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In the rectangular coordinate system above, the area of tria
r08
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Greetings,
Forgive me if this seems too simple.
I searched for the question in the forums but could not find it.
Could you help explain the answer?
(I calculated 14 but I knew it was wrong)
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There are several ways to do this and I'm going to show you the two minute solution.

The question is asking us to calculate the different sides of the triangle, identify the base and height, and calculate.

These problems can get time consuming so the important thing is to work with what they give you.




By looking at the points of the yellow triangle, we see that it contains the points (0,3) and (4,0). This means its a 3,4,5 triangle and we know one side of the mystery triangle = 5

By drawing an imaginary line to the point (7,0) from (7,4), we can see that we have another 3,4,5 triangle and the other side of the mystery triangle is 5.

(BxH)/2 = 25/2 = 12.5


You can always use the distance formula too which is
In the rectangular coordinate system above, the area of tria
r08
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Thank you!
Great explanation.
Ron Purewal
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Joined: 08 Oct 2007
Posts: 1949

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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
Priyanka
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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)
Ron Purewal
MGMAT STAFF

Joined: 08 Oct 2007
Posts: 1949

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Priyanka wrote:
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)


yes, this works too, if you have this sort of artillery in your arsenal.
In the rectangular coordinate system above, the area of tria
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