In the rectangular coordinate system above, the area of tria

[jyothi h]

the solution is beautiful, but it also relies on the fortuitous fact that the lines QP and PR are perpendicular. you didn't explicitly acknowledge this fact in your otherwise wonderfully detailed solution, so i'm not sure you're aware of it; the only reason that the formula (1/2)(b)(h) works in this case is because PQ just so happens to be perpendicular to PR.
for instance, if you move R from (7, 4) to (8, 3), then your green triangle would still be 3-4-5, but in that case the formula (1/2)(5)(5) would yield the wrong area.

here's a much more universally applicable way to find the area of triangles like these:
surround the triangle with a rectangle, and then subtract out the unwanted area.
this will always work, regardless of the shape of your triangle, because the unwanted parts of the rectangle will always be either rectangles or right triangles with sides perpendicular to the axes (both of which have areas that are extremely easy to calculate).
in this problem:
surround the triangle with a rectangle whose vertices lie at O(0, 0), A(7, 0), R, and B(0, 4).
the area of the rectangle is (7)(4) = 28.
the area of unwanted triangle QBR is (1/2)(1)(7) = 3.5.
the area of unwanted triangle RAP is (1/2)(3)(4) = 6.
the area of unwanted triangle POQ is (1/2)(4)(3) = 6.
28 - 3.5 - 6 - 6 = 12.5

========================
Hi Ron,

I guess the reason he(first reply post) is multiplying (1/2 *b*h) above , is because the ratio is the sides is 5:5: 5sqrt(2) ... implies it is a right isoceles triangle. with right angle at point p , which gives us the base and height as 5 and 5. Also you are right about the point that we cannot apply the same method when u move the point R to (8,3) , cos this would not give us the QR length to be 5 sqrt(2) , but rather gives us the length of 8. This would still make it possible to calculate the area of triangle, cos R and Q have the same y axis which gives us the base of triangle 8 and height 3.
Not sure if I am missing something. correct me if I am wrong , cos many a times i tend to mess up with my concepts, esp in geometry.

Thanks,

[jyothi h]

the solution is beautiful, but it also relies on the fortuitous fact that the lines QP and PR are perpendicular. you didn't explicitly acknowledge this fact in your otherwise wonderfully detailed solution, so i'm not sure you're aware of it; the only reason that the formula (1/2)(b)(h) works in this case is because PQ just so happens to be perpendicular to PR.
for instance, if you move R from (7, 4) to (8, 3), then your green triangle would still be 3-4-5, but in that case the formula (1/2)(5)(5) would yield the wrong area.

here's a much more universally applicable way to find the area of triangles like these:
surround the triangle with a rectangle, and then subtract out the unwanted area.
this will always work, regardless of the shape of your triangle, because the unwanted parts of the rectangle will always be either rectangles or right triangles with sides perpendicular to the axes (both of which have areas that are extremely easy to calculate).
in this problem:
surround the triangle with a rectangle whose vertices lie at O(0, 0), A(7, 0), R, and B(0, 4).
the area of the rectangle is (7)(4) = 28.
the area of unwanted triangle QBR is (1/2)(1)(7) = 3.5.
the area of unwanted triangle RAP is (1/2)(3)(4) = 6.
the area of unwanted triangle POQ is (1/2)(4)(3) = 6.
28 - 3.5 - 6 - 6 = 12.5

========================
Hi Ron,

I guess the reason he(first reply post) is multiplying (1/2 *b*h) above , is because the ratio is the sides is 5:5: 5sqrt(2) ... implies it is a right isoceles triangle. with right angle at point p , which gives us the base and height as 5 and 5. Also you are right about the point that we cannot apply the same method when u move the point R to (8,3) , cos this would not give us the QR length to be 5 sqrt(2) , but rather gives us the length of 8. This would still make it possible to calculate the area of triangle, cos R and Q have the same y axis which gives us the base of triangle 8 and height 3.
Not sure if I am missing something. correct me if I am wrong , cos many a times i tend to mess up with my concepts, esp in geometry.

Thanks,

I see you have acknowledged that it is right angled and which is why the method works and went about explaining the universal way.... Sorry I jus read the main explanation , in a rush . My mistake.

[tim]

no problem. let us know if there are any further questions on this one..