This problem comes in many guises. One can be found here:
http://math.stackexchange.com/questions/105084/a-line-which-bisects-two-sides-of-a-triangle-is-parallel-to-the-third
Naturally, the area of the small triangle CBE is 1/4 the area of CAD, since it has half the base and half the altitude.
The diagram: it is a large triangle labelled ACD(AC a line horizontal, then AD and CD joining at the tip going down-so the triangle is "upside down" with the base up) with a line drawn that connects point E, which intersects segment CD at the middle(a median), with point B, which intersects line AC at the middle.
1. Question: In the figure above, line AD is parallel to line BE, and the area of quadrilateral ABED is three times the area of triangle BCE. What is the value of AD/BE(aka the two sides parallel)?
(I am inquiring about a general concept/formula/rule about a certain type of question through this question that I am stuck on- but I am unable to provide an image of the problem's diagram on this website so I hope someone can understand what I am trying to ask through my descriptions.
I think it is something about the ratio of sides,in this case 1/2(the square root of the area- a rule?) being the square root of the ratio of area?(which is 1/4 says my teacher, but why is it 4 and not 3 when one triangle's area is x and the other 3x?)The answer is two but I do not understand it.
3 answers
Wait, I have a logic problem. How does it equal 1/4 again? Could you please explain Steve?
the large triangle has base b and height h. So, its area is bh/2
The smaller triangle has base b/2 and height h/2, so its area is (b/2)(h/2)/2 = bh/8 = 1/4 (bh/2)
The smaller triangle has base b/2 and height h/2, so its area is (b/2)(h/2)/2 = bh/8 = 1/4 (bh/2)
2 answers