Problem
A's perimeter is 26 cm and width 8 cm - working backwards, the two heights total 10 cm, so each height is 5 cm.
Undoing the perimeter to find a missing side is a natural Grade 3 'work backwards' with perimeter.
3.MD.D.8Work BackwardsB's left side runs down from inside A; A's bottom edge splits it into a 3 cm overlap piece and a 4 cm piece below A.
Drawing where A's bottom edge cuts across B shows the one side broken into two labeled pieces I can simply add.
3.OA.D.8Draw A DiagramSquare B's left side has length 3 + 4, because rectangle A's bottom edge splits that one side into a 3 cm top piece inside the overlap and a 4 cm piece below A.
B's left side is one straight segment, and rectangle A's bottom edge meets it at a single point, so it is divided into exactly two pieces stacked in a line: the 3 cm piece inside the overlap on top and the 4 cm piece below A on the bottom.
Those two pieces share only that one meeting point, with no gap between them and no overlap, so laid end to end they cover the whole side and nothing more.
Because B is a square, the side made of the 3 cm and 4 cm pieces is one full side length of B.
Adding the overlap part and the below part to get the whole side is straightforward Grade 3 addition.
4.MD.A.3Identify SubproblemsFind where one shape's edge cuts across the other, then add the two pieces of that side - it's just 3 + 4!