Seems like the winds of luck are blowing your way - pity it's not exam season now!

It appears you're implying in your workings that:

1. 7m is the hypotenuse (with 2m and the rectangle's breadth being the opposite and adjacent sides of a right-angled triangle respectively)
2. x is the hypotenuse (with 6m and the rectangle's breadth being the opposite and adjacent sides of a right-angled triangle respectively)

But bearing in mind this is a 2-D diagram (not a 3-D pyramid as many students mistakenly assumed), Miss still couldn't find your right-angled triangles despite a prolonged period of staring at the diagram in a Zen-like state.

Please excuse Miss Loi if she has failed to see something that's very, very obvious as she's been pretty tired from all the late-night JB petrol pumping for this couple of days

In any case, this (being a Sec 2 question) is an exercise on Pythagoras' Theorem and simultaneous equations.

So the way to solve this is to form 4 simultaneous equations using each of the leashes as hypotenuses of right-angled triangles (please click on Someone's link for his well-crafted diagram & workings - don't worry there's no virus there!)

Hopefully younger readers can see from this example that simultaneous equations don't only come in pairs like what you normally see in your textbook. Sometimes situations might require you to form > 2 initial equations based on whatever info that's given in the question (don't be lazy!).

1)Breadth of rectangular area is found by:

=
=m

Therefore, length of rope needed=

=
= 9m

Whew, pmath is driving me crazy... :/

This working is WRONG (but answer's miraculously correct)! Please see correct workings in this link in comment #2 above! - Miss Loi

Here is the link.