<p>ABC is an equilateral traingle with sidees of 2 cm. BC is extended its own length to point D and poiint E is the midpoint of AB. ED meets AC at F. find the area of quadrilateral BEFC in square cm in simplest radical form.</p>
<p>Can you PM me the answer?</p>
<p>My guess would be three square root of 3 divided by four , or 3[3^(1/2)]/4. I'm probably wrong though, as I'm no math genius and I am terrible at hard math problems. Just draw a picture.</p>
<p>damn...i'm stumped..that's a tough problem, i feel like i'm missing something, some type of geometry rule that fills the hole..hm</p>
<p>You extend BA by 2 to some point (lets call it x). Then connect x and D. This should give you an equilateral triangle with side 4. Notice that XD and AC are parallel. By some random geomety theorem, angle EFA must equal angle BDX, which is part of an equilateral triangle and therefore 60 degrees. Since you now know two angles of triangle AXF to be 60, the next must be 60 and it is equilateral, and it has a side 1. Use area formulas for equilateral triangles and some subtraction to get the answer. area of AZF equals sqrt3/4. triangle ABC equals sqrt3. Subtract to get quad ZBCF.</p>
<p>Dude, where the **** did you get this problem from?</p>
<p>maybe i'm drawing this wrong but doesn't it all depend on where D is?
it seems like there isn't enough information</p>
<p>it would be 1.5 cm because F is the midpoint of AC. If you then draq 4 triangles within the triangle each having sides of 1 cm then three would make up BEFC. Thus equaling 1.5. (1/2)bh or .5cm for each triangle *3</p>
<p>your formula for area is wrong. 1 isn't the height of the triangle. I'm pretty sure my solution is right, except the last two words should say "quad. BCEF". The position of D can't vary since the problem says it must be a one length extension of BC. This means that you pretty much just stick another 2cm to the end of BC. Regardles of which of the two sides you put it on, the drawing will look the same and you will get the same area.</p>