We can vary the shift between the two tilings, to get other equivallent dissections, for instance a symmetric dissection, still with 4 pieces.

However it is less obvious to construct the side of equivallent triangle.

The cross area is 5a² and naming s the triangle side, triangle area is (s/2)²√3.

We deduce (s/2)² = 5a²/√3 and a construction :

Copy AS = 5 AB, and draw line ST at 30°, AT = 5 AB/√3.

The circle with diameter HT intersects line AB in P, AP² = AH.AT = (s/2)²

Construct the equilateral triangle AMN, then the parallel UV to AM in N.

Another dissection, more complex and mainly using a piece "holding at a thread" is
obtained by the stripe method. The stripe from dissecting the cross is less obvious :

The cross is cut in half by UV, and the two half cross build the stripe.

We then overlay these two stripes in such a way that the "center point" of the cross stripe is at midpoint M of triangle's side AB, and the vertex S of the cross lies on the edge of the triangle stripe :

This gives a construction of S :

The cross area equals that of a square with side MS

Triangle area is s.h/2. Hence MS² = MM'.h/2

Copy MH = h/2. The perpendicular in H and the circle with diameter MM' intersect in K.

MK is the searched value, construct MS = MK.

Construct V with SV = MS/2 and SV_|_MS and deduce the drawing of the cross, then the construction of the cuts.

These are defined solely by the common parts of the two stripes. All that is outside is irrelevant.

Really, just one half of this common part is used, as the common part involves two triangles and two crosses.

Note that for tiling the whole plane, adjacent stripes should be shifted : this can be seen on the triangle dissection.

The important point is here the location of P with respect to triangle side AB !

We have to prove that angle AMV is > 45°. P is then strictly inside the triangle,
and the two purple pieces are only one.

<)AMV = AMS + SMV = AMM' - SMM' + SMV

AMM' is of course 60°

SMM' = arcsin(h/(2MS)) = arcsin(^{ 4}√3 / 2) ≈ 41.15°

finally SMV = arctan(1/2)

We deduce AMV = 45.41°. QED.

But the thin thread joining the two parts of this piece
makes this dissection impracticable. It's just of theoretical interest...

Remains the red cut to get a cross instead of a square.

Construct BC = AB/2, perpendicular to AB. I midpoint of AB.

Triangles ABC and AUI are similar, the zigzag AUVB through I is then the searched cut (AU = UV = VB and perpendicular)

Dissections of greek cross into other polygons are more difficult, mainly because of weird
dissections of these polygons into stripes.

Try with the pentagon, for which the dissection into a stripe is easy,
so that dissection of greek cross ↔ pentagon requires just 7 pieces.

Let A and C the midpoints of the large sides.

A line at 30° intersects AB in H, AH = a√3

Circle with diameter AB and perpendicular in H intersect at M.

AM² = AB.AH = (2a) × (a√3) = (s/2)²

Construct the equilateral triangle AED, then the parallel in D to AE.

Note : this parallel goes through vertex U.

Let AB = 5a/2. Circle with diameter AB and circle centered in B with radius BK = a/2 intersect in M and : AM² = (5a/2)² - (a/2)² = 6a²

Complete the square AMNP.

This results into a very small piece JNU.

Consider then the rotation centered in I transforming CD into EF, and center of the circumcircle to CNE.

This circle defines then a larger piece.

But why so complicated, but the pleasure to use curved cuts ?

Dissection of the latin cross into a 2a wide strip is immediate, by cutting the large branch,
and results into the following much simpler dissection :

Copy CM = CB = a √2. Then AM² = AC² + CM² = 4a² + 2a²
gives the side of the equivallent square AM = a√6.

The other cuts from the cross are CP // AM, and KH _|_ AM through C.

The large branch cut is then put in AEFG to build the strip from crosses, and this strip is overlayed by the one built of repeating the square AMUV, then defining this dissection.

Other dissections with latin cross are not much harder than with the greek cross.

Let's give the beautifull dissection by H. Lindgren :

Construct the equilateral triangle, with C = BF ∩ AA'

D is deduced at once from I I' ∩ HH"

HE = AB on HH' then complete the dissection of dodecagon into : an equilateral triangle ABC, a weird piece ACDEFGHI and the remaining, these three pieces tiling the plane (magenta outlines)

We don't really need to effectively tile the plane to continue the dissection :

Vertex K of the cross is constructed by HH" ∩ B'I'

The parallels to AI in K and E give from intersections with BB' and HH' the points N,J and R.

Complete the square NJQP

Finally, the cut IM comes from the diagonal IB'.

Regarding other dissections with crosses, we have also other cross shapes :
svastika, Malta cross, "globular cross" (with arcs of circles) etc...

Before the svastika got a taboo historical meaning, it was quite usual and
Sam Loyd (1841-1911 !) gave several problems and dissections using this cross.
Specifically a "stretched" cross made of 17 little squares on a 7×7 grid,
and as 17 = 4² + 1², the side of the equivallent square is
found at once !

The more compact variant on a 5×5 grid (10×10 smaller squares) is not much harder.

Another cross by Sam Loyd is the "Iron cross", in the "cyclopedia of 5000 puzzles".

A large freedom is let to that cross.

The applet offers to construct the cross from the square and any point P (draggable) in triangle ABC.

M is midpoint of AP, MN = MP and perpendicular.

NQ = BP and perpendicular to BP

QF // MNE (or DF = AE)

If P is outside triangle ABC, the construction shows just garbage (negative areas).

If angle APB obtuse, it is even not so a cross... rather a square with 4 bumps !

Specifically if APB = 135° (P on the black circle tangent to AC), BPN are in line and we get
a square with 4 isosceles right triangles (tans).

