noboru wrote:
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
Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
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(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.
(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.
Answer: A.
Hope it heps.
From the 2nd statemen, the hypotenuse is 5 units in length since the triangle is a right angle triangle that is 90 degree at point C.. then that basically means that the other 2 sides(base and height of the triangle) will be be either 3 and 4 since 3,4 and 5 forms a Pythagorean triple.. that means area which is half * base * height will always be (3*4)/2 .. That would mean that 2nd statement is also sufficient.. I do get your point about drawing a circle but how would you justify the 3 lengths according to Pythagoras theorem ?