From the stem, we know for sure that it's a right triangle, which doesn't mean it is not isosceles, but we aren't for sure that it is isosceles.

From Statement 1, it could be isoceles, but not necessarily. I'll draw a picture to explain...headed to Photoshop.

In statement 1, the same color X's go together. Because x = y. As you can see, the area of the triangle if the BLUE X is used is 1.5. if the Purple/Pink X is used, the area is 2, and if the Orange X is used, the area is back to 1.5 S1 cannot be sufficient because we've found different values of area when we also satisfy the rule presented in S1

In Statement 2, we do (as freshina12 says) have an isosceles triangle. This can only be true when we use the pink X's as this will create two 45 degree angles. The area here is 2, but there are infinite number of other options. For the X along y=1, x can be anything upto the value of 5. As for the line x=5, the y value can be anything as long as it is > 1. So as long as we have 2-45 degree angles, S2 will be satisfied, but we will get different values for the area of the triangle formed.

Only when we consider the statements together do we get a single value, which is seen at the PINK X in the picture below. I'm not sure what you mean the area could be 8. I don't see that. I do see how it could be 2 though.

I say answer is C.




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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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JM..thanks I realized my mistake..and its a stupid one..i mis-read stmnt 1 to say (5-x)=(y-1)..

which is true but we have stmnt 1 saying point x is equal to point y..

I can draw an isoceles triangle when x-5=4 and y-1=4

Since the angle at vertex (x,1)=(5,y)...this infers that the sides opposite to those angles are equal...so it is a Right Isosceles Triangle and hence we get 5-x=y-1 -> x+y=6

Since x=y, the value of x & y is 3 and the length of each side is 2.

So Area of triangle = 1/2 * 2 * 2 = 2 sq. units

stmt1 : x=y

stmt2 : angle (x,1) = angle (y,1)
this means sides opposite to that angle are equal

to identifty the distance b/w the sides (y2-y1)^2+(x2-x1)^2

so (5-x)^2 +(1-1)^2=(5-5)^2+(y-1)^2

(5-x)^2=(y-1)^2 taking square root ob both sides

5-x=y-1 so x+y=6

not sufficient

by combining both 1 & 2 we get

x=3 and y=3

so ans is C

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.

Answer: C.
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