Math by First Principles

Areas of Quadrilaterals

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Learning goals

  • Derive the trapezoid area formula by joining two congruent copies
  • Calculate rhombus and kite areas using perpendicular diagonals
  • Select an area formula whose geometric conditions are satisfied
  • Decompose a convex quadrilateral into triangles with known heights
  • Recover missing bases, heights, and diagonal lengths from area

Two trapezoids make one parallelogram

A trapezoid has at least one pair of parallel sides; choose such a pair as its bases, with lengths b1b_1 and b2b_2. Its height hh is the perpendicular distance between their supporting lines, and all lengths here are positive.

In Area from Rectangles to Triangles, we established that parallelogram area is bhbh and triangle area is 12bh\tfrac12 bh, with perpendicular height. We can use those results and addition of nonoverlapping areas here.

Joining a trapezoid to a turned copy makes its area visible inside a parallelogram.

The trapezoid area formula#

Take a convex trapezoid with bases 1111 and 55 units, separated by 33 units. Turn a congruent copy halfway around the midpoint of one leg and join the copies along that leg.

Two congruent trapezoids make a parallelogramAn original solid blue trapezoid with bases 11 and 5 units and height 3 units is joined to a red dashed congruent copy turned half a turn. Their common slanted side is an interior seam. The outside is a parallelogram: its bottom comprises lengths 11 and 5, its top comprises 5 and 11. A dashed perpendicular height 3 has a right-angle mark.5111153OriginalTurned copy
The second whole trapezoid is a half-turn copy; the joined area is twice the original area.

The top edges are collinear because the bases are parallel and a half turn reverses their directions. Each outside slanted side is the half-turn image of the other, so the joined outside boundary is a parallelogram. The copies lie on opposite sides of the joining leg, with no overlap of their interiors.

The parallelogram has base 11+511+5 units and height 33 units. Its area is

2A=(11+5)⋅3,2A=48,A=24.\begin{aligned} 2A&=(11+5)\cdot3,\\ 2A&=48,\\ A&=24. \end{aligned}

Nothing in the arrangement depended on these particular lengths. For any convex trapezoid with chosen parallel bases b1,b2b_1,b_2 and perpendicular height hh, the two congruent copies make a parallelogram with base b1+b2b_1+b_2 and height hh:

2A=(b1+b2)h,A=b1+b22h.\begin{aligned} 2A&=(b_1+b_2)h,\\ A&=\frac{b_1+b_2}{2}h. \end{aligned}

The fraction is the average of the base lengths. A parallelogram is included in our trapezoid definition: its chosen bases have equal length, and averaging them gives the familiar base-times-height formula.

Worked example 1 Choose the perpendicular measurement

Problem. A convex trapezoid has parallel bases 1313 cm and 77 cm, perpendicular height 44 cm, and a slanted leg of 55 cm. Find its area.

Parallel bases and perpendicular heightSchematic trapezoid, not to scale. Parallel top and bottom bases are labeled 7 cm and 13 cm. A dashed perpendicular between them is labeled 4 cm, with a right-angle square. The right slanted side is labeled 5 cm.7 cm13 cm4 cm5 cmNot to scale
Use the square-marked separation, not the slanted leg.

Solution. The 44 cm distance is perpendicular to the bases. The 55 cm leg is not the required height.

A=13+72⋅4,A=40 cm2.\begin{aligned} A&=\frac{13+7}{2}\cdot4,\\ A&=40\text{ cm}^2. \end{aligned}

Check your understanding

Two congruent copies of a convex trapezoid join along a leg to form a parallelogram. Each trapezoid has parallel bases 1111 m and 66 m and perpendicular height 55 m. If AA is the area of one trapezoid in square meters, which equation describes the joined parallelogram?

Work backward from a trapezoid’s area

An area equation can determine a missing measurement. First identify the parallel bases and their perpendicular separation; then solve for the unknown length and check it in the original equation.

Worked example 2 Recover a parallel base

Problem. A convex trapezoid has area 9595 square cm, perpendicular height 55 cm, and one parallel base of 1212 cm. Find the other parallel base.

