Prove or disprove the existence of a dissection of a square of side 21 into a rectangle of sides 13 and 34 according to the diagram provided.

Consider a square of side length 21. Decompose the triangle as constructed above and use isometries to move these pieces into a rectangle. [br][br]Proof: Consider a square of length 21 units and a rectangle of length 13 units by 34 units. The area of the square can be found by adding the sums of the pieces. The area of a trapezoid( LMNO and ONPQ) can be found using the formula [math]\frac{1}{2}\left(top+bottom\right)\left(height\right)[/math]. The area of a triangle (MRP and RPS) can be found using the formula [math]\frac{1}{2}\left(base\right)\left(height\right)[/math]. Using these formulas the area of LMNO is [math]\frac{1}{2}\left(LO+MN\right)\left(LM\right)[/math], ONPQ is [math]\frac{1}{2}\left(OQ+NP\right)\left(QP\right)[/math], [math]\triangle MRP[/math] is [math]\frac{1}{2}\left(MP\right)\left(RM\right)[/math], and [math]\triangle PSR[/math] is [math]\frac{1}{2}\left(PS\right)\left(RS\right)[/math]. Note that by construction [math]LO=NP,QO=MN=LM=QP,MP=MN+RS=RS,RM=PS=LO=NP[/math]. With this knowledge we can find the sum of the area of the pieces as follows: [math]\frac{1}{2}\left[\left(LO+MN\right)\left(LM\right)+\left(OQ+NP\right)\left(QP\right)+\left(MP\right)\left(RM\right)+\left(PS\right)\left(RS\right)\right][/math] and using algebra and the knowledge of the construction, we can simplify this expression to [math]2\left(LO\right)\left(LM\right)+2\left(LM^2\right)[/math].[br][br]Notice that using the area formula for a rectangle([math]\left(base\right)\left(height\right)[/math]), we can find the area of the rectangle to be [math]\left(MN\right)\left(LM+MP\right)[/math]. Using the same knowledge of construction from above, we see that this simplifies to [math]2LM^2+\left(LM\right)\left(LO\right)[/math].[br][br]Notice that these two are formulas are not equivalent. Therefore, we can conclude that although this decomposition appears to have the qualities necessary to fulfill this statement, the areas are in fact different. This makes this dissection not existent.[math]\diamondsuit[/math]