Quadrilaterals and Polygons: The Angle Family

The square that would not close

I was sketching a shape I was sure existed: a quadrilateral with four fat, blunt corners, each one a little past 90 degrees. Squares have 90, rectangles have 90, why not all four past 90? I drew it, and the left edge kept sliding off the paper before it met the bottom edge. Not carelessness. It simply could not close.

Four corners of 100 degrees need 400 degrees of turning, and a quadrilateral only ever hands out 360. Forty degrees over budget. My pencil was right and my shape was not, and the reason turned out to be a rule that covers every polygon ever drawn. Corners do not cost 90 each. They cost whatever you make them cost, out of a fixed allowance. Spend too many wide ones and the sides cannot meet.

Most angle teaching stops at the sum, then moves on. Knowing the total is how you find a missing angle, which is covered in how triangles hold themselves rigid. But the total also caps how many corners of a given kind a shape can possibly have. That is a different question, and the answers are stranger than the formula suggests.

The budget: (n − 2) × 180

A polygon with n sides has an interior angle total of (n − 2) × 180. A triangle gets 180, a quadrilateral 360, a pentagon 540, a hexagon 720. The proof is the one from the triangle article: cut a corner off and you turn an n-gon into an (n−1)-gon, and the piece you removed is a triangle carrying 180 degrees. Peel until you reach a triangle and you have counted (n−2) triangles worth of turning.

That total is a fixed allowance, not a guideline the shape tries to respect. Which gives a counting question nobody usually asks: if the budget is fixed and each corner draws from it, how many corners can I afford to be wide?

The headcount rule

Every angle wider than 90 spends more than its share. So take k angles that are each over 90, and the rest still have to be real corners, each one under 180. For the wide ones to fit, their combined share cannot reach the whole budget:

k × 90 < (n − 2) × 180, so k is at most 2n − 5

Then cap it at n, because a shape cannot have more corners than it has sides. Together they give the table I wish a textbook had printed:

• Triangle: at most 1 angle over 90. Two would need more than 180, leaving nothing for the third corner.
• Quadrilateral: at most 3.
• Pentagon: at most 5, and all five can be obtuse — the regular pentagon does it at 108 each.
• Hexagon and beyond: the formula allows 2n − 5, but the shape runs out of corners first, so every corner can be obtuse. A regular hexagon sits at 120 each.

Watch what happens between those rows. A triangle cannot afford two blunt corners. A quadrilateral can afford three but not four. A pentagon can afford all of them. The limit climbs by two with every new side, until it hits the ceiling.

That doubling is not a coincidence. Adding a side adds 180 degrees to the budget, and a wide corner consumes more than 90, so one extra 180 of room buys two more wide corners. It is also why the triangle feels so tight — 180 to spend, and anything wide eats more than half of it.

Wide corners stop being special

Add sides generously and obtuse angles become ordinary. On a hexagon every corner can be 120 degrees and nothing is remarkable. On a triangle a single 120 corner is already the most extreme thing the shape can do. Same angle, opposite situations, entirely because of how much budget surrounds it.

The quadrilateral's three impossibilities

The four-sided case earns its own list, because these three catch people out.

The naming ladder

Once you see the budget, the names stop being a memorisation list. Each answers one question: how many pairs of sides promise to stay parallel?

• Trapezoid: exactly one pair of parallel sides. The most corners going spare.
• Parallelogram: two pairs, so opposite angles match. Two obtuse corners, two acute, always.
• Rectangle: a parallelogram whose angles reach 90 all round.
• Rhombus: a parallelogram with all four sides equal. Angles can be any pairs, so it may have zero, two, or four right angles.
• Square: where the two ladders meet.
• Kite: two pairs of adjacent equal sides.

The kite is the one that gets mis-sorted. In a parallelogram, opposite angles match. In a kite, the two angles between the unequal sides do not — put the long pairs meeting at the top and bottom, and the left and right corners come out equal while the top and bottom are free to differ. That single unequal pair is why a kite can land on zero, one, or two blunt corners.

This is not the tiling question

It is easy to confuse this with a related one. When polygons cover a floor, the angles at a shared meeting point must total 360, and that decides which shapes can tile at all — the rule behind why only three regular shapes cover a floor. Two questions, two different 360s. Tiling asks whether corners from several shapes can meet; the corner budget asks whether one shape's corners can pay for themselves. A pentagon that tiles beautifully still cannot have two blunt corners of its own.

Try it on these three

Do not calculate anything. Just draw and see when the edges refuse to meet.

1. Draw a quadrilateral and push all four corners past 90. Notice exactly when the shape stops closing.
2. Draw a pentagon and make all five corners obtuse. The regular pentagon does it at 108 each — find that yourself and the constraint stops feeling arbitrary.
3. Give a quadrilateral three right angles. You cannot. Give it a fourth and you have drawn a rectangle without deciding to.

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