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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?
(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees
(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees
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02 Sep 2010, 05:45
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Vertices of a triangle have coordinates \((-2, 2)\), \((3, 2)\), and \((x, y)\). What is the area of the triangle?
Given two points \(A(-2, 2)\) and \(B(3, 2)\). The question asks to find the area of triangle \(ABC\), where \(C(x,y)\). Refer to the diagram below:
(1) \(|y-2|=1\).
Either \(y=3\) or \(y=1\), hence vertex \(C\) could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). In any case, the area of \(ABC\) will be the same. Using the formula \(area=\frac{1}{2}*base*height\), we have \(base=AB=5\), and the height would be 1 for any point \(C\) (see two possible locations of \(C\): \(C_1\) and \(C_2\). The heights of \(ABC_1\) and \(ABC_2\) are the same and equal to 1). Therefore, the area is \(area=\frac{1}{2}*base*height=\frac{5}{2}\). This statement is sufficient.
(2) The angle at the vertex \((x, y)\) equals 90 degrees.
This statement implies that \(ABC\) is a right triangle with a hypotenuse \(AB\). Consider \(AB\) as the diameter of a circle. In this case, \(C\) could be anywhere on the circle, and it will form a right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle). However, the height of \(ABC\) will be different for different locations of point \(C\), resulting in different areas (see two possible locations of \(C\): \(C_3\) and \(C_4\). The heights of \(ABC_3\) and \(ABC_4\) are different). This statement is not sufficient.
Answer: A
Hope it helps.
render.php (1).png [ 17.42 KiB | Viewed 55335 times ]
Given two points \(A(-2, 2)\) and \(B(3, 2)\). The question asks to find the area of triangle \(ABC\), where \(C(x,y)\). Refer to the diagram below:
(1) \(|y-2|=1\).
Either \(y=3\) or \(y=1\), hence vertex \(C\) could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). In any case, the area of \(ABC\) will be the same. Using the formula \(area=\frac{1}{2}*base*height\), we have \(base=AB=5\), and the height would be 1 for any point \(C\) (see two possible locations of \(C\): \(C_1\) and \(C_2\). The heights of \(ABC_1\) and \(ABC_2\) are the same and equal to 1). Therefore, the area is \(area=\frac{1}{2}*base*height=\frac{5}{2}\). This statement is sufficient.
(2) The angle at the vertex \((x, y)\) equals 90 degrees.
This statement implies that \(ABC\) is a right triangle with a hypotenuse \(AB\). Consider \(AB\) as the diameter of a circle. In this case, \(C\) could be anywhere on the circle, and it will form a right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle). However, the height of \(ABC\) will be different for different locations of point \(C\), resulting in different areas (see two possible locations of \(C\): \(C_3\) and \(C_4\). The heights of \(ABC_3\) and \(ABC_4\) are different). This statement is not sufficient.
Answer: A
Hope it helps.
Attachment:
render.php (1).png [ 17.42 KiB | Viewed 55335 times ]
04 Aug 2010, 08:42
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I think i understand why (2) is not sufficient. If we take the points (-2,2), (3,2) as a diameter of a circle and the point (x,y) lies on the perimeter of the circle, then any triangle formed is a right-angled triangle but with different height, therefore producing different areas. What do you think?
