gmat prep coordinat system

After a few seconds of looking at the picture I have the strong feeling that we are dealing with a right triangle, so I'm going to try an follow this lead. How do you prove that a triangle is right? Well, you demonstrate that a^2 = b^2 + c^2. So let's find out the lengths of PQ, PR and RQ.
The distance between two points (A and B) can be calculated using the following formula: sqrt[(ya - yb)^2 + (xa - xb)^2], with ya and xa the coordinates of point A and xb and yb the coordinates of point B.
Using this very long but logical formula we can calculate that QP = PR = 5 and QR = 5sqrt(2). Since these three lengths satisfy Pythagoras's equation, then my hunch was right: we are dealing with a right triangle. Its surface will be QP*PR/2 = 25/2 = 12.5.

Answer: A.

Answer is A.

Sides are 5,5, Sqrt(50). This is a right triangle.
SO

Area = 1/2 (5*5 ) = 12.5

The right triangle formed by the x axis, y axis and line QP can give you the length of QP

0,0 to Q = 3
o,o to P = 4
so the hypotenuse QP = 5 and angle 0,0 P Q = 30 (opposite the smallest side of a 30/60/90 right triangle.

Now, we can do the same for PR by drawing a straight line from R to the X axis to create a right triangle with P R and point 7,0.
P to 7,0 = 3
7,0 to Q = 4
We have another 30/60/90 right triangle, so PR = 5
and the angle 7,0 P R = 60 (across for the longer side of a 30/60/90 right triangle).

So, now we can calculate the angle of QPR as it is 180 -30 -60 = 90
So QPR is a 45, 45, 90 right triangle, making side QR=5 sqrt2

But this math is looking messy, so realize that a 45-45-90 triangle is just a perfect square cut in half. The area of a square with side 5 = 25 so the area of triangle PQR = 25 / 2 = 12.5

Answer is A

-Carrie

Brent Hanneson wrote:It was fortunate that the triangle in question was a right triangle. If, for example, point P were at (5,0) then we would need another strategy.

One such strategy would be to draw a rectangle around the triangle (as shown below) and then subtract from the rectangle's area (28) the areas of the 3 right triangles that surround the tirangle in question.

You can see that this strategy will work with any triangle, right triangles and non-right triangles ("wrong" triangles? :D )

Great Solution Brent!

Though its not accurate but the easier way to solve this question would be

Area of triangle = 1/2 (b*h)

Now since the slant of the traingle is only 1 unit (difference between the y cordinates of point Q and R) we can consider the length of RQ as base and the height to be the average of y co-ordinates of Q and R

from above base = 7 and height = (4+3)/2 = 3.5

So Area = 1/2*7*3.5
= 12.25 (aaprox 12.5)