Area of Triangular Region

I don't get 1/2, but I also don't get any of the answer choices. Where is the question from?

It helps to draw the diagram to scale. Triangle BCE and triangle DCE are actually exactly the same size (both have two sides of length 1, and one side of length DE = BE).

The angle at BCE must therefore equal the angle at DCE, and since these must add to 270 (because BCD is 90, and the angles in a circle add to 360), each is 135.

From here, there are a few different solutions, all easier to see if you draw the diagram:

Solution 1: To find the area of BCE, we need the base and the height. If we use BC = 1 as the base, we need to find the height - the distance from E at 90 degrees to the line we make by extending the base BC to the right. Extend BC to the right to a point F directly below E, and draw the triangle CEF, where the angle at F is 90 degrees. FE should be perpendicular to CF. This must be a 45-45-90 triangle, since the angle at DCE was 135, and we have divided it into a 90 degree angle and another angle. Since the hypotenuse of CEF is 1, each side is 1/sqrt(2), and 1/sqrt(2) is therefore the height of BCE. So the area of BCE should be

(1/2)*b*h = (1/2)*(1)*(1/sqrt(2)) = sqrt(2)/4

Solution 2: Look at triangle BCE. The angles, from the above, must be 135, 22.5 and 22.5 (it's isosceles). Reflect the triangle in the line BE: that is, draw an identical triangle BEG, with G above the line BE. Notice now that we have a parallelogram BCEG, with angles 45/135/45/135. The area of the parallelogram is base*height. The base is 1. The height can be found by drawing a line from G to BC, to make a 45-45-90 triangle; the hypotenuse BG is 1, and the height is 1/sqrt(2). Thus the area of BCEG is 1/sqrt(2) = sqrt(2)/2, and the area of triangle BCE is half of that, or sqrt(2)/4.

Solution 3: Connect BD. Consider the area of triangle BDE. Take the base to be BD; the length of BD is sqrt(2), by Pythagoras. The height is the distance from the centre of the diagonal (the centre of the square) to E. That's just CE + half of one diagonal of the square = 1 + sqrt(2)/2. So the area of BDE = (1/2)(sqrt(2))(1 + sqrt(2)/2) = 1/2 + sqrt(2)/2. But the area of BDE is just the area of half the square + the area of BCE + the area of CDE. The area of half the square is 1/2, and BCE and CDE are equal, so the area of BCE = (1/2)(sqrt(2)/2) = sqrt(2)/4.

malolakrupa wrote:Ian ,

I got this problem from the archives . The reaosn I say that the area is 1/2 is due to the fact that BC is 1 and CE is 1 . Hence BCE is an isoceles right angle triangle having base (BC) = 1 and height as CE(1) . Hence the answer 1/2

Let me know if i am wrong.

Thanks

Be very careful about making assumptions about angles. If two sides of a triangle are 1 and 1, then it's certainly isosceles, but it doesn't need to be a right triangle. Indeed, as you can probably judge from the diagram, the angle at C will be quite a bit larger than 90 degrees.

parallel_chase wrote: I drew a perpendicular from point C to line BE. These will give two right angled triangles and with hypotenuse of 1.

Angles will be 30, 60, 90, therefore following would be the sides of the triangle 1/2, sqrt3/2, 1.

And again, take care when making assumptions about angles. BCE should be 135, and when you draw the perpendicular from BE to C, you'll be cutting that angle precisely in half (since you're cutting the isosceles triangle BCE in half), so you should get a 22.5-67.5-90 triangle, and not a 30-60-90 triangle. Actually, if you use trigonometry, this leads to a fourth solution: the base BE is just 2*sin(67.5) and the height is cos(67.5), so the area is sin(67.5)*cos(67.5). Not a convenient solution for the GMAT though.

I'd add, as well, that at first glance I thought this problem was about 3-dimensional geometry, because of the perspectival diagram, which makes the square look like one face of a cube. It's not, of course- everything is intended to be in the 2-dimensional plane (and if it were in three dimensional space, there would be more than one correct answer, incidentally). I only mention this in case anyone else initially sees the diagram the same way I did at first.