Area and Perimeter

Double every length: what doubles, and what quadruples?

Worked example — Area and Perimeter

Perimeter	26.00 u
Area	40.00 u²
Scaled perimeter	52.00 u
Scaled area	160.00 u²
Compactness Q	0.744
Original	26.00 u · 40.00 u²

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Perimeter is linear; area is square 🖖

Scale every length by k and every boundary segment gains one factor of k, so perimeter scales by k. Area is a product of two lengths, so it gains k².

A larger pizza exposes the square 🖖

A 12-inch pizza is not twice a 6-inch pizza. Its radius is doubled, therefore its area is multiplied by four while its circumference is only doubled.

Area does not determine perimeter 🖖

The same area can have many perimeters. Stretch a rectangle while keeping ab fixed and its boundary grows, which is why area alone cannot tell you how much fencing is needed.

Problem solved in full

What 24.94 metres of edging could have held 6 steps
Load the trapezium preset: parallel sides of 10 m and 6 m, 4 m from one to the other, with the tool's u read as your metre. Work out the bed's area and the edging it takes. Then decide whether it is the shape or the size that is costing you planting area, and whether building the bed half again as large would fix it.
1. The two parallel sides are different lengths, so the first job is to find the single width that would do the same work. Cut the right-angled triangle off one end and set it against the other: the notch it leaves is exactly the shape it fills, and what is left is a rectangle 8 m wide and 4 m tall. The slant never enters the area at all.
  
  $\frac{a + b}{2} = \frac{10 + 6}{2} = 8 m, A = 8 \times 4 = 32.00 m^{2}$
2. The edging is all four sides, and two of them lean. The bed loses 4 m of width over its height, 2 m at each end, so each leaning side is the hypotenuse of a triangle 4 m tall and 2 m across: √20, a little under 4.5 m. The perimeter card reads 24.94 m.
  
  $\frac{a - b}{2} = 2, c = 4^{2} + 2^{2} = 20 = 4.4721, P = 10 + 6 + 2 c = 24.94 m$
3. The third card, compactness, reads 0.646. Q is defined as 4πA/P², which looks like a score with nothing behind it until you ask what P²/4π is on its own.
  
  $Q = \frac{4 π A}{P ^{2}} = \frac{4 π \times 32.00}{24.9 4 ^{2}} = 0.646$
4. It is the area of a circle made from the same edging. Unbend the 24.94 m, close it into a ring, and the radius is P/2π = 3.9700 m, enclosing 49.51 m². So Q is this bed's area divided by that circle's, and 0.646 says the trapezium keeps under two thirds of what its own edging can hold.
  
  $r_{\circ} = \frac{P}{2 π} = \frac{24.9443}{2 π} = 3.9700 m, A_{\circ} = π r_{\circ}^{2} = \frac{P ^{2}}{4 π} = 49.51 m^{2}$
5. The gap is 17.51 m² of bed you have paid for and cannot plant. Turn it round and it is the same sentence: a circular bed holding your 32.00 m² needs 20.05 m of edging, 4.89 m less, about a fifth of the roll.
  
  $A_{\circ} - A = 49.51 - 32.00 = 17.51 m^{2}, P_{\circ} = 2 π A = 2 32 π = 20.05 m$
6. Size does not help. Set k to 1.5 and the bed reaches 72.00 m² behind 37.42 m of edging, and Q does not move, because the k² the area gains is cancelled by the k² that comes from squaring the perimeter. Scaling this shape until it matched the circle's 49.51 m² would take k = 1.2439 and 31.03 m of edging, 6.08 m more than the circle needs for the same area.
  
  $k = 1.5 : P^{'} = 37.42, A^{'} = 72.00, Q^{'} = \frac{4 π k ^{2} A}{( k P ) ^{2}} = Q = 0.646; k = \frac{49.51}{32.00} = 1.2439 \Rightarrow P^{'} = 31.03 m$
Answer

32.00 m² behind 24.94 m of edging, and it is the shape that is costing you. The same edging closed into a circle holds 49.51 m², so 17.51 m² of bed is being bought and not received. Building it half again as large fixes nothing: at k = 1.5 the area reaches 72.00 m² and Q sits at 0.646, exactly where it started.

Because Q cancels every factor of k, it is the one card here that reports the shape rather than the size. It also has a ceiling. Nothing on this tool reaches Q above 1: the circle encloses more area than any other figure of the same perimeter. Jakob Steiner published the argument for that in 1838, by showing that any non-circular shape can be improved. It leaves open whether a best shape exists at all, and Weierstrass supplied the missing existence proof later. A regular polygon climbs the gap one side at a time: 0.605 at three sides, 0.785 at four, 0.865 at five, 0.907 at six, and never quite a circle.

$Q = \frac{A}{P ^{2} /4 π} = \frac{A}{A _{\circ}} \leq 1, Q (k) = Q \forall k > 0$

Learning path

The triangle rules, and where they come from

Leads to Similar triangles

By the 42Math team · Updated 2026-09-04

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

8 × 5 rectangle - Perimeter 26.00 u and area 40.00 u², beside the doubled copy at 52.00 u and 160.00 u². Both panels are drawn on the same unit grid, so the 40 squares in one and the 160 in the other are there to be counted rather than believed.
6 to 12 inch pizza - Radius 3 is the 6-inch pizza: 18.85 u of crust around 28.27 u² of topping. The scaled row is the 12-inch, 37.70 u and 113.10 u². Two 6-inch pizzas use that same 37.70 u of crust and give you half as much pizza.
trapezium - Parallel sides 10 and 6 with a height of 4 give an area of 32.00 u², because only the average of the two parallel sides does any work. The perimeter is 24.94 u, and compactness Q = 0.646 is the lowest of the three presets: this shape buys the least area per unit of edge.