Problem
Slice where the narrow top meets the wide bottom, giving two rectangles that are easy to measure.
The area of a whole shape equals the sum of the areas of the non-overlapping pieces it is cut into, a Grade 3 area idea.
3.MD.C.7Draw A DiagramCutting the figure along the join gives two rectangles whose areas add back to the whole figure.
The cut divides the figure with no part left out and no part counted twice, so the two pieces put back together are the original figure.
Each piece is a rectangle, so its area is one side length taken as many times as the other, counting equal rows of unit squares.
The top rectangle is 11.8 cm wide and 8 cm tall, so multiply the two side lengths.
Area of a rectangle is length times width; this is a single multiplication once the piece is isolated.
4.MD.A.3Identify SubproblemsThe bottom stretches 12.3 to the left, 11.8 across the middle and 7.5 to the right: 31.6 cm.
Adding the overhangs and the shared middle rebuilds the whole base; lining up the decimal points keeps tenths with tenths.
5.NBT.B.7Identify SubproblemsThe bottom rectangle is 31.6 cm wide and 13.5 cm tall, so multiply these decimal lengths.
Multiplying decimals to hundredths gives the area of the bigger piece; the product is around 30 times 13, near 400, which feels right.
5.NBT.B.7Identify SubproblemsThe total area is the top rectangle's area plus the bottom rectangle's area.
Joining the two pieces back together means adding their areas to get the whole figure's area.
5.NBT.B.7Identify SubproblemsCut a tricky shape into simple rectangles, find each area with length x width, then add them up.