The quarter of circle with diameter AB defines <)P = 90°, an "Iron cross" is then only for
P outside this quarter circle.

The Malta cross on a grid is dissected easy, the cuting lines going through grid points.

The "standard" cross in a 5×5 grid has a 17 grid square area.
Side c of equivallent square is then c² = 17u² = (4u)² + u².
Hence is the diagonal of a 1×4 rectangle.
The sequel of the dissection gives weird pieces, but naturally deduced from parallels and grid points.

The Malta cross in a 8×8 grid suggested by P. Berloquin has an area
40 = 6² + 2². The side of the equivallent square is then the diagonal of a
6×2 rectangle.

The dissection is much simpler.

Finally the dissection of the "standard" Malta cross into a greek cross uses only points of a 4 times smaller grid (green points), on a 20×20 grid, shown here just 10×10 to simplify the drawing.

The potence cross of 25 unit squares on a 9×9 grid is obviously dissected into a square
with 6 pieces.

Is there a better dissection, may be with slanted cuts ? or from 25 = 3² + 4² ?

The globular cross will be seen in the curved dissections.

To end up with the crosses, we show the dissection (by Sam Loyd) of a greek cross
into two greek crosses (of half area then).

This dissection is given because it shows the relation
arctan(1) = arctan(1/2) + arctan(1/3).

Consider the diagonal AB of the greek cross, diagonal of a 1×3 rectangle

The diagonal of the half area cross is in ratio √2, that is the side of an
isosceles right triangle (tan) ABC (45° angle)
whose hypothenuse is the initial diagonal AB.

The above mentioned relation proves that this diagonal of the half area cross is diagonal BC
of a 1×2 rectangle,
hence idealy fits in MN, M and N being the midpoints of the sides of the initial cross.

Hence the vertices of the half area cross are the midpoints M, N, K, Q of the sides.

We then construct this cross from MK, using a a previous construction, seen when dissecting the
greek cross ↔ dodecagon :

TK perpendicular to MK and equals to MK/2 gives the direction of MP.

The perpendicular from N to MT gives P and the remaining of the cros with side MP.

Note : NP goes through I, midpoint of MK.

Let's go on with star polygons. The hexagram is simpler than pentagram, as it is easy dissected into a strip.

Copy AH = h. Circle with diameter AD intersects the perpendicular from H in point K, and AK² = AH.AD. AK is then the side of the equivallent square.

The proof that DM = AK is also equals to this side is immediate, considering equal triangles AHK and DIM (right angled, other angle A = D, and AH = DI)

Note : M is **NOT** the midpoint of EF.

Because IM² = DM² - ID² = h.AD - h² = h²√3 - h²
and (AD/2)² = (h√3/2)² = 3h²/4, hence IM ≠ AD/2
(at 1.2 %)

The cut of the orange piece is without any comment.

Considering the red and yellow pieces results, because from angles in O,
into a red piece being an isosceles triangle and cut lines of these pieces parallel to the pentagram sides.

Points M,N,Q,F are used as hinges for pieces, hence HM and KQ equal half the side a of the square.

Even more, M and N are midpoints of segments.

Finally NH + NK = a as well as FH + FK = a, hence 2NF = 2a.

Therefore NF = a

We certainely could construct the side a of the equivallent square, then deduce at once line NF.

We shall contruct instead directly H,K,F,Q from the previous property,
without first constructing a.

First care of the red and yellow pieces and point N.

We get at once ZR // BD and AR = ZR = ZY.

AR (yellow) comes into SZ (violine) and ZR (yellow + red pieces)
comes into UZ (violine piece), hence the constructions ZS = ZU = ZY
and UV // ST // AC.

Consider now constructing H from M and N.

Angle H = 90°, H is then on the circle with diameter MN.

We have to find H so that NF = 2MH.

Midpoint P of NF is on a line parallel to CE, homothetic of CE from center N and ratio 1/2.

Then NP = MH, hence NP is transformed of MH from a rotaion by 90° which transforms M into N.

This rotation has the vertex I of the MIN tan as center,
that is midpoint of arc MN, intersection of circle with diameter MN and perpendicular bisector of MN.

The locus of H is then transformed by this rotation into a circle, locus of P.

P is then defined as intersection of the two loci of P, line and circle.

We deduce line NF, then H, then Q (parallel to NF at distance = MH) hence K.

Note : P is midpoint of HK.

Other dissections of star polygons are generally complicated.

Just give :

Copy OH = OA, then D is intersection of OB with AH.

This gives the starting point of the construction of the small hexagram EDFetc. that we cut into three identical parts.

Construction of D ensures that DB = DE, and gives then the side of the star dodecagon.

The parallel to OI from M intersects IJ in N, and NP is parallel to IU.

Similarily for N' and K.

All small segments constructed in that way are then equal, and the angles have the suitable value to build the star dodecagon.

Let O the pentagram center (intersect point of DV and EU.

CM // EU, VN // OB.

Let's finish with the star polygons and a miraculous multiplication :

Ignoring the peaks, we can then "see" the disection of a square into five squares, and

The square ABCD is dissected into 5 squares by the lines connecting one vertex (A) to a midpoint (M). and B to midpoint of CD, → B'C', C to midpoint of AD, → C'D' and finally D to midpoint of AB, → A'D'.

The central star octogon is then completed using the same method from other vertices of the initial octogon (EH etc.)

The other cut lines are deduced from extending and parallels to sides of the central octogon : PF // EH → P and similarily for N and J.

Finally we complete the two small cyan and green squares.

This construction is repeated 3 times more from other vertices, and results into 4 identical sets of pieces, giving 4 star octogons. Only one is shown here. The 5th octogon is of course the central one in blue.