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# M08-06

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Math Expert
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15 Sep 2014, 23:36
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95% (hard)

Question Stats:

48% (01:42) correct 52% (01:20) wrong based on 125 sessions

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If the vertices of a triangle have coordinates $$(x, 1)$$, $$(5, 1)$$, and $$(5, y)$$ where $$x \lt 5$$ and $$y \gt 1$$, what is the area of the triangle?

(1) $$x = y$$

(2) Angle at the vertex $$(x, 1)$$ is equal to angle at the vertex $$(5, y)$$

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15 Sep 2014, 23:36
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Official Solution:

Look at the diagram below:

Notice that vertex (x,1) will be somewhere on the green line segment and the vertex (5,y) will be somewhere on the blue line segment. So, in any case our triangle will be right angled, with a right angle at vertex (5, 1). Next, the length of the leg on the green line segment will be $$5-x$$ and the length of the leg on the blue line segment will by $$y-1$$. So, the area of the triangle will be: $$\text{area}=\frac{1}{2}*(5-x)*(y-1)$$.

(1) $$x=y$$. Since $$x \lt 5$$ and $$y \gt 1$$ then both $$x$$ and $$y$$ are in the range (1,5): $$1 \lt (x=y) \lt 5$$. If we substitute $$y$$ with $$x$$ we'll get: $$\text{area}=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(x-1)$$, different values of $$x$$ give different values for the area (even knowing that $$1 \lt x \lt 5$$). Not sufficient.

(2) Angle at the vertex $$(x,1)$$ is equal to angle at the vertex $$(5,y)$$. We have an isosceles right triangle: $$5-x=y-1$$. Again if we substitute $$y-1$$ with $$5-x$$ we'll get: $$\text{area}=\frac{1}{2}*(5-x)*(y-1)=\frac{1}{2}*(5-x)*(5-x)$$, different values of $$x$$ give different values for the area. Not sufficient.

(1)+(2) $$x=y$$ and $$5-x=y-1$$. Solve for $$x$$: $$x=y=3$$, so $$\text{area}=\frac{1}{2}*(5-3)*(3-1)=2$$. Sufficient.

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31 Dec 2015, 10:16
Hello Bunuel,

I have a concern regarding this question. I think OA should be E not C.

How can x=y translates into 5−x=y−1 ??

Please see the attached pic, using which we can see that even when X = Y areas could be different.

Thanks
>> !!!

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03 Jan 2016, 10:29
WillGetIt wrote:
Hello Bunuel,

I have a concern regarding this question. I think OA should be E not C.

How can x=y translates into 5−x=y−1 ??

Please see the attached pic, using which we can see that even when X = Y areas could be different.

Thanks

In your example with red hypotenuse how is it true that x=y?...
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12 Jan 2016, 22:34
1
WillGetIt wrote:
Hello Bunuel,

I have a concern regarding this question. I think OA should be E not C.

How can x=y translates into 5−x=y−1 ??

Please see the attached pic, using which we can see that even when X = Y areas could be different.

Thanks

Hi WillGetIt,

With both conditions being true, you will get only a single isosceles triangle ((3,1), (5,1), (5,3)). Only these points will satisfy both requirements.

Hope this helps.

Regards,

Renin
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13 Jan 2016, 03:48
Already explained well. For the area, we need both height and base. So, here it is only possible, if we can know all vertices.
Can you get the answer using (1) only, no, since many possibilities such as, (2,1),(5,2);(3,1),(5,3);(4,1),(5,4)..etc.Insufficient
Forget about (1), think on (2), but still many possibilities when equal angles give us equal sides,such as (4,1),(5,2);(3,1),(5,3);(1,1),(5,1)..etc, Insufficient.
So, A,B,D all cancelled since not at least one sufficiency.
Now combine both, ...(2) requires that both distance from 5 will be equal. Make it narrower using (1)..so that finally x=y. Only one case is possible here which is (3,1),(5,3). Because we can calculate the area combinedly, C is the answer.
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14 Jul 2016, 10:28
I think this is a high-quality question and I agree with explanation. Great question. Any similar co-ordinate geometry questions for practice outside GMATCLUB tests please suggest such questions are awesome man!
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15 Jul 2016, 06:33
1
Senthil7 wrote:
I think this is a high-quality question and I agree with explanation. Great question. Any similar co-ordinate geometry questions for practice outside GMATCLUB tests please suggest such questions are awesome man!

