SOLUTION: A right triangle in the first quadrant is bounded by lines y = 0, y = x, and y = -x+5. Find its area

Question 175673: A right triangle in the first quadrant is bounded by lines y = 0, y = x, and y = -x+5. Find its area
Found 2 solutions by Mathtut, gonzo:
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
(5/2,5/2) is where y=x and y=-x+5 meet
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we know that the hypothenuse is 5 units
we also know that the 2 legs are equal. lets call each leg x
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so
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A=1/2bh
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we know the base and height are the length of the legs
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Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
the lines y = x and y = -x + 5 are perpendicular to each other since their slopes are negative reciprocals of each other.
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the line y = 0 means that one of the sides of the triangle will be on the x-axis.
that line forms the bottom of the triangle.
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the intersection of the lines y = x, and y = -x + 5, will form the top of the triangle.
that point of intersection is found by solving y = x, and y = -x + 5 simultaneously.
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since y = x, you can substitute in the second equation to get x = - x + 5 which solves to be x = 5/2.
since x = 5/2, then y also equals 5/2, so the intersection point is (5/2,5/2)
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since the line y = x intersects the line y = 0 at the point (0,0), then (0,0) is also a point on the triangle.
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since the line y = -x + 5 intersects the line y = 0 at the point (5,0), then (5,0) is the third point of the triangle.
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if you label the triangle ABC starting from (0,0) and going around clockwise, then the 3 points on the triangle are:
A (0,0)
B (5/2,5/2)
C (5,0)
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angle B is the right angle because it is the intersects of the two perpendicular lines AB and BC.
recall that AB is on the line with the equation y = x, and BC is on the line with the equation y = -x+4, and that these lines are perpendicular to each other because the slopes of their equations are negative reciprocals of each other.
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to find the area, you need to use the formula A = 1/2 base * height.
since the right angle is at B, the base can be either AB or BC. we'll use AB.
since the base is AB, then the height is BC.
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all we need to do is find the lengths of AB and BC and we can then plug into the equation to get the area.
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the length of the line segment AB is square root of ((5/2 - 0)^2 + (5/2-0)^2).
this becomes square root of ((5/2)^2 + (5/2)^2).
this equals square root of (25/4 + 25/4) equals square root of (50/4) = 3.535533906....
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the length of the line segment BC is square root of ((5/2-0)^2 + (5/2-5)^2)
this becomes square root of ((5/2)^2 + (-5/2)^2).
this equals square root of (25/4 + 25/4) equals square root of (50/4) = 3.535533906....
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area of the triangle is 1/2 bh which equals 1/2 of .....
square root of (50/4) * square root of (50/4) which equals 1/2 of .....
(50/4) which equals ...
50/8 which equals 6.25.
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area of the triangle is 50/8 = 6.25.
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a graph of the triangle is shown below.
look below the graph for further comments.

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