another triangle qn

many thanks for trying to help

sorry to disturb u guys.

Hi,

Please let us know what your difficulty is, rather than expect a full solution. Thanks.

Cheers,
Wen Shih

hi thanks for replying

im just completely lost on what to do. Sorry im really weak.

my method is 25 divide by 9 and then find each of the line segment. e.g for the second line segment = 25 divide by 9 multiply by 8

Is this correct?

similarly (no pun intended, please), make use of similar triangles.

the answer should be 125.

can give some clue or not pls SBS261P

Hi,

Your diagram is no longer there.

I'll try to remember as follows:

1. We have a triangle with vertices A, B and C.

2. Join the points along AB to the points along AC such that each line segment is parallel to BC.

3. BC has a length value, say x.

This is a non-routine problem, so it is difficult.

Take the first line segment, say D1 E1. By similar triangles,

BC/AC = (D1 E1)/(A E1) from which we obtain

D1 E1 = (BC/AC) x (A E1).

Note that any subsequent line segment Di Ei has this form.

Now we have to find the sum of all line segments, i.e.,

(D1 E1) + (D2 E2) + ... + BC = ?

Try it. Thanks.

Cheers,
Wen Shih

my solution is actually

first length is 1/9 x 25.

second length is 2/9 x 25.

so on and so forth.

so add together you have 45/9 x 25 = 125.