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# In the figure above, if the length of MO is 10, is MNO an equilateral

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In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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05 Apr 2017, 11:15
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25% (01:31) correct 75% (01:45) wrong based on 85 sessions

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In the figure above, if the length of MO is 10, is MNO an equilateral triangle?

(1) The length of MN is 10.
(2) h = 5$$\sqrt{3}$$

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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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06 Apr 2017, 05:06
1
1
In the figure above, if the length of MO is 10, is MNO an equilateral triangle?

(1) The length of MN is 10.
(2) h = 5$$\sqrt{3}$$

Hi,

6 people answered it till now and all 6 wrong..

It just tells us that we get carried away with sketches and believe what is not stated

Here we believe that the altitude 'h' is in center of MN which is nowhere mentioned.
Believing the altitude to be in middle of BASE and with two statements, we think the answer can be found.

But we don't know if the point is in center, so we cannot answer
Combined insufficient

E
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06 Apr 2017, 06:26
In the above question if two sides are 10 each and height as 5root3 then the height is not in the middle, shifting the line anywhere else will not make the angle 90 which is a requirement for height of triangle which has been mentioned in the question.

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In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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06 Apr 2017, 09:20
I agree that H is not in middle of the base, But How to identify that its not making 90 degrees on the base? Its given (H) which means Height... Can you please help me?

and if its a 90 degree , then we can evaluate the two portions of the base independently using both Options, Correct me if I am wrong.
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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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06 Apr 2017, 11:14
If the angle is 90 degrees then we can evaluate it .

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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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06 Apr 2017, 12:09
chetan2u wrote:
In the figure above, if the length of MO is 10, is MNO an equilateral triangle?

(1) The length of MN is 10.
(2) h = 5$$\sqrt{3}$$

Hi,

6 people answered it till now and all 6 wrong..

It just tells us that we get carried away with sketches and believe what is not stated

Here we believe that the altitude 'h' is in center of MN which is nowhere mentioned.
Believing the altitude to be in middle of BASE and with two statements, we think the answer can be found.

But we don't know if the point is in center, so we cannot answer
Combined insufficient

E

Is the angle 90 degrees? Yes or No? Why? Because height (H) is given and its always perpendicular to the base...If its 90 degrees then we can evaluate it
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In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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06 Apr 2017, 12:28
B) 30 60 90 triangles follow 1: root 3: 2 format

H is given as 5 root 3 so that makes the base 5 and the outside line 10.

A is not correct because the triangle could be an isosceles

Edit: upon further inspection, h is not given as the height. Without knowing that, the answer becomes E

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In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 07:58
1
I think it is 'C' assuming h is height.

Statement 1 is Insuff.
Because,
MNO could be isosceles with MN=MO=10 and NO is some other value.

Statement-2 is Insufficient (assuming h is indeed the height).
Example: triangle MNO is a right angled triangle with angle NMO = $$90^o$$ and MN = h = 5$$\sqrt{3}$$

Statements 1+2 together: Sufficient

Under height assumption for h we have,
h = MN. sine(of angle MNO)
=> $$5\sqrt{3}$$ = 10.sine(MNO)
=>sine(MNO) = $$\sqrt{3}/2$$
=> angle MNO = $$60^o$$

Since MN=MO=10 hence other 2 angles are $$60^o$$ each - Hence Equilateral triangle.
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In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 08:59
shakun wrote:
I think it is 'C' assuming h is height.

Statement 1 is Insuff.
Because,
MNO could be isosceles with MN=MO=10 and NO is some other value.

Statement-2 is Insufficient (assuming h is indeed the height).
Example: triangle MNO is a right angled triangle with angle NMO = $$90^o$$ and MN = h = 5$$\sqrt{3}$$

Statements 1+2 together: Sufficient

Under height assumption for h we have,
h = MN. sine(of angle MNO)
=> $$5\sqrt{3}$$ = 10.sine(MNO)
=>sine(MNO) = $$\sqrt{3}/2$$
=> angle MNO = $$60^o$$

Since MN=MO=10 hence other 2 angles are $$60^o$$ each - Hence Equilateral triangle.

