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| In the rectangular coordinate system above, the area of tria |
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r08
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| In the rectangular coordinate system above, the area of tria |
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r08
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Thank you!
Great explanation. |
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Ron Purewal
MGMAT STAFF
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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 |
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Priyanka
Guest
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Hi the area can be calculated by solving a 3 x 3 matrix
= 1/2 | 1 x1 y1 | | 1 x2 y2 | | 1 x3 y3 | where A (x1,y1) , B(x2,y2) and C (x3,y3) |
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Ron Purewal
MGMAT STAFF
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yes, this works too, if you have this sort of artillery in your arsenal. |
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| In the rectangular coordinate system above, the area of tria |
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