Hi, pls assist us. Thank you

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# Question

# Answer

a) Trivial. Each bulb is 2m apart from another bulb, the total distance must be an even number. Hence it’s a multiple of 2m.

b) If 24m distance, then corner A has a bulb, then every 2m to corner B, total is 13 bulbs. (At 0m, 2m, 4m, 6m… 20m, 22m, 24m.)

c) If you project AF and CD so that they cross at a point E’, you’ll find that the perimeter of ABCE’ is still the same as ABCDEF. It’s easier to move E to that corner E’ to make it a rectangle for counting purposes.

That means AB and CE’ (or AB and CD+EF) both have 13 light bulbs. That means 26 light bulbs are on the long edges of the rectangle ABCE’. The short edges (excluding corners) have the remaining 14 bulbs. So BC has 7 bulbs plus 2 corner bulbs already counted on the long edges. That’s 9 bulbs.

So the count is this: AB = 13 (incl 2 corner bulbs), BC = 9 bulbs (incl 2 corner bulbs), CE’ = 13 bulbs (incl 2 corner bulbs), E’A = 9 bulbs (incl 2 corner bulbs). But we shouldn’t count corner bulbs twice. So it’s 13+7+13+7 if we count the corner bulbs only on long edges, or 11 + 9 + 11 + 9 if we count the corner bulbs only on short edges.