Decompose Polygons on a Coordinate Grid

Sort the sides. The bottom side joins $(1, 1)$ and $(7, 1)$; the top side joins $(1, 4)$ and $(5, 4)$. Each pair shares a $y$-coordinate, so both sides are horizontal and parallel. That makes the shape a trapezoid: a quadrilateral with at least one pair of parallel sides. The left side is vertical ($x = 1$). The side from $(7, 1)$ to $(5, 4)$ is slanted: both coordinates change.

Choose the cut. The upper end of the slanted side is $(5, 4)$. Cut straight down from there to $(5, 1)$, which is on the bottom side because its $y$-coordinate is $1$ and its $x$-coordinate, $5$, is between $1$ and $7$. Both ends of the cut have $x = 5$, so the cut is vertical. It meets the horizontal top and bottom sides at right angles.

Name the pieces as the figure below labels them. Piece 1 has vertices $(1, 1)$, $(5, 1)$, $(5, 4)$ and $(1, 4)$. Its sides go horizontal, vertical, horizontal, vertical, so it has four right angles: a rectangle. Piece 2 has vertices $(5, 1)$, $(7, 1)$ and $(5, 4)$. Its horizontal side and its vertical side meet at $(5, 1)$, so it is a right triangle.

Read each base and height from the piece's own vertices. Let $b$ be the base and $h$ the perpendicular height, both in units. Piece 1: $b = 5 - 1 = 4$ and $h = 4 - 1 = 3$. Piece 2: its base lies on $y = 1$, so $b = 7 - 5 = 2$. Its height is the vertical side from $(5, 1)$ to $(5, 4)$, which is perpendicular to that base, so $h = 4 - 1 = 3$.

Find each area in square units. Piece 1 holds $4 \times 3 = 12$ grid squares. Piece 2's slanted side is a diagonal of the rectangle with vertices $(5, 1)$, $(7, 1)$, $(7, 4)$ and $(5, 4)$, which is 2 units wide and 3 units tall. So piece 2 is half of that rectangle.

Add the two areas. Piece 1 lies left of the cut and piece 2 lies right of it, so they touch only along the cut and do not overlap. Their outer sides are exactly the four sides of the trapezoid, so there are no gaps. Both conditions hold, so the areas add.

Find the slanted sides. Both coordinates change from $(7, 4)$ to $(3, 8)$ and from $(3, 8)$ to $(1, 4)$, so those two sides are slanted. The other three sides are horizontal or vertical. Cut between the lower ends of the slanted sides, from $(1, 4)$ to $(7, 4)$. Both points have $y = 4$, so the cut is horizontal.

Name what the cut makes, using the labels in the figure below. Below it is piece 1, with vertices $(1, 1)$, $(7, 1)$, $(7, 4)$ and $(1, 4)$. Its sides go horizontal, vertical, horizontal, vertical, so it is a rectangle. Above it is a triangle with vertices $(1, 4)$, $(7, 4)$ and $(3, 8)$. Its base is the cut, on $y = 4$, but it has no vertical side to be the height.

Cut straight down from the top vertex $(3, 8)$ to $(3, 4)$ on the base. Both points have $x = 3$, so the cut is vertical and perpendicular to the base. It makes piece 2 with vertices $(1, 4)$, $(3, 4)$, $(3, 8)$ and piece 3 with vertices $(3, 4)$, $(7, 4)$, $(3, 8)$. In each one, a horizontal side meets the vertical cut at $(3, 4)$, so both are right triangles.

Read each base $b$ and perpendicular height $h$, in units, from the piece's own vertices. Piece 1: $b = 7 - 1 = 6$, $h = 4 - 1 = 3$. Pieces 2 and 3 share the vertical cut as their perpendicular height, so $h = 8 - 4 = 4$ for both. Piece 2: $b = 3 - 1 = 2$. Piece 3: $b = 7 - 3 = 4$. Check: $2 + 4 = 6$, the whole length of the first cut, so the triangles leave no gap along it.

Find each area in square units. Piece 1 is a rectangle, so its area is $6 \times 3 = 18$. Pieces 2 and 3 are right triangles with a horizontal base and a vertical height, so each is half of a $b$-by-$h$ rectangle. For piece 3 that rectangle is 4 by 4. It is a square, and a square is a rectangle, so the same rule applies.

Add the three areas. Piece 1 lies below the first cut. Pieces 2 and 3 lie above it, on opposite sides of the second cut. So the pieces touch only along the cuts and do not overlap. Their outer sides are exactly the five sides of the pentagon, so there are no gaps.