Perimeter

Relevant wiki: Challenging Composite Figures

Divide the figures into small squares with side length $a$ same length as the shorter edge as shown above.

We note that there are 33 (see notes) of these small squares making up the figure. Therefore, we have $33a^2 = 528$ , $\implies a = 4$ .

The perimeter is made up of $36a = 36 \times 4 = \boxed{144}$ (see notes).

Notes:

1. Each of the four "staircases" has 6 small square and the center block is a $3 \times 3$ . Therefore, the total number of small squares is $4\times 6 + 9 = 33$ .
2. Each staircase contributes $9a$ of perimeter. Therefore, the total perimeter is $9a \times 4= 36a$ .

Relevant wiki: Composite Figures

The area of the polygon is area of the yellow region plus area of the blue region. Since there are four yellow regions and one blue region, we have

$A=4[x^2+x(2x)+x(3x)]+(3x)^2$

$528=4(x^2+2x^2+3x^2)+9x^2$

$528=24x^2+9x^2$

$528=33x^2$

$x^2=16$

$x=4$

Therefore, the perimeter is

$P=4(9x)=4(9)(4)=$ $\color{plum}\large{\boxed{144}}$

Relevant wiki: Composite Figures

The perimeter can be anything from the interval (91.9, 145.3], and values from (128.7, 145.3) are reached in two cases each. The second interval can be extended to (112.6, 145.3) if we allow the "long" sides to be shorter than the "short" sides.

The obvious solution (and the only one where the sides line up as the picture suggests, making b=3a) is a=4, b=12, perimeter = 144. But its not stated anywhere that the sides do actually line up, and a perfectly acceptable solution is be, for example, a=2, b=12×sqrt(3), perimeter = 48(1+sqrt(3)), which is about 131.1. Jimmy Orioottis's example with picture illustrates the idea.

a has too be positive, but in the deformed case a=0 we get a square with side b = sqrt (528) = 4 sqrt (33), perimeter 16 sqrt (33) which is about 91.9.

As a grows, the perimeter grows and at a = sqrt (66/5) (about 3.63) it reaches maximum: b=4a and the perimeter is 40a, about 145.3.

As a grows further, the perimeter goes down. After just slight increase, a=4, we get the known solution of 144.

If we assume a<b (as they are referred to as short and long), the boundary case a = b = 4 sqrt (33) / 5 (about 4.60) takes the perimeter down to 28a = 112 sqrt (33) / 5, which is about 128.7.

Of course, we can increase a further, but then the "short" a will be longer than the "long" b. The perimeter goes down further. The deformed boundary case is b=0, which happens when a = sqrt (22) (about 4.69), then the perimeter is 24a = 24 sqrt (22), which is about 112.6.

Why does everyone assume that 3 short sides = 1 long side.

The diagram implies it but the question does not mention that.

Using a ratio of 4 to 1, I get a different perimeter of 145.3272...

So the perimeter is a function of the ratio of short to long.

We divide the figure and find we have three long bars each with 7 squares, that gives us $3 \times 7 =21$ squares. The top and bottom each consists of one block of six so that gives us another $6\times 2=12$ squares. Totally, there are 33 squares so each square has area $\frac{528}{33}=16$ . Then the side of each square is $\sqrt{16}=4$ . It can also be seen that the perimeter of the figure is equivalent to that of a square of side 9 times one small square so the perimeter is $9\times 4=36$ times the side of one small square which is $36\times4 =\boxed{144}$ .

Let shorter side be $y \implies$ longer side is $3y \implies 528 = 4(3y^2 + 2y^2 + y^2) + 9y^2 \implies 33y^2 = 528 \implies y = 4 \implies$

longer side $3y = 12 \implies$ Perimeter $P = 4 * (9y) = 4 * (36) = \boxed{144}$