What is the area of triangle ABC plotted above?

What would be the quickest way of solving this? I created two additional right triangles as shown.


Based on the drawing I made blow, the red (3-4-5) and green (6-8-10) are both derivatives of 3-4-5 triangles.
Thus, you get the hypotenuses, can multiple them together and then by 1/2 to get the triangle area formula of 1/2 * base * height.
This yields 25. Answer D.

When I read the OA, however, I didn’t understand this necessarily: It was written that before even calculating the area, one must test if the original triangle is a right triangle. They did that somehow with the slope (appreciate ideas/answers), which I didn’t understand.

I just assumed it looks like a right triangle, but surely it is not explicitly stated nor is there a small square in the corner to indicate it is (as I indicated for the red and green triangle).
Thanks.

The area of the triangle when the vertices are specified is given by (In determinant form):

If the points are provided as (x1, y1), (x2,y2) and (x3,y3) -> the formula is attached along with the post.

If you expand the determinant

1/2 [ x1 (y2 * 1 - y3*1) - y1 (x2 * 1 - x3*1) + 1 (x2*y3 - x3 * y2)] and take the +ve value of it. (The value might come in -ve, but ignore that and take Modulus (||) of the value)

P.S. I have never seen this kind of question in GMAT, but this appeared in previous CAT

thanks kinjiGC, seems like a nice formula to have on tap

maybe not best best approach as it could backfire, but I just calculated the slope of the two apparent legs (AB and BC), confirmed they were negative inverses (perpendicular), then used those #'s to calculate distances and finally 1/2bh

I think this was, what I was originally asking about. And you are right. It actually takes only seconds to calculate the slopes of AB and BC.

In case they they are perpendicular, they should be negative inverses (reciprocals) as you say. So

Slope: (y2-y1)/(x2-x1)

Slope BC: (3-(-3)) / ((4-(-4)) = 6/8 = 3/4

Slope AB: 3-7/4-1 = -4/3


I also realized the following:
From point C, if you move three points up, four points to the right. You get to zero. If you move again three points up, four to the right you get to point B.
From point B, if you move four points up, and three points backwards you get to point A.
Four points up, three points left is the negative reciprocal of three points up and four points right. So it has to be perpendicular....(in practice this is the same as calculating the slopes but it may be a quicker way of seeing it in this particular case).

Hey guys, I like your solution. I did it the same way as you smurf, I also didn't test the angle.

Could anyone explain what would've happened if this wasn't a right triangle?

How would we calculate the sides / height?

Thanks!!

The original question above is a copy of GMAT Club's problem:

Check the following topic for different approaches possible to solve it: what-is-the-area-of-a-triangle-with-the-following-vertices-125983.html

Hope it helps.
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You can also quickly calculate the base and height by finding the distance between the points B & A and B & C:

B&A = Sqrt((4-1)^2 + (3-7)^2) = Sqrt(25) = 5
B&C = Sqrt((4-(-4))^2 + (3-(-3))^2) = Sqrt(100) = 10

Area = (5*10)/2

I calculated the lengths of all sides using the distance formula.

AB = 5
BC = 10
AC = \sqrt{125}

Since, AB^2 + BC^2 = AC^2, B must be a right angle.

So, area of the triangle = 1/2 * AB * BC = 1/2 * 5 * 10 = 25 (OPTION: D)

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