Practice Maths

Area of Rhombuses, Kites and Trapeziums

Key Ideas

Key Terms

rhombus
a quadrilateral with all four sides equal and diagonals that bisect each other at right angles; A = ½ × d1 × d2.
kite
a quadrilateral with two pairs of adjacent equal sides; its diagonals cross at right angles; A = ½ × d1 × d2.
trapezium
a quadrilateral with exactly one pair of parallel sides (a and b); A = ½ × (a + b) × h.
diagonal (d1, d2)
a line segment connecting opposite vertices of a shape; for rhombuses and kites, the two diagonals are needed for the area formula.
parallel sides
the two sides of a trapezium that are parallel to each other; labelled a and b in the area formula.
perpendicular height
the distance between the two parallel sides of a trapezium, measured at right angles; used as h in the area formula.
Hot Tip Rhombus and kite both use ½ × diagonal 1 × diagonal 2. For a trapezium you need the two parallel sides (a and b) and the perpendicular height — not a slant side.

Worked Examples

Rhombus: diagonals 8 cm and 6 cm
A = ½ × 8 × 6 = 24 cm²

Trapezium: parallel sides 7 m and 5 m, perpendicular height 4 m
A = ½ × (7 + 5) × 4 = ½ × 12 × 4 = 24 m²

Rhombuses and Kites: The Diagonal Formula

Both rhombuses and kites have a formula based on their two diagonals (d1 and d2).

Area of rhombus = (d1 × d2) ÷ 2
Area of kite = (d1 × d2) ÷ 2

The diagonals of these shapes cross at right angles. If you draw those two diagonals, they divide the shape into four right-angled triangles. The four triangles together make a rectangle with length d1 and width d2, and the rhombus/kite covers exactly half that rectangle. So area = (d1 × d2) / 2.

Example (rhombus): diagonals are 10 cm and 6 cm. Area = (10 × 6) / 2 = 30 cm2.

Example (kite): diagonals are 14 cm and 9 cm. Area = (14 × 9) / 2 = 63 cm2.

The Trapezium Formula

A trapezium has one pair of parallel sides (called the parallel sides or bases, often labelled a and b) and a perpendicular height h between them.

Area of trapezium = ½(a + b) × h

Why? Think of the trapezium as two triangles — or better, as the average of the two parallel sides times the height. The formula averages the two base lengths: (a + b)/2 is the average width of the trapezium. Multiply by height to get area.

Example: A trapezium has parallel sides 5 cm and 9 cm, and height 4 cm.
Area = ½(5 + 9) × 4 = ½ × 14 × 4 = 28 cm2.

Visual Proof: Decomposing into Simpler Shapes

Another way to see why these formulas work is by cutting the shapes apart:

Trapezium: Cut it into one rectangle and two right-angled triangles (or one parallelogram and one triangle). Adding these areas gives the same result as the formula.

Rhombus: Draw both diagonals. You get 4 right-angled triangles, each with legs d1/2 and d2/2. Area of each = ½ × (d1/2) × (d2/2) = d1d2/8. Four of them = 4 × d1d2/8 = d1d2/2. Same formula.

Composite Area Problems with These Shapes

Often you'll need to find the area of a shape that is made up of a combination of these quadrilaterals, plus perhaps rectangles or triangles.

Strategy: identify each part, label the known dimensions, find each area separately, then add or subtract.

Example: A garden is in the shape of a trapezium (parallel sides 8 m and 12 m, height 5 m) with a kite-shaped pond cut out of it (diagonals 3 m and 2 m). Garden area = ½(8 + 12) × 5 = 50 m2. Pond area = (3 × 2)/2 = 3 m2. Remaining garden = 50 − 3 = 47 m2.

Identifying the Height in a Trapezium

The height of a trapezium is the perpendicular distance between the two parallel sides — always measured at right angles, never along the slant side. The slant sides are used only for perimeter calculations, not for area.

Check: if a problem gives you the slant side and you're finding area, you need to use the perpendicular height instead. Sometimes you'll need Pythagoras' theorem to find the perpendicular height from the slant side.

Key tip: For rhombuses and kites, you need the two diagonals, not the side lengths. For trapeziums, you need the two parallel sides and the perpendicular height. Read the question carefully to identify which measurements you've been given and which formula applies.

Mastery Practice

  1. Calculate the area of each rhombus. Both diagonals are given. Fluency

    1. d1 = 10 cm, d2 = 6 cm
    2. d1 = 14 m, d2 = 8 m
    3. d1 = 9 cm, d2 = 9 cm
    4. d1 = 20 mm, d2 = 15 mm
    5. d1 = 5.6 m, d2 = 4 m
    6. d1 = 18 cm, d2 = 11 cm
    7. d1 = 7.4 m, d2 = 6 m
    8. d1 = 30 cm, d2 = 22 cm

  2. Calculate the area of each kite. Both diagonals are given. Fluency

    1. d1 = 12 cm, d2 = 7 cm
    2. d1 = 16 m, d2 = 9 m
    3. d1 = 8 cm, d2 = 5 cm
    4. d1 = 24 mm, d2 = 10 mm
    5. d1 = 11 m, d2 = 8 m
    6. d1 = 6.4 cm, d2 = 5 cm
    7. d1 = 3.6 m, d2 = 2.5 m
    8. d1 = 15 cm, d2 = 12 cm