If h is the height, it must be perpendicular to the base, meaning that B would be the answer

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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 09:55
gmathopeful19 wrote:
shakun wrote:
I think it is 'C' assuming h is height.

Statement 1 is Insuff.
Because,
MNO could be isosceles with MN=MO=10 and NO is some other value.

Statement-2 is Insufficient (assuming h is indeed the height).
Example: triangle MNO is a right angled triangle with angle NMO = $$90^o$$ and MN = h = 5$$\sqrt{3}$$

Statements 1+2 together: Sufficient

Under height assumption for h we have,
h = MN. sine(of angle MNO)
=> $$5\sqrt{3}$$ = 10.sine(MNO)
=>sine(MNO) = $$\sqrt{3}/2$$
=> angle MNO = $$60^o$$

Since MN=MO=10 hence other 2 angles are $$60^o$$ each - Hence Equilateral triangle.

If h is the height, it must be perpendicular to the base, meaning that B would be the answer

- Agreed, h will be perpendicular - that doesn't guarantee B. see my example against statement 2.
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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 10:00
@
gmathopeful19 wrote:
shakun wrote:
I think it is 'C' assuming h is height.

Statement 1 is Insuff.
Because,
MNO could be isosceles with MN=MO=10 and NO is some other value.

Statement-2 is Insufficient (assuming h is indeed the height).
Example: triangle MNO is a right angled triangle with angle NMO = $$90^o$$ and MN = h = 5$$\sqrt{3}$$

Statements 1+2 together: Sufficient

Under height assumption for h we have,
h = MN. sine(of angle MNO)
=> $$5\sqrt{3}$$ = 10.sine(MNO)
=>sine(MNO) = $$\sqrt{3}/2$$
=> angle MNO = $$60^o$$

Since MN=MO=10 hence other 2 angles are $$60^o$$ each - Hence Equilateral triangle.

If h is the height, it must be perpendicular to the base, meaning that B would be the answer

- Agreed that h will be perpendicular but that doesn't guarantee B. See my example against statement 2.
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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 10:24
shakun wrote:
@
gmathopeful19 wrote:
shakun wrote:
I think it is 'C' assuming h is height.

Statement 1 is Insuff.
Because,
MNO could be isosceles with MN=MO=10 and NO is some other value.

Statement-2 is Insufficient (assuming h is indeed the height).
Example: triangle MNO is a right angled triangle with angle NMO = $$90^o$$ and MN = h = 5$$\sqrt{3}$$

Statements 1+2 together: Sufficient

Under height assumption for h we have,
h = MN. sine(of angle MNO)
=> $$5\sqrt{3}$$ = 10.sine(MNO)
=>sine(MNO) = $$\sqrt{3}/2$$
=> angle MNO = $$60^o$$

Since MN=MO=10 hence other 2 angles are $$60^o$$ each - Hence Equilateral triangle.

If h is the height, it must be perpendicular to the base, meaning that B would be the answer

- Agreed that h will be perpendicular but that doesn't guarantee B. See my example against statement 2.

NMO could never be 90 in that case

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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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24 Apr 2017, 21:13
Ans should be C. If the height is 5*root(3), MN = 10 and since the height is always perpendicular to base, so the height must intersect the base MO at midpoint. Then since, MO=MN=10 and height = 5*root(3) and it intersects the base at midpoint, so triangle MNO has to be an equilateral triangle.
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Re: In the figure above, if the length of MO is 10, is MNO an equilateral  [#permalink]

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11 Aug 2019, 17:58
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Re: In the figure above, if the length of MO is 10, is MNO an equilateral   [#permalink] 11 Aug 2019, 17:58
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