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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)
(1) x=y
(2) Angle at the vertex (x,1) is equal to angle at the vertex (5,y)
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25 Oct 2009, 14:49
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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?
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.
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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