Areas of Trapezoids and Rhombuses

In the figure, one trapezoid has vertices (corner points) $(1, 1)$, $(8, 1)$, $(6, 5)$ and $(3, 5)$. Its bases lie on the horizontal grid lines $y = 1$ and $y = 5$, so they are parallel. Their lengths are $8 - 1 = 7$ units and $6 - 3 = 3$ units. Vertical grid lines cross both bases at right angles, so the height is $5 - 1 = 4$ units.

The other trapezoid, with vertices $(8, 1)$, $(11, 1)$, $(13, 5)$ and $(6, 5)$, is a copy of the first, turned halfway around. Its $3$-unit base is on the bottom and its $7$-unit base is on top.

Together they form a parallelogram with vertices $(1, 1)$, $(11, 1)$, $(13, 5)$ and $(3, 5)$. Its top and bottom sides are both horizontal and both $7 + 3 = 10$ units long. Its height is still $4$ units, so its area is $10 \times 4 = 40$ square units. The trapezoid is half of that: $40 \div 2 = 20$ square units.

This works for any trapezoid. Two copies make a parallelogram with base $b_1 + b_2$ and height $h$, and the trapezoid is half of it. So $A = \frac{1}{2}(b_1 + b_2)h$. Here $b_1$ and $b_2$ are the lengths of the parallel sides, and $h$ is the perpendicular distance between them.

Now for the rhombus. A rhombus is a quadrilateral with four sides of equal length. A square is one kind of rhombus. In every rhombus the opposite sides are parallel, so a rhombus is also a parallelogram. That gives you one way to find its area right away. Use any side as the base $b$. Multiply it by the height $h$, the perpendicular distance from that side to the opposite side. So $A = bh$. For example, a rhombus with sides $7$ cm long and a perpendicular height of $6$ cm has an area of $7 \times 6 = 42$ square centimeters.

Sometimes you know the diagonals instead of a height. A diagonal is a line segment that joins two opposite corners. In every rhombus, the two diagonals cross at right angles and cut each other in half. Call their lengths $d_1$ and $d_2$. The figure below shows how the diagonals give the area.

The figure shows a rectangle with corners $(1, 1)$, $(9, 1)$, $(9, 7)$ and $(1, 7)$, cut into eight triangles. The four triangles that meet at $(5, 4)$ make a rhombus with vertices $(5, 1)$, $(9, 4)$, $(5, 7)$ and $(1, 4)$. Each of its sides goes $4$ units across and $3$ units up or down, from corner to corner of a $4$-by-$3$ rectangle. So all four sides are the same length.

One diagonal lies on the grid line $y = 4$, from $(1, 4)$ to $(9, 4)$, so $d_1 = 9 - 1 = 8$ units. The other lies on $x = 5$, from $(5, 1)$ to $(5, 7)$, so $d_2 = 7 - 1 = 6$ units. One is horizontal and one is vertical, so they cross at right angles. They meet at $(5, 4)$, the middle of each.

Now compare the rhombus with the rectangle around it. The two diagonals split the big rectangle into four $4$-by-$3$ rectangles. Each side of the rhombus cuts one of them from corner to corner into two triangles of equal area, one inside the rhombus and one outside. So the rhombus fills exactly half of the big rectangle.

The big rectangle is $8$ units by $6$ units, the same as the diagonals. Its area is $8 \times 6 = 48$ square units, so the area of the rhombus is $\frac{1}{2} \times 48 = 24$ square units.

Every rhombus has diagonals that cross at right angles and cut each other in half, so this works for any rhombus: $A = \frac{1}{2}d_1 d_2$.

Takeaway. A trapezoid is half of the parallelogram that two copies of it make. So add its two parallel bases, multiply by the perpendicular height, and take half: $A = \frac{1}{2}(b_1 + b_2)h$. A rhombus is a parallelogram with four equal sides. Multiply a side by the perpendicular height, $A = bh$, or take half the product of its diagonals, $A = \frac{1}{2}d_1 d_2$. Always measure a height at a right angle to the base, never along a slanted side.

Next, you will use these formulas to find a missing dimension of a quadrilateral, solve area word problems, and spot mistakes in area work.