Area of PQR

[alex.gellatly]

In the rectangular coordinate system below, the are of triangular region PQR is

12.5
14
10srt2
16
25

[theCEO]

Calculate the area of the parallelogram (draw line from R vertically down to the x axis. Call interception X)
Calcuate area of trainge - QOP
Calcualte area of triangle - RPX
Area of parallelogram - (area of the 2 triangles) = area of triangle PQR

[Anurag@Gurome]

If you are good with calculation and remembering formulas, then easiest way to solve this problem is to to remember the following problem.

The area of triangle formed by the coordinates (x1, y1), (x2, y2), and (x3, y3) is given by

• A = |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|/2

Hence, area of the triangle = |0*(4 - 0) + 7*(0 - 3) + 4*(3 - 4)|/2 = |0 - 21 - 4|/2 = 25/2 = 12.5

The correct answer is A.

[GMATGuruNY]

In the rectangular coordinate system below, the are of triangular region PQR is

12.5
14
10srt2
16
25

The area of the rectangle drawn around triangle PQR = 7*4 = 28.
Since triangle PQR takes up less than half the rectangle, PQR < 14.

The correct answer is A.

[alex.gellatly]

In the rectangular coordinate system below, the are of triangular region PQR is

12.5
14
10srt2
16
25

The area of the rectangle drawn around triangle PQR = 7*4 = 28.
Since triangle PQR takes up less than half the rectangle, PQR < 14.

The correct answer is A.

Is an inscribed triangle always less than half of the area like this?

Also, if there were more answer choices less than 14 is there a way to solve this with out the use of a formula?
Thanks

[GMATGuruNY]

Is an inscribed triangle always less than half of the area like this?

Also, if there were more answer choices less than 14 is there a way to solve this with out the use of a formula?
Thanks

The area of a rectangle = bh.
The area of a triangle = (1/2)bh.
A triangle WITHIN a rectangle cannot have a greater bh than the rectangle.
Thus, the greatest possible area of a triangle inscribed inside a rectangle = (1/2)(base of the rectangle)(height of the rectangle) = 1/2 the area of the rectangle.
Here is such a triangle:

Since the triangle and the rectangle have the same base and height, the triangle = 1/2 the rectangle.

The problem at hand does not offer such a triangle:

Since ∆PQR does not have the same base and height as rectangle STRU, the area of ∆PQR must be LESS than 1/2 the area of the rectangle STRU.

Here's an easy way to calculate the area of ∆PQR:

Area of PQR = rectangle STRU - (∆SQP + ∆QTR + ∆RUP).

Rectangle STRU = 7*4 = 28.
∆SQP = (1/2)(4*3) = 6.
∆QTR = (1/2)(1*7) = 3.5
∆RUP = (1/2)(3*4) = 6.

Thus, ∆PQR = 28 - (6 + 3.5 + 6) = 28 - 15.5 = 12.5.