: M28-40 [#permalink] New post   28 Mar 2015, 07:47
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bigzoo wrote:
could help me please what does it mean: "The length of the median BD" ... first I thought it is the midpoint of BC, but it is not ... and since english is not my mother tounge I do not know what it means


The median of a triangle is a line from a vertex to the midpoint of the opposite side. Check for more HERE.
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aaigla
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Re: M28-40 [#permalink] New post   02 May 2019, 17:17
Please correct my reasoning. I do not understand why A is not correct. Let's name each side of our equilateral triangle "x." AB=BC=AC=x. Since we are told BD is a median, it means AD=DC=x/2. In an equilateral triangle, a line drawn from one of the vertices to the midpoint of the opposite line will form a 90 degree angle. So we would have two equal triangles, AB-BD-DA and CB-BD-DC that each have one side, BD, length 12, one side, DC or DA, length x/2, and one side, BC or BA, length x. Since these are right triangles, (x/2)^2+12^2=x^2. We have an equation with one unknown. Why is this not sufficient? Bunuel can you help?
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Bunuel
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Re: M28-40 [#permalink] New post   02 May 2019, 21:33
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aaigla wrote:
Please correct my reasoning. I do not understand why A is not correct. Let's name each side of our equilateral triangle "x." AB=BC=AC=x. Since we are told BD is a median, it means AD=DC=x/2. In an equilateral triangle, a line drawn from one of the vertices to the midpoint of the opposite line will form a 90 degree angle. So we would have two equal triangles, AB-BD-DA and CB-BD-DC that each have one side, BD, length 12, one side, DC or DA, length x/2, and one side, BC or BA, length x. Since these are right triangles, (x/2)^2+12^2=x^2. We have an equation with one unknown. Why is this not sufficient? Bunuel can you help?



(1) says that ABC is isosceles, not equilateral.
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Re: M28-40 [#permalink] New post   29 May 2020, 00:43
Hi Bunuel,

Could you please elaborate (as in how does it come about) on the geometrical property mentioned for the sufficiency of Statement 2 : "Important property: median from right angle is half of the hypotenuse"?

It would just make it easier to remember.

Thanks.
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Re: M28-40 [#permalink] New post   24 Aug 2020, 15:32
Sriparnacj I had the same thought. Draw a line from midpoint D to line BC on the triangle, and call the point "E" on line BC. You can then prove that triangle CDE and BDE are identical. Or, you could prove that BDE and ABC are congruent.

Annoyingly I can't post URLs but I would google the triangle midsegment theorem for more info. Hope this helps
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Bunuel
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Re: M28-40 [#permalink] New post   14 Aug 2023, 08:15
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Duplicate of M04-24. Unpublished.
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Re: M28-40 [#permalink]
14 Aug 2023, 08:15
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