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Triangle ABC is inscribed in a rectangle ABEF forming two right triang

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Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 28 Feb 2017, 02:58
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Question Stats:

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Q. Triangle ABC is inscribed in a rectangle ABEF forming two right triangles: AFC and BEC. Is triangle ABC an equilateral triangle?

    (1) BE = (\(\sqrt{3}AB)/2\)

    (2) Point C is the midpoint of EF


Answer Choices :

    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.


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Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post Updated on: 27 Mar 2017, 04:06
The official solution has been posted. Looking forward to a healthy discussion..:)
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Originally posted by EgmatQuantExpert on 28 Feb 2017, 02:59.
Last edited by EgmatQuantExpert on 27 Mar 2017, 04:06, edited 1 time in total.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 09 Mar 2017, 10:42
EgmatQuantExpert wrote:
Q. Triangle ABC is inscribed in a rectangle ABEF forming two right triangles: AFC and BEC. Is triangle ABC an equilateral triangle?

    (1) BE = (\(\sqrt{3}AB)/2\)

    (2) Point C is the midpoint of EF


Answer Choices :

    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.


Thanks,
Saquib
Quant Expert
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The altitude of equilateral triangle is √3/2 a, where a is the length of the equilateral triangle. Based on this rule, lets look at the two statements.

1) BE = (\(\sqrt{3}AB)/2\) - The altitude is defined, however we don't know if BE is the altitude.
2) Point C is the midpoint of EF - Insufficient, does not define the altitude.

Combining 1+2, the altitude has a length of √3a/2. Hence sufficient. C.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 12 Mar 2017, 05:09
1
SVP482 wrote:
EgmatQuantExpert wrote:
Q. Triangle ABC is inscribed in a rectangle ABEF forming two right triangles: AFC and BEC. Is triangle ABC an equilateral triangle?

    (1) BE = (\(\sqrt{3}AB)/2\)

    (2) Point C is the midpoint of EF


Answer Choices :

    A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
    B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
    C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
    D. EACH statement ALONE is sufficient to answer the question asked.
    E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.


Thanks,
Saquib
Quant Expert
e-GMAT

The altitude of equilateral triangle is √3/2 a, where a is the length of the equilateral triangle. Based on this rule, lets look at the two statements.

1) BE = (\(\sqrt{3}AB)/2\) - The altitude is defined, however we don't know if BE is the altitude.
2) Point C is the midpoint of EF - Insufficient, does not define the altitude.

Combining 1+2, the altitude has a length of √3a/2. Hence sufficient. C.


Here in this explanation, in (1) we not assuming that C lies on BF. But if C does not lie on BF, then triangle ABC can not form two RATs in rectangle ABEF. So, it goes without saying that C lies on EF and thus BE is its altitude. (since AB is the base). So statement 1 is alone sufficient.
Please correct me if i am wrong.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 12 Mar 2017, 08:11
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1
santoshparsewar wrote:
Here in this explanation, in (1) we not assuming that C lies on BF. But if C does not lie on BF, then triangle ABC can not form two RATs in rectangle ABEF. So, it goes without saying that C lies on EF and thus BE is its altitude. (since AB is the base). So statement 1 is alone sufficient.
Please correct me if i am wrong.


Hey Santhosh,

You're partly right! C has to lie on EF. Is that alone sufficient to say whether the triangle is equilateral? Certainly no! So this is not sufficient. See if the below explanation helps.

To prove the triangle is an equilateral, we need to prove all 3 sides of triangle are equal to the length of the rectangle.

Is AC=AB?

(1) BE = ( 3‾√AB)/2 - Nothing can be said as we don't know where the point C lies on EF. Not sufficient.

(2) Point C is the midpoint of EF

By Pythagoras Theorem, AC^2 = (AB/2)^2 + (BE)^2

AC^2 = AB^2/4 + BE^2 - We still can't tell whether AB=AC. So not sufficient.

