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gmat prep co-ordinate geometry

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by limestone » Mon Oct 11, 2010 12:30 am
Well, so big a picture.

My suggestion:

First, calculate the rectangle that enclose triangle QRP; rectangle OARB, where A is (0,4); B is (7,0). You can draw A and B in the coordinate plane for a clearer illustration.

You can figure out that the rectangle length and width is 7 and 4 respectively.

Area of rectangle OARB = Area of 4 triangles : QAR ,RBP,OQP and QPR

S(QAR) = 1*7 /2 =3.5
S(RBP) =4*3/2 = 6
S(OPQ) =3*4/2 = 6
S(OARB) = 4*7 = 28
Then S(QPR) =28- (6+6+3.5) =12.5
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by limestone » Mon Oct 11, 2010 12:53 am
there's also formulas to calculate triangle ABC area from its vertexes :

S = |Ax*(By -Cy)+ Bx(Cy- Ay) + Cx(Ay-By)| /2

Where Ax is the x coordinate of A vertex, Ay is the y coordinate for A vertex.

So on for B, C.

Apply to the above example: R(7,4) ;Q(0,3); P(4,0)

S = |Px(Qy-Ry) + Qx(Ry-Py) + Rx(Py-Qy)|/2

= |4(3-4) + 0(4-0) + 7(0-3)| /2 = |-4 + -21|/2 |-25|/2 = 12.5
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by lokesh r » Mon Oct 11, 2010 2:50 am
Using distance formula one can solve for the lengths of individual sides.

Triangle in the picture has sides 5,5,5sqrt(2).

Use area of triangle formula to get area as = 12.5
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by Rahul@gurome » Mon Oct 11, 2010 3:10 am
Area of the triangle when coordinates of the vertices are (x1, y1), (x2, y2), (x3, y3)
= {|x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)|}/ 2

= {4(4 - 3) + 7(3 - 0) + 0(0 - 4)}/2
= (4 + 21)/2
= 12.5 sq units

[spoiler]The correct answer is (A).[/spoiler]
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by hitech78 » Mon Oct 11, 2010 4:03 am
I am not very good with formala. Here is my approach.

When I look at the diagram it makes me suscpicious that it is a right angle traingle.
I confirmed that by quickly calculating slopes of QP and RP
slope of QP = 3-0 / 0-4 = -3/4
slope of RP = 4-0/7-4 = 4/3
Slope of QP * Slope of RP = -1 hence they are perpendicular.

by pathagores theorem calculate QP = 5 and PR = 5 (perfect pythagorean triplets 3,4,5)

Area = 1/2*base*height = 1/2*5*5= 12.5
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by GMATGuruNY » Mon Oct 11, 2010 4:03 am
Do only as much calculating as is necessary.

Estimating is the quickest way to solve this problem. See the attached .pdf. The area of the rectangle drawn around triangle PQR is 28. PQR takes up less than half the rectangle, so PQR < 14. Only answer choice A works: 12.5 < 14.

Onto the next question!
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BTG_area PQR.pdf
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by sumit.sinha » Tue Oct 12, 2010 5:40 am
limestone wrote:Well, so big a picture.

My suggestion:

First, calculate the rectangle that enclose triangle QRP; rectangle OARB, where A is (0,4); B is (7,0). You can draw A and B in the coordinate plane for a clearer illustration.

You can figure out that the rectangle length and width is 7 and 4 respectively.

Area of rectangle OARB = Area of 4 triangles : QAR ,RBP,OQP and QPR

S(QAR) = 1*7 /2 =3.5
S(RBP) =4*3/2 = 6
S(OPQ) =3*4/2 = 6
S(OARB) = 4*7 = 28
Then S(QPR) =28- (6+6+3.5) =12.5
OR
as an alternative, we can calculate the area of the trapezoid containing the triangle PQR. The sides OQ and Rx (x on x axis) will be the parallel sides of the trapezoid.
Area of trapezoid = 49/2 =24.5
Area of the two triangles to be subtracted = 6 + 6 = 12
Area of figure = 12.5
Cheers,
Sumit
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