Geometry Question

[nitrocan]

Sorry for my lousy drawing. Here's the question:

ABCD is a square.
The area of the triangle DEF is 2. Then what's the area of the triangle EFC?

The FEC angle is 45.

Important (Forgot to draw):
-|DE| = |EC|-

 

[Chuberchuckee]

Is this drawn to scale?

 

[nitrocan]

Don't even try measuring it with a ruler
No, I just drew it right away.

 

[Poke]

I wish I could learn this NOW... I haven't taken geometry yet, but it seems simple give a little more info on how to do.

I wish there was an online math class. Or book.

 

[Karthik]

Answer = sqrt(2)

Not a very elegant one, but still a solution.
Highlight for solution:
angle(CAB)=angle(BEC)=45°. So points A,B,C and E lie on a circle with AC as the diameter. Thus DB is also a diameter of the circle.

Since C and E are on the circle with DB as diameter, angle(DCB)=angle(DEB)=90°

Now once we know that angle(DEB)=90°, the question is pretty simplified.

Give that area(triangle DEF)=2. So
(1/2)*DE*EF=2.
=>EF=4/DE=4/EC

Now drop a perpendicular from point F to EC. Call it FG.

Area(triangle CEF)=(1/2)*EC*FG
Also, FG=EF*sin(45°)=(4/EC)*(1/sqrt(2))
So area(triangle CEF)=(1/2)*EC*(4/EC)*(1/sqrt(2))=sqrt(2)

 

[nitrocan]

Answer = sqrt(2)

Not a very elegant one, but still a solution.
Highlight for solution:
angle(CAB)=angle(BEC)=45°. So points A,B,C and E lie on a circle with AC as the diameter. Thus DB is also a diameter of the circle.

Since C and E are on the circle with DB as diameter, angle(DCB)=angle(DEB)=90°

Now once we know that angle(DEB)=90°, the question is pretty simplified.

Give that area(triangle DEF)=2. So
(1/2)*DE*EF=2.
=>EF=4/DE=4/EC

Now drop a perpendicular from point F to EC. Call it FG.

Area(triangle CEF)=(1/2)*EC*FG
Also, FG=EF*sin(45°)=(4/EC)*(1/sqrt(2))
So area(triangle CEF)=(1/2)*EC*(4/EC)*(1/sqrt(2))=sqrt(2)

That solution is the one I had in mind, but you don't have to do the perpendicular point thing.

Area of an ABC triangle: |AB|*|AC|*sinA / 2

Well time for another one? This one didn't take as long as I thought it would.

 

[qqwref]

Here's an alternate, arguably more efficient one based on Karthik's solution.

Highlight: Observe that angles BAC, BDC, and BEC are all 45 degrees; this means that points A, B, C, D, and E all lie on a circle. Call the center (which is also the center of the square) O.

Now since |DE| = |EC|, the triangle on top is isosceles, so E is actually midway between C and D on the circle's circumference. This means angle EOB is 135 degrees, and because EOB is isosceles we have angle EBO is (180-135)/2 = 22.5 degrees. Similarly, angle COB is 90 degrees, and because COB is also isosceles we have angle CBO is (180-90)/2 = 45 degrees. This means that angle EBC is 22.5 degrees.

We're almost done. The ratio |FC|/|CB| is tan(22.5) = sqrt(2) - 1; since CB = CD, the ratio |DF|/|FC| is |CD|/|FC| - |FC|/|FC| or 1/(sqrt(2) - 1) - 1 = sqrt(2).

Finally, the ratio of the areas of triangle DEF to triangle FEC is the same as the ratio of the lengths |DF| to |FC|. Therefore the area of triangle FEC is
2 * (|FC|/|DF|) = 2/(sqrt(2) = sqrt(2).

 

[TheBB]

Well time for another one? This one didn't take as long as I thought it would.

Leave the geometry out, I might join.