Cross-multiply down, cross-multiply up, subtract and halve — any polygon's area falls out of its vertex list, no height required.
The problem
What is the area of the triangle with vertices (1, 1), (5, 2) and (3, 6)?
Shoelace: downs 1·2 + 5·6 + 3·1 = 35, ups 1·5 + 2·3 + 6·1 = 17, area = ½|35 − 17| = 9. A forgets the ½ — the single most common shoelace slip.
Go around the triangle one way — clockwise or counter, either works.
(1, 1) → (5, 2) → (3, 6) → (1, 1)
the loop closes back on the start
The name comes from the picture: writing the pairs in two columns and drawing the cross-multiplications makes a lacing pattern, like a shoe. The method works for any polygon, not just triangles — same loop, more vertices.
downs: 1·2 + 5·6 + 3·1 = 35
each x times the NEXT point's y
ups: 1·5 + 2·3 + 6·1 = 17
each y times the next point's x
area = ½ |35 − 17| = ½ (18)
the absolute value forgives a wrong direction
= 9
the triangle's area
Going around the other way just swaps which sum is bigger — the absolute value absorbs it. What it cannot absorb is a forgotten ½.
Prefer pictures? Draw the smallest axis-aligned box around the triangle — here 4 wide by 5 tall, area 20 — and cut away the right triangles in the corners: 2, 4 and 5. What remains is 20 − 11 = 9, the same area. The box method is slower but self-explanatory; shoelace is the exam-speed tool.
The triangle with vertices (1, 2), (5, 4) and (3, k) has area 6. Find k.
downs: 1·4 + 5·k + 3·2 = 10 + 5k
carry k like a number
ups: 2·5 + 4·3 + k·1 = 22 + k
same loop, other diagonal
½ |4k − 12| = 6 → |4k − 12| = 12
clear the ½ first
4k − 12 = 12 or 4k − 12 = −12
an absolute value opens two ways
k = 6 or k = 0
one vertex on each side of the base
Dropping the minus case loses an answer, and the question may ask for the other one ("k < 1"). Forgetting the ½ before solving gives |4k − 12| = 6, a third value that is never right.
Area 0 is a test for collinear points: here 4k − 12 = 0 gives k = 3, and (3, 3) does lie on the line through (1, 2) and (5, 4). Three points with downs = ups make no triangle at all.
Negatives everywhere. The triangle (−4, −1), (2, −3), (−2, 3).
downs: (−4)(−3) + 2·3 + (−2)(−1) = 12 + 6 + 2 = 20
each x times the next y
ups: (−1)(2) + (−3)(−2) + 3(−4) = −2 + 6 − 12 = −8
each y times the next x
area = ½ |20 − (−8)| = ½(28)
minus a negative adds
= 14
box check: 36 − 6 − 12 − 4 = 14 ✓
What is different: a sum can come out negative. Keep its sign, then subtract it.
Treating the ups as 8 gives ½|20 − 8| = 6. The ups total is −8, and 20 − (−8) = 28.
Corners listed out of order. A quadrilateral has corners P(1, 0), Q(3, 4), R(5, 2), S(0, 3). Find its area.
sketch: the boundary runs P → R → Q → S
go round the edge, not P, Q, R, S
downs: 1·2 + 5·4 + 3·3 + 0·0 = 31
(1, 0) → (5, 2) → (3, 4) → (0, 3)
ups: 0·5 + 2·3 + 4·0 + 3·1 = 9
the other diagonal
area = ½ |31 − 9| = 11
box check: 20 − 1.5 − 4 − 2 − 1.5 = 11 ✓
What is different: the letters are not in boundary order. In the printed order P, Q, R, S the loop crosses itself and gives ½|25 − 23| = 1.
Worked: the quadrilateral (0, 1), (4, 0), (5, 4), (1, 5), taken in order.
(0, 1) → (4, 0) → (5, 4) → (1, 5) → (0, 1)
four vertices, first one repeated
downs: 0·0 + 4·4 + 5·5 + 1·1 = 42
four products now
ups: 1·4 + 0·5 + 4·1 + 5·0 = 8
and four the other way
area = ½ |42 − 8| = 17
it is a square of side √17 ✓
The vertices must follow the boundary. If a figure labels them out of order, sketch them first; looping across a diagonal gives a bow-tie and a wrong area.
Worked: the triangle bounded by y = 2x, x + y = 6 and the y-axis.
y = 2x and x + y = 6: 3x = 6 → (2, 4)
each pair of lines meets at a vertex
y = 2x meets the y-axis at (0, 0); x + y = 6 at (0, 6)
put x = 0
base 6 (along the y-axis), height 2
the height is the x of (2, 4)
area = ½ · 6 · 2 = 6
shoelace on the three points agrees
Expect a quadrilateral or pentagon from a figure, an unknown vertex from a given area, three points that turn out collinear, or a region cut off by lines. Shoelace handles all of them once you have the corners in order; when one side lies on an axis, ½ × base × height is a ten-second cross-check.
Find the area
What is the area of the triangle with vertices (−2, 1), (3, −1) and (4, 3)?
Loop (−2, 1) → (3, −1) → (4, 3) → (−2, 1). Downs: (−2)(−1) + 3·3 + 4·1 = 15. Ups: 1·3 + (−1)·4 + 3·(−2) = −7. Area = ½|15 − (−7)| = ½(22) = 11. B forgets the ½. A loses the sign of the ups total and works ½|15 − 7| = 4. C is the whole 6-by-4 box around the triangle, before the corner triangles 5, 2 and 6 are cut away — and 24 − 13 = 11 confirms the answer.