Similar questions:
in-the-diagram-above-coordinates-are-given-for-three-of-the-vertices-194197.html
if-vertices-of-a-triangle-have-coordinates-87344.html
if-vertices-of-a-triangle-have-coordinates-2-2-3-2-and-82159.html
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29 Aug 2016, 20:44
I'm guessing we can't assume x=y to be an equation of a line? Can you confirm?
Tx.
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30 Aug 2016, 01:31
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wmichaelxie wrote:
I'm guessing we can't assume x=y to be an equation of a line? Can you confirm?
Tx.

Yes it is. Check below:
Attachment:
Untitled.png

>> !!!

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Updated on: 09 Oct 2016, 19:43
This just clicked. Thank you.

Originally posted by wmichaelxie on 30 Aug 2016, 06:02.
Last edited by wmichaelxie on 09 Oct 2016, 19:43, edited 1 time in total.
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01 Sep 2016, 23:22
1
1
Bunuel wrote:
If the vertices of a triangle have coordinates $$(x, 1)$$, $$(5, 1)$$, and $$(5, y)$$ where $$x \lt 5$$ and $$y \gt 1$$, what is the area of the triangle?

(1) $$x = y$$

(2) Angle at the vertex $$(x, 1)$$ is equal to angle at the vertex $$(5, y)$$

Responding to a pm:

Angle at (5, 1) will be 90 degrees.

(1) $$x = y$$
The co-ordinates are (x, 1), (5, 1) and (5, x).
Not sufficient alone since we don't know x.

(2) Angle at the vertex $$(x, 1)$$ is equal to angle at the vertex $$(5, y)$$[/quote]
This means the two sides are equal. So
5 - x = y - 1
Not sufficient alone.

Using both statements,

5 - x = x - 1
x = 3 = y

We get all co-ordinates so we will be able to find the area.
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12 Oct 2016, 08:39
I am able to understand line distance as y-1 is because of sqrt((5-5)^2+(y-1)^2) = y-1
but, why the other distance is 5-x and not x-5 as the calculation would be sqrt((x-5)^2/+(1-1)^2) = x-5

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07 Nov 2016, 23:42
hello,
as per the graph, we can obtain all three vertices of the triangle with statement 1 itself.
please tell me where am i thinking wrong?
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07 Jan 2017, 10:24
Sumanth8492, as per statement 1, vertices of x and y can be 2,3, or 4.
We can not obtain the area of the triangle unless we know exactly which one.
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27 Oct 2017, 09:50
In this case, from the problem statement itself we can deduce that x & y are nothing but the two sides of the right angled triangle. So, doesn't x = y itself imply that the triangle is iso. rt angled triangle?In that case we already get to know that the angles are equal. Please let me know if I am missing something/understanding is incorrect.

TIA.
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27 Oct 2017, 09:57
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tejasa07 wrote:
In this case, from the problem statement itself we can deduce that x & y are nothing but the two sides of the right angled triangle. So, doesn't x = y itself imply that the triangle is iso. rt angled triangle?In that case we already get to know that the angles are equal. Please let me know if I am missing something/understanding is incorrect.

TIA.

The length of the leg on the green line segment will be 5−x and the length of the leg on the blue line segment will by y−1, NOT x and y. So, x = y does NOT mean that we have a right isosceles triangle.
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27 Oct 2017, 10:04
My bad. Thank you for the prompt response.
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08 Oct 2018, 04:33
Hi Bunuel

Can't we get the answer using only statement 2?
Consider the following:

Stat 2: Angle at the vertex (x,1) is equal to angle at the vertex (5,y)

Thus,
As proven earlier, the triangle is isosceles.
So, 5-x=y-1 ==> x+y=6
Also, x<5 and y>1
thus, (x,y) can take: (4,2); (3,3); (2,4); (1,5) etc
but, since the triangle is isosceles, we can only take the value as (3,3)...and hence the area can be found.
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Thanks,
Rnk

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09 Nov 2018, 09:57
Bunuel chetan2u

how did you get the length of side as 5-x ?? please explain ... are we subtracting 1 x coordinate from 2nd x coordinate. and if so ...how to decide which vertice to consider ?? and whatever you temme will that be applicable to all shapes??
Re: M08-06 &nbs [#permalink] 09 Nov 2018, 09:57
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# M08-06

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