  3. Calculate the area of each trapezium. The parallel sides (a and b) and perpendicular height (h) are given. Fluency

    1. a = 8 cm, b = 4 cm, h = 5 cm
    2. a = 12 m, b = 6 m, h = 7 m
    3. a = 10 cm, b = 10 cm, h = 4 cm (special case — what shape is this?)
    4. a = 15 mm, b = 9 mm, h = 8 mm
    5. a = 7.5 m, b = 4.5 m, h = 6 m
    6. a = 20 cm, b = 14 cm, h = 9 cm
    7. a = 11 m, b = 5 m, h = 4.8 m
    8. a = 18 cm, b = 12 cm, h = 6.5 cm

  4. For each shape, name it, choose the correct formula, and calculate its area. Fluency

    1. A quadrilateral with all sides equal, diagonals 16 cm and 10 cm
    2. A quadrilateral with one pair of parallel sides of length 9 m and 5 m, height 4 m
    3. A quadrilateral with two pairs of adjacent equal sides, diagonals 14 cm and 8 cm
    4. A quadrilateral with parallel sides 11 mm and 7 mm, height 6 mm
    5. Diagonals are 9 m and 9 m, all sides equal — find area
    6. Parallel sides 20 cm and 8 cm, perpendicular height 7 cm
    7. A kite with diagonals 18 m and 13 m
    8. Trapezium: parallel sides 6.5 cm and 3.5 cm, height 4 cm

  5. Find the missing measurement for each shape. Understanding

    1. A rhombus has area 48 cm² and one diagonal of length 12 cm. Find the other diagonal.
    2. A kite has area 35 m² and one diagonal of 10 m. Find the other diagonal.
    3. A trapezium has area 54 cm², parallel sides 10 cm and 8 cm. Find the perpendicular height.
    4. A trapezium has area 66 m², height 6 m, and one parallel side of 9 m. Find the other parallel side.
    5. A kite has area 90 mm² and equal diagonals. Find the length of each diagonal.
    6. A rhombus has area 120 cm². If d1 = 2 × d2, find both diagonal lengths.
    7. A trapezium has area 80 m². The parallel sides are in the ratio 3 : 5 and the height is 8 m. Find both parallel sides.
    8. Two identical trapeziums are placed together to form a parallelogram of area 96 cm² with base 12 cm. Find the height and the area of one trapezium.

  6. Solve each problem. Show all working and include units. Problem Solving

    1. A diamond-shaped (rhombus) traffic sign has diagonals of 60 cm and 45 cm. Find its area in cm² and m².
    2. A kite-shaped garden bed has diagonals 8.4 m and 5 m. Garden edging costs $12 per metre. If the perimeter of the kite is 26 m, find the cost of edging.
    3. A block of land is trapezium-shaped. The parallel boundaries are 48 m and 36 m. The perpendicular distance between them is 25 m.
      1. Find the area in m².
      2. Convert to hectares.
      3. If the land sells for $850 per m², find the total sale price.
    4. A composite shape is made by joining a rectangle 10 cm × 6 cm with a trapezium on top. The trapezium has parallel sides 10 cm (the shared edge) and 6 cm, with a height of 4 cm. Find the total area of the composite shape.

  7. Find the area of each shape shown. Understanding

    1. Find the area of the trapezium below. 10 cm 16 cm h = 8 cm
    2. Find the area of the rhombus below. d₁ = 14 cm d₂ = 9 cm
    3. Find the area of the kite below. d₁ = 12 cm d₂ = 18 cm

  8. Complete the table. Understanding

    ShapeFormulaGiven valuesArea
    RhombusA = ½ d1 d2d1 = 8 cm, d2 = 5 cm___
    KiteA = ½ d1 d2d1 = 10 m, d2 = 7 m___
    TrapeziumA = ½(a + b)ha = 9 cm, b = 5 cm, h = 4 cm___
    TrapeziumA = ½(a + b)ha = 14 mm, b = 10 mm, h = 6 mm___
    RhombusA = ½ d1 d2d1 = 24 m, d2 = 18 m___

  9. Each student below made an error. Find the mistake and correct it. Understanding

    1. Taylor calculated the area of a trapezium with parallel sides 8 cm and 6 cm, height 5 cm as: A = (8 + 6) × 5 = 70 cm². What did Taylor do wrong?
    2. Lee calculated the area of a rhombus with diagonals 10 m and 4 m as: A = 10 × 4 = 40 m². What is the correct answer?
    3. Morgan said: “A kite with diagonals 6 cm and 4 cm has area A = ½ × (6 + 4) = 5 cm².” What should it be?
    4. For a trapezium with parallel sides 12 m and 8 m, Sam used the slant height of 6 m instead of the perpendicular height of 5 m. Calculate the correct area.

  10. Extended response: Show all steps clearly. Problem Solving

    1. A mosaic tile is kite-shaped with diagonals 12 cm and 8 cm.
      1. Find the area of one tile.
      2. How many tiles are needed to cover a wall panel of area 3 600 cm²?
      3. Each tile costs $1.45. Find the total cost.
    2. A garden is divided into two sections: a rhombus with diagonals 6 m and 4 m, and a trapezium with parallel sides 6 m and 4 m, height 3 m. Both sections are to be fertilised.
      1. Find the area of the rhombus section.
      2. Find the area of the trapezium section.
      3. Fertiliser costs $8 per m². Find the total fertiliser cost for both sections.
See Answers ➔