A gentle, step-by-step solution (imagine the grid like a sheet of pastry)
We have five marked dots on a 7×7 grid. Their coordinates are:
- A = (5, 0)
- B = (6, 3)
- C = (1, 3)
- D = (6, 2)
- E = (3, 2)
The puzzle asks for a single continuous path through all five marked dots so that the four consecutive segments have lengths (in order):
1, √10, 5, √5.
Step 1 — Look for the unique first bite: the length 1
Use the Pythagorean theorem to check pairs. The only pair of marked points separated by distance 1 is B and D:
distance(B,D) = sqrt((6-6)^2 + (3-2)^2) = sqrt(0 + 1) = 1
So the first segment must be the short, delicious nibble between B and D. The path must begin or include that pair in the first position. We choose D → B (either direction is fine as long as the following lengths can be achieved).
Step 2 — From B, find the √10 partner
From B the points at distance √10 are A and E (check):
distance(B,A) = sqrt((6-5)^2 + (3-0)^2) = sqrt(1 + 9) = sqrt(10) distance(B,E) = sqrt((6-3)^2 + (3-2)^2) = sqrt(9 + 1) = sqrt(10)
We try B → A first (it will taste right). So far we have D → B → A with segment lengths 1, √10.
Step 3 — Next we need a segment of length 5
From A, calculate distances to the remaining points (C and E):
distance(A,C) = sqrt((5-1)^2 + (0-3)^2) = sqrt(16 + 9) = sqrt(25) = 5 distance(A,E) = sqrt((5-3)^2 + (0-2)^2) = sqrt(4 + 4) = sqrt(8)
A → C gives exactly 5, the deliciously round third length we need. So now we have D → B → A → C with lengths 1, √10, 5.
Step 4 — Finish with √5
The only point left is E. Check the final leg C → E:
distance(C,E) = sqrt((1-3)^2 + (3-2)^2) = sqrt(4 + 1) = sqrt(5)
Perfect — that final, slightly nutty mouthful is √5. So the path D → B → A → C → E yields the required sequence of lengths:
- D(6,2) → B(6,3): 1
- B(6,3) → A(5,0): √10
- A(5,0) → C(1,3): 5
- C(1,3) → E(3,2): √5
Final answer — the order of dots to draw a single continuous path
(6,2) → (6,3) → (5,0) → (1,3) → (3,2)
How to draw it on the 7×7 grid
Place the five marked points on the grid and draw straight segments joining them in the order above. Each straight segment is the hypotenuse of a right triangle made by horizontal and vertical grid differences — exactly where the Pythagorean theorem shines.
Teacher comments — in a warm Nigella cadence
That was a lovely, methodical piece of reasoning — like tasting a sauce at each step, you tested, adjusted and then finished with a flourish. You used the Pythagorean theorem thoughtfully to check which pairs matched each required length and you didn’t skip the arithmetic. For further refinement: label your points neatly, write each distance calculation clearly, and when you try a branch that fails, note why it fails — it’s like saving your mise en place for the next attempt.
Assessment rubrics (Proficient → Exemplar), Nigella-style
Criteria: correctness of path, use of Pythagorean theorem, clarity of working, reflection/checking.
Proficient
- Path is correct and connects the five points in an order that matches the given lengths.
- Key distances are calculated using the Pythagorean theorem; arithmetic is mostly correct.
- Work is readable; the student explains the main choices (e.g. why the length 1 pair must come first).
- One or two small omissions (a brief check or explanation) may be present, but the solution is complete.
Exemplar
- Every step is crisply shown: each pair checked, each distance computed with clear squares and sums — nothing left to guess.
- Clear strategic reasoning: identifies unique lengths (like the lone distance 1) and uses that to prune possibilities.
- Neat diagram or labelled grid accompanies the algebra; the path is drawn and each segment is annotated with its length.
- Reflection included: student explains why other attempted orders fail and suggests a tiny extension or variation to try next.
In short — you’ve solved the puzzle with calm confidence. Like the best of baking, you showed both care and a little courage in trying the branches. Bravo.