GMAT PREP COORDINATE GEOMETRY ??

Consider Triangle QOP
OQ=3 OP=4 and angle QOP=90
So this is a 3-4-5 right triangle which gives QP=5.

Similarly drop a line from R on the x-axis. Name the point R'
In triangle RPR', again a 3-4-5 triangle gives RP=5.

Also, looking at these two triangles together will show that anle RPQ is a right triangle. Area with RP as height and QP as base = 0.5 * 5 * 5 = 12.5

There might be a easier approach that I am missing here.

Ans is A=12.5

let X be the point on the x-axis for point R [which has coordinates(7,0)].

Calculate area of trapezoid OQRX = 49/2

Calculate area of triangles OQP=6 and PRX=6.

Area of PQR=49/2 - 12 = 25/2.

STILL CONFUSED...

can u show how area of trapezoid=49/2....?

Let's use the trapezoid method.



First, see that the leftside triangle is a 3-4-5 right triangle.

The area is then:

1/2 * b * h
1/3 * 4 * 3 = 6



Second, see that the rightside triangle is also a 3-4-5 right triangle.

The area is then:

1/2 * b * h
1/3 * 3 * 4 = 6



Third, see that the overall quadrilateral is a trapezoid.

The area of a trapezoid is

[(b1 + b2)/2] * h

Where

b1: left side from (0,0) to (0,3); length = 3
b2: right side from (7,0) to (7,4); length = 4
h: bottom from (0,0) to (7,0); length = 7

So
[(3 + 4)/2]*7 = 49/2 = 24.5


Area of the triangle we want:

(Area of trapeziod) - (Area of leftside triangle) - (Area of rightside triangle)

24.5 - 6 - 6 = 12.5

Let's visualize a rectangle bounded by (0,0), (0,4), (7,4) and (7,0). This rectangle is 7 by 4 so has an area of 28.

The rectangle is made up of 4 triangles: the one we care about (PQR) plus 3 right triangles, each of whose hypotenuse is a side of triangle PQR.

One way to calculate the area of PQR is to subtract the areas of the 3 right triangles from that of the rectangle.

The triangle with hypotenuse PQ has legs of 3 and 4, so the area is 1/2(3)(4) = 6.

The triangle with hypotenuse QR has legs of 1 and 7, so the area is 1/2(1)(7) = 3.5.

The triangle with hypotenuse PR has legs of 3 and 4, so the area is again 6.

Therefore , the area of triangle PQR is 28 - (6 + 3.5 + 6) = 28 - 15.5 = 12.5

Well... When you combine coordinate geometry with algebra life becomes easy.

In this problem we are given the coordinates of all the three points. One can easily use the determinats to arrive at the answer.

Area = 1/2 | 0 3 1 |
| 4 0 1 |
| 7 4 1 |

=> 1/2 * [ 0 * ( 0 - 4 ) -3 *(4 - 7) +1(16 - 0) ]
=> 12.5

thats your answer.

The answer is 12.5

cjiang16 wrote:The answer is 12.5

First use the distance formula to calculate the length of the 3 sides. You will get the ratios of the sides to be 1:1: root 2. This is a 45-45-90 triangle.

The distance between PQ = Sq rt of (4-0)^2 + (0-3)^2 = 5.
Similarly distance PR = 5 and distance QR =root 50 = 5rt2
Therefore the ratios of the sides are 1:1:rt2
This is true for a 45-45-90 triangle with the hypotenuse at rt2.
Hence QR is the hypotenuse and angle QPR = 90
Now are = 1/2*PR*PQ = 1/2*5*5 = 12.5

ash g wrote: Also, looking at these two triangles together will show that anle RPQ is a right triangle.

How do you deduce this? QOP and RPR' may very well be right, but at slightly different angles, and then QRP isn't going to be a right one. What am I missing?