From (1) lets substitute the value of BE in above equation and see where it takes us, we end up with AC^2 = AB^2. Hence AC=AB. If we take the other side of the triangle BC, we'll end up in the same result. Hence AC=BC=AB.

So the triangle is an equilateral one.

Hence C it is!

Cheers!
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 27 Mar 2017, 04:05
1
1

Solution




Steps 1 & 2: Understand Question and Draw Inferences

Given:

The given information corresponds to the following figure:

Image



To find: Is triangle ABC equilateral?



Step 3: Analyze Statement 1 independently

    • \(\mathrm{BE}=\frac{\sqrt3}2\mathrm{AB}\)
      Let’s drop a perpendicular CD on side AB.

Image

    • CD is parallel to and equal to BE.
      o So, CD = \(\frac{\sqrt3}2\mathrm{AB}\)
    • If D is the mid-point of AB,
      o Then \(\mathrm{AD}\;=\frac{\mathrm{AB}}2\)
      o So, \(\tan⁡(\angle\mathrm{CAD})=\frac{\mathrm{CD}}{\mathrm{AD}}\;=\;\frac{\frac{\sqrt3}2\mathrm{AB}}{\frac{\mathrm{AB}}2}=\;\sqrt3=\tan60^\circ\)
      o Thus, \(\angle\mathrm{CAD}=60^\circ\)
    • Similarly, we can prove that \(\angle\mathrm{CBD}=60^\circ\)
    • By Angle sum property therefore, \(\angle\mathrm{ACB}=60^\circ\)
    • Thus, the triangle ABC is an equilateral triangle
    • But the question is, is D the mid-point of AB?
    • We do not know.

Therefore, Statement 1 is not sufficient to answer the question.



Step 4: Analyze Statement 2 independently

    • Point C is the midpoint of EF
    • Let’s drop a perpendicular CD on side AB

    • CD is parallel to and equal to BE.
    • Since C is the mid-point of EF, D will be the mid-point of AB.
      o Therefore, \(\mathrm{AD}\;=\frac{\mathrm{AB}}2\)
      o But, we don’t know the magnitude of either AB or CD or AC. So, we cannot find the angles of the triangle.
    Therefore, Statement 2 is not sufficient to answer the question.



Step 5: Analyze Both Statements Together (if needed)

    • From Statement 1: If D is the mid-point of AB, then triangle ABC is an equilateral triangle
    • From Statement 2: D is the mid-point of AB


    Thus, the two statements together are sufficient to answer the question.

Answer: Option C

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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 19 Jun 2017, 04:18
1
EgmatQuantExpert wrote:

Solution




Steps 1 & 2: Understand Question and Draw Inferences

Given:

The given information corresponds to the following figure:

Image



To find: Is triangle ABC equilateral?



Step 3: Analyze Statement 1 independently

    • \(\mathrm{BE}=\frac{\sqrt3}2\mathrm{AB}\)
      Let’s drop a perpendicular CD on side AB.

Image

    • CD is parallel to and equal to BE.
      o So, CD = \(\frac{\sqrt3}2\mathrm{AB}\)
    • If D is the mid-point of AB,
      o Then \(\mathrm{AD}\;=\frac{\mathrm{AB}}2\)
      o So, \(\tan⁡(\angle\mathrm{CAD})=\frac{\mathrm{CD}}{\mathrm{AD}}\;=\;\frac{\frac{\sqrt3}2\mathrm{AB}}{\frac{\mathrm{AB}}2}=\;\sqrt3=\tan60^\circ\)
      o Thus, \(\angle\mathrm{CAD}=60^\circ\)
    • Similarly, we can prove that \(\angle\mathrm{CBD}=60^\circ\)
    • By Angle sum property therefore, \(\angle\mathrm{ACB}=60^\circ\)
    • Thus, the triangle ABC is an equilateral triangle
    • But the question is, is D the mid-point of AB?
    • We do not know.

