Problem
Area of the rectangle
The rectangle MBCD is 12 cm wide and 7 cm tall. Multiply width by height to get its area.
Area of a rectangle is just rows times columns of unit squares, a basic 4th-grade formula.
4.MD.A.3Draw A DiagramArea of the parallelogram
The parallelogram ABCF sits on the same base of 12 cm and has the same height of 7 cm as the rectangle. Base times height gives its area.
A parallelogram can be slid into a rectangle of the same base and height, so its area is base times height, the same as the rectangle here.
3.MD.C.7Draw A DiagramFind the overlap area by working backwards
Adding both figures counts the overlap twice, so 84 + 84 - overlap = 132 leaves overlap = 36.
When two regions stack, the part they share is inside both, so adding both totals double-counts it once; peeling that back is plain subtraction.
3.MD.C.7Work BackwardsThe overlapping triangle's area is the rectangle plus the parallelogram minus the shaded region.
Why?
Adding the two figures counts the overlap once from the rectangle and once from the parallelogram, so it is counted twice, while the shaded region counts it only once.
Why?
The gap between that doubled count and the true shaded area is therefore exactly one extra copy of the overlap.
Why?
The shaded region splits into the rectangle-only part, the parallelogram-only part, and the overlap, with no area missed and none counted twice.
Use the triangle area to find GF
The overlap is triangle GBC on base 12, so 12 x GF divided by 2 = 36 makes GF = 6 cm.
A triangle is half of a rectangle with the same base and height, so reversing the half-of formula turns a known area back into the missing height.
3.MD.C.7Identify SubproblemsWhen two shapes overlap, add their areas and take away the shared part once, then reverse the triangle formula to find the missing height.
- Area of the rectangle
- Area of the parallelogram
- Find the overlap area by working backwards
- Use the triangle area to find GF