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# If the vertices of a triangle have coordinates (x,1), (5,1),

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If the vertices of a triangle have coordinates (x,1), (5,1), [#permalink]

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03 Nov 2009, 13:49
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If the vertices of a triangle have coordinates (x,1), (5,1), and (5,y) where x<5 and y>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)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Apr 2012, 02:20, edited 2 times in total.
Edited the question and added the OA

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03 Nov 2009, 13:56
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If the vertices of a triangle have coordinates (x,1), (5,1), and (5,y) where x<5 and y>1, what is the area of the triangle?

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: $$area=\frac{1}{2}*(5-x)*(y-1)$$

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

(1)+(2) $$x=y$$ and $$5-x=y-1$$ --> solve for $$x$$: $$x=y=3$$ --> $$area=\frac{1}{2}*(5-3)*(3-1)=2$$. Sufficient.

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11 Nov 2009, 06:35
Good explanation Bunnel!

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16 Dec 2009, 07:31
hi tania..ill try.. firstly the 3 set of coord tells us that it is a right angled triangle(right angle is at vertex(5,1))
SI tells us that x=y....x and y can be any values...not suff..
SII tells us the angles at two other vertices are 45...so the other two sides are equal...can have many values.. eq we get from it is 5-x=y+1.... not suff
combined 5-x =x+1... so x=y=3.. suff.. C ans
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21 Dec 2009, 18:35
Distance formula for area:
A = 1\2\sqrt{(5-x)^2 (x-1)^2}

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06 Mar 2014, 11:45
Dear Bunuel

Can we have B as the answer, since we know that angles are equal, the only coordinate that will give same the distance will be 3. Hence, we dont need option A for any support. Kindly help!

Regards

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07 Mar 2014, 02:14
ShantnuMathuria wrote:
Dear Bunuel

Can we have B as the answer, since we know that angles are equal, the only coordinate that will give same the distance will be 3. Hence, we dont need option A for any support. Kindly help!

Regards

That's not correct. Any $$x<5$$ and $$y>1$$, which satisfy $$5-x=y-1$$ ($$x+y=6$$) will give equal legs and thus equal angles. For example, $$x=4$$, $$y=2$$, or $$x=3.5$$, $$y=2.5$$, ...

Hope it's clear.
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Re: If the vertices of a triangle have coordinates (x,1), (5,1), [#permalink]

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30 Oct 2017, 16:08
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Re: If the vertices of a triangle have coordinates (x,1), (5,1),   [#permalink] 30 Oct 2017, 16:08
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