Therefore, Statement 1 is not sufficient to answer the question.



Step 4: Analyze Statement 2 independently

    • Point C is the midpoint of EF
    • Let’s drop a perpendicular CD on side AB

    • CD is parallel to and equal to BE.
    • Since C is the mid-point of EF, D will be the mid-point of AB.
      o Therefore, \(\mathrm{AD}\;=\frac{\mathrm{AB}}2\)
      o But, we don’t know the magnitude of either AB or CD or AC. So, we cannot find the angles of the triangle.
    Therefore, Statement 2 is not sufficient to answer the question.



Step 5: Analyze Both Statements Together (if needed)

    • From Statement 1: If D is the mid-point of AB, then triangle ABC is an equilateral triangle
    • From Statement 2: D is the mid-point of AB


    Thus, the two statements together are sufficient to answer the question.

Answer: Option C

Thanks,
Saquib
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Do we really need trigonometry here to solve this? I think not :)
We can easily conclude that either of 2 given statements is not sufficient. Combining 1 and 2, we get in triangle AFC:
(root3/2*AB)^2 +1/2 (AB)^2 = AC^2
This gives AC = AB. From there on we can prove ABC is equilateral.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 28 Jun 2018, 14:28
Why couldn't answer be (B)

1) Not Sufficient

2) C is the midpoint of EF.
Let's suppose EC = AB = b,
and AF = BE = a

Since rectangle ABEF forming two right triangles: AFC and BEC

So CA = CB = root(a2 + (b2/4)

it means the triangle would be an isosceles triangle. Sufficient.

So , the answer would be 'B'
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Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 05 Jul 2018, 10:00
Diwakar003 wrote:
santoshparsewar wrote:
Here in this explanation, in (1) we not assuming that C lies on BF. But if C does not lie on BF, then triangle ABC can not form two RATs in rectangle ABEF. So, it goes without saying that C lies on EF and thus BE is its altitude. (since AB is the base). So statement 1 is alone sufficient.
Please correct me if i am wrong.


Hey Santhosh,

You're partly right! C has to lie on EF. Is that alone sufficient to say whether the triangle is equilateral? Certainly no! So this is not sufficient. See if the below explanation helps.

To prove the triangle is an equilateral, we need to prove all 3 sides of triangle are equal to the length of the rectangle.

Is AC=AB?

(1) BE = ( 3‾√AB)/2 - Nothing can be said as we don't know where the point C lies on EF. Not sufficient.

(2) Point C is the midpoint of EF

By Pythagoras Theorem, AC^2 = (AB/2)^2 + (BE)^2

AC^2 = AB^2/4 + BE^2 - We still can't tell whether AB=AC. So not sufficient.

From (1) lets substitute the value of BE in above equation and see where it takes us, we end up with AC^2 = AB^2. Hence AC=AB. If we take the other side of the triangle BC, we'll end up in the same result. Hence AC=BC=AB.

So the triangle is an equilateral one.

Hence C it is!

Cheers!

Mehn...! This explanation is phenomenal. Thanks a lot for sharing.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang  [#permalink]

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New post 05 Jul 2018, 20:54
With Time a luxury, such a long approach as given in official solution is unnecessary.

Condition 1: BE= Sqrt(3)/2 AB is insufficient to prove that AB=AC=BC( equilateral triangle)
Condition 2: C is the midpoint of EF. This doesn't tell anything about length of BE. So again insufficient to prove AB=AC=BC.

Combine 1 & 2: BE= Sqrt(3)/2 AB and EC=AB/2 we get, BC= AB
similarly, AC=AB.
Therefore, AB=AC=BC (equilateral triangle)
Therefore Answer is C.
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Re: Triangle ABC is inscribed in a rectangle ABEF forming two right triang &nbs [#permalink] 05 Jul 2018, 20:54
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Triangle ABC is inscribed in a rectangle ABEF forming two right triang

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