Advanced Math/Similar Triangles
Expert: Paul Klarreich - 7/12/2009
QuestionHi,
I've got a maths problem:
Let ABC be a triangle. Two straight lines which are parallel to the side AC divide the triangle into three figures of equal area. In what parts is the side AB divided by the lines, if AB = 10 cm?
I know it has something to do with similarity, but I really have no idea as to where to start the problem.
Thankyou.
Answer
Questioner: Thomas
Country: Australia
Category: Advanced Math
Private: No
Subject: Similarity Problem
Question: Hi,
I've got a maths problem:
Let ABC be a triangle. Two straight lines which are parallel to the side AC divide the triangle into three figures of equal area. In what parts is the side AB divided by the lines, if AB = 10 cm?
I know it has something to do with similarity, but I really have no idea as to where to start the problem.
Thankyou.
....................................
Hi, Thomas,
Yes, it certainly does have something to do with similarity. In the attached figure, there are three triangles, black, purple, green and they are similar. (Yuk, I could have chosen better colors, but..)
There is this number, called the 'ratio of similitude' (r-s) that gives the ratios of corresponding sides. For example, if two triangles have r-s = 3, then each side of the big one is 3 times the corr. side of the little one.
AND THE AREAS ARE IN THE RATIO = the square of this r-s. So if r-s = 3, the big one is 9 times the area.
Now in the picture:
Triangle Black = Area I
Purple = Area I + II = 2*Black.
Green = Area I + II + III = 3*Black.
Now the three triangles are similar, with 'ratio of similitude' (r-s) equal to:
Purple side = sqrt(2) * Black side.
That means, each side of Purple = sqrt(2) * corresponding side of Black.
Green side = sqrt(3) * Black side.
That means, each side of Green = sqrt(3) * corresponding side of Black.
So AD/AC = sqrt(2) and AB/AC = sqrt(3)
And we have AB = 10. So I think now you can finish up.