GMATPrep 1 - Geometry question

Hey Amir,

Great question. One of my favorite things about coordinate geometry is that you can always find a right triangle, so that's the paradigm I use to approach these, and it usually works out pretty efficiently. My first take on this one was that, if that bottom angle (P) looks like a right angle, and if I could prove that it is I'll be able to use that base-height combination of those two lines that create it.

The advantage there is that those two lines both form right triangles with the x and y axes, so those lengths will be easy to calculate. They're actually both 3-4-5 triangles, so the length of each is 5.

Now, you can use the right triangle of that third side, which has a height of 1 and a length of 7, making its Pythagorean Theorem application:

1^2 + 7^2 = side^2
50 = side^2
Side = sqrt 50

Remember, my goal here was to prove that angle P was a right angle, which would make this problem significantly easier. If the isosceles sides are each 5, and the opposite side is sqrt 50, that fits with the 45-45-90 triangle ratio (x, x, sqrt2 * x), then we've proven that angle P is a right angle.

Because that's a right angle, we can use 5 as the base, 5 as the height, and the formula:

A = 1/2 bh = 1/2*5*5 = 12.5


When you're using coordinate geometry, use the fact that the whole coordinate grid is full of right angles - and therefore right triangles - to your advantage!

amirp wrote:Hi guys,

Can anyone tell me how you find 12.5 for this question? I can get it by estimation...

for example:

I estimated that 10.5 < area < 14 but is there a faster or more straight forward solution to this?

Thanks

I will give you a hint it has to do with slope of line..and you can calculate the perpendicular of the triangle...and then you can calculate the area...

Both approaches make total sense. Thank you guys.

amirp wrote:Hi guys,

Can anyone tell me how you find 12.5 for this question? I can get it by estimation...

for example:

I estimated that 10.5 < area < 14 but is there a faster or more straight forward solution to this?

Thanks

The area of the rectangle drawn around triangle PQR is 28. Since triangle PQR takes up less than half the rectangle, PQR < 14.
Only answer choice A works.

The correct answer is A.

GMATGuruNY wrote:

amirp wrote:Hi guys,

Can anyone tell me how you find 12.5 for this question? I can get it by estimation...

for example:

I estimated that 10.5 < area < 14 but is there a faster or more straight forward solution to this?

Thanks

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.

Great Explanation..

Quickest way to solve without much calculation!!

Thanks GMATGuruNY

Area of triangle when vertices are given,

area=| [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]/2 |

(x1,y1)=(4,0)

(x2,y2)=(0,3)

(x3,y3)=(7,4)

=[4(-1)+0+7(-3)]/2=|-12.5|=12.5