A geometry puzzle with several solutions

If the sides of the rectangle are $a$ and $b$ , the area of the rectangle is $A_{\color{#D61F06}\square}=ab$ .

$\angle RCQ=360^\circ-60^\circ-90^\circ-60^\circ=150^\circ$

The area of one of the blue triangles is $A_{\color{#3D99F6}\triangle}=\frac 12 ab \sin(150^\circ)=\frac 14 ab$

The four blue triangles have a combined area $A_{\color{#3D99F6}4\triangle}=4\times A_{\color{#3D99F6}\triangle}=ab$

$A_{\color{#D61F06}\square}=A_{\color{#3D99F6}4\triangle}$

The blue triangles are concident, because they all have two common sides and a common angle: $PA=PB=RC=RD, AS=BQ=CQ=SD$ and $\angle PAS=\angle PBQ=\angle QCR=\angle RDS=360°-90°-2*60°=150°$ .

I give two solutions:

First solution: Let's look at the next figure.

We reflected $S$ . We can see that $T_{ABCD}=T_{SABQCD}$ (because $T_{ASD}=T_{BCQ}$ ). The $ABQ, AQS, SQD, DCQ$ triangles are condicent with each other and with the purple triangles (because $AS=BQ=DS=CQ, PB=AB=CD=CR$ , and $\angle PBQ=\angle ABQ=\angle ASQ=\angle DSQ=\angle DCQ)$ .

So the red rectangle's area is the same as the sum of the blue triangles' areas.

Second solution: Let's look at only one blue triangle (from the original figure). The $MSDR$ is a paralelogram. We know that $\angle SDR=150°$ , so $\angle MSD$ have to be equal $30°$ . We will calculate the area of the paralelogram. We draw the "height' of $MSDR$ in the next figure. If $X$ is the foot point, then the $DXS$ triangle is a half-regular triangle, so $XS*2=SD$ . If the original triangle's sides are $a$ and $b$ $(a>b)$ , then $T_{MSDR}=DR*SX=a*(b/2)=2*T_{SDR}$ . Now we know that $T_{PBQ}=T_{QCR}=T_{RDS}=T_{SAP}=ab/4$ .

So the red rectangle's area is the same as the sum of the blue triangles' areas.

Let $a = AB, b = AD$ . Then $SQ = b + 2(\tfrac12\sqrt 3 a) = b + \sqrt 3a,\ \ PR = a + \sqrt 3 b,$ giving the rhombus $PQRS$ an area $A_{\text{total}} = \tfrac12 (a + \sqrt 3 b)(b + \sqrt 3 a) = 2ab + \tfrac12\sqrt 3(a^2 + b^2).$

The grey triangles have areas $|PAB| = |QCD| = \tfrac12(b\cdot \tfrac12\sqrt 3 b) = \tfrac14\sqrt 3 b^2,\ \ \ |RBC| = |SAD| = \tfrac14\sqrt 3 a^2,$ with a total of $A_{\text{grey}} = \tfrac12\sqrt 3 (a^2 + b^2).$

Now clearly the red rectangle has an area of $A_{\text{red}} = ab,$ leaving a blue area of $A_{\text{blue}} = A_{\text{total}} - A_{\text{grey}} - A_{\text{red}} \\ = 2ab + \tfrac12\sqrt 3(a^2 + b^2) - \tfrac12\sqrt 3 (a^2 + b^2) - ab = ab.$ Thus the blue area equals the red area.

I added angles and named sides to the picture to my explanation was clearer. The area of one blue triangle is: $A= \frac{b\times h}{2}$ .

From picture we can see that $h= \sin 30^{\circ} \times a = \frac{1}{2}\times a$ .

Back to area, we now have $A= \frac{a\times b}{4}$ .

The area of the four blue triangles is $4A= 4\times \frac{a\times b}{4}= a\times b$ which is exactly the same as area of ABCD rectangle.

if (sideDC == 4 && sideCB == 2), because Why Not?

sideRC = 4; sideCQ = 2;

Since the grey triangles are equilaterals, their angles are 60, while the red triangle is 90. So angleRCD, plus the angleDCB, plus angleBCQ are 60 + 90 + 60 = 210, leaving 150 degrees for angleRCQ.

With sideRC, sideCQ, and angleRCQ, you can use trig to discover areaRCQ

sin(angleRCQ) / 2 * RC * CQ

sin(150) / 2 * 4 * 2

0.5 / 2 * 4 * 2

areaRCQ = 2

Four equal blue triangles = 4 * areaRCQ = 8

Area of Red Rectangle = DC * CB = 4 * 2 = 8