Can Two Polygons Form a Triangle, Quadrilateral, and Pentagon?

Yes. The pieces can be a triangle and a quadrilateral. The key is to choose them so that they have more than one pair of equal-length edges to join. Depending on which edges meet and how their endpoints are matched, the joined outline can have three, four, or five sides.

Here is a concrete construction. It uses ordinary rigid pieces: no stretching, bending, or overlapping.

Make the two pieces

Start with a triangle $PQR$ whose angles are $40^\circ$ at $P$, $60^\circ$ at $Q$, and $80^\circ$ at $R$. Let $X$ be the midpoint of $PQ$, and let $Y$ be the midpoint of $PR$. Cut along $XY$.

The cut produces:

• the small triangle $PXY$; and
• the quadrilateral $XQRY$.

Because $X$ and $Y$ are midpoints, $XY$ is parallel to $QR$ and has half its length. Also, $PX=XQ$ and $PY=YR$. Those equalities provide alternative edges for joining the pieces.

The angles of the small triangle at $P,X,Y$ are $40^\circ,60^\circ,80^\circ$. The quadrilateral’s angles at $X,Q,R,Y$ are $120^\circ,60^\circ,80^\circ,100^\circ$. In particular, the angle at $X$ in the quadrilateral is supplementary to the small triangle’s angle at $X$.

Join them to make a triangle

Put the pieces back together along the cut edge $XY$. At $X$, their interior angles add to $60^\circ+120^\circ=180^\circ$. At $Y$, they add to $80^\circ+100^\circ=180^\circ$. Thus the boundary is straight through both $X$ and $Y$: neither point is a corner of the outside shape. The remaining outline is the original triangle $PQR$.

This illustrates how joining two pieces can result in fewer boundary sides than the pieces appear to have. Where the outside edges line up in a straight line, the meeting point is not counted as a polygon vertex.

Join them to make a quadrilateral

The small triangle has an edge $PX$ equal in length to the quadrilateral’s edge $XQ$, since $X$ is the midpoint of $PQ$. Join these edges, placing the pieces on opposite sides of the joined edge. Match the small triangle’s endpoint $X$ to the quadrilateral’s endpoint $X$, and its endpoint $P$ to $Q$.

At the joined point corresponding to $X$, the angles add to $60^\circ+120^\circ=180^\circ$. That point becomes straight along the outside boundary. At the other joined point, the angles add to $40^\circ+60^\circ=100^\circ$, so it remains a corner. The pieces have five exposed edge segments in total after joining, and the straight point merges two of them. The result is a convex quadrilateral.

Join them to make a pentagon

Use the same equal-length edges, $PX$ and $XQ$, but match the endpoints the other way: match $P$ to the quadrilateral’s $X$, and the small triangle’s $X$ to $Q$. Again place the pieces on opposite sides of the joined edge.

The two outside angles at the join are now $40^\circ+120^\circ=160^\circ$ and $60^\circ+60^\circ=120^\circ$. Both are less than $180^\circ$, so both joined points remain corners. No exposed edges merge into a straight line. The outside boundary therefore has five sides, and the result is a convex pentagon.

Join	What happens at the endpoints	Outside shape
Along $XY$	Both endpoint angle sums are $180^\circ$	Triangle
Along $PX$ and $XQ$, first alignment	One sum is $180^\circ$	Convex quadrilateral
Along $PX$ and $XQ$, reversed alignment	Both sums are less than $180^\circ$	Convex pentagon

Why the endpoint angles matter

When two convex pieces are joined along a full edge, their boundary edges remain exposed except for the joined edge. The angle at each end of the join is the sum of the two pieces’ angles there. If that sum is less than $180^\circ$, the point is a corner of the combined outline. If the sum is exactly $180^\circ$, the outline is straight there, so adjacent boundary segments form one side instead of two.

In this construction, the same pieces use those two outcomes deliberately: two straight points give a triangle, one gives a quadrilateral, and none gives a pentagon.

Common mistakes

• Counting every piece edge as an outside side. The joined edge is internal, and straight boundary points do not count as corners.
• Assuming equal edge lengths are enough. The pieces must be placed on opposite sides of the joining edge, and the endpoint angles must leave a convex outline.
• Ignoring endpoint order. Joining the same two edges with their endpoints matched in the opposite order can change the angle sums, and therefore the number of sides.