Parallel bases and perpendicular heightSchematic trapezoid, not to scale. Parallel top and bottom bases are labeled 12 cm and x cm. A dashed perpendicular between them is labeled 5 cm, with a right-angle square. 12 cmx cm5 cmNot to scale
The unknown labels one of the two parallel bases.

Solution. Let xx be the missing base length in cm.

95=12+x2⋅5,38=12+x,x=26.\begin{aligned} 95&=\frac{12+x}{2}\cdot5,\\ 38&=12+x,\\ x&=26. \end{aligned}

The other base is 2626 cm. Substitution checks the area:

12+262⋅5=95.\frac{12+26}{2}\cdot5=95.

Check your understanding

A convex trapezoid has area 6666 square cm and parallel bases 88 cm and 1414 cm. What is its perpendicular height?

Rhombi and their diagonals

A rhombus has perpendicular diagonals that bisect each other, as proved in Chapter 8. This gives four right triangles whose areas add to half the product of the full diagonal lengths.

The rhombus diagonal area formula#

For a rhombus with diagonals 1818 and 88 units, each right triangle has legs 99 and 44 units. Each has area 12⋅9⋅4\tfrac12\cdot9\cdot4, so all four together have area 7272 square units.

Perpendicular diagonals partition the areaSchematic rhombus ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 18. AC is labeled 8. A single tick on every side marks all four sides congruent.ABCDO188Not to scale
Four right triangles partition the rhombus; the labels give full diagonal lengths in units.

Now let d1d_1 and d2d_2 be the full diagonal lengths of any nondegenerate rhombus. Perpendicularity supplies the right angles, and bisection supplies the half lengths. The four triangles cover the rhombus without overlapping interiors:

A=4(12⋅d12⋅d22),A=12d1d2.\begin{aligned} A&=4\left(\frac12\cdot\frac{d_1}{2}\cdot\frac{d_2}{2}\right),\\ A&=\frac12 d_1d_2. \end{aligned}

The same partition worked for the numerical lengths; their values were not part of the reason it worked.

Worked example 3 Use full diagonal lengths

Problem. The diagonals of a rhombus measure 1414 m and 1010 m. Find its area.

Perpendicular diagonals partition the areaSchematic rhombus ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 14 m. AC is labeled 10 m. A single tick on every side marks all four sides congruent.ABCDO14 m10 mNot to scale
The two full diagonals meet perpendicularly.

Solution. The formula uses the full lengths, not their halves.

A=12⋅14⋅10,A=70 m2.\begin{aligned} A&=\frac12\cdot14\cdot10,\\ A&=70\text{ m}^2. \end{aligned}

Check your understanding

A rhombus has diagonals 1111 cm and 88 cm. What is its area?

Kites need not split into four equal areas

A kite here is a convex quadrilateral with two disjoint pairs of adjacent congruent sides, each side in one pair. Rhombi and squares qualify; a kite’s diagonals are perpendicular, but in a general kite only the diagonal joining the equal-side vertices must bisect the other.

The kite diagonal area formula#

In the kite below, the horizontal diagonal is 1212 units and is bisected into two 66 unit parts. The vertical diagonal is 99 units, split into 22 and 77 units. The four right triangles are visible, but the two upper triangles are smaller than the two lower ones.

Perpendicular diagonals partition the areaSchematic convex kite ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. The two horizontal halves are each 6. The upper vertical part is 2 and the lower vertical part is 7. Single side ticks pair AB with AD; double ticks pair BC with CD.ABCDO6627Not to scale
The four right triangles have two different sizes; all lengths are in units.

Combine the upper pair into triangle ABDABD, using base BDBD. Then do the same with triangle CBDCBD:

A=12⋅12⋅2+12⋅12⋅7,A=54.\begin{aligned} A&=\frac12\cdot12\cdot2+\frac12\cdot12\cdot7,\\ A&=54. \end{aligned}

For the general argument, let the perpendicular diagonals ACAC and BDBD of a convex quadrilateral meet at OO. Let the full length of BDBD be d1d_1, and let the parts AOAO and OCOC have lengths xx and yy. Then the full diagonal length d2d_2 of ACAC is x+yx+y. The triangles on the two sides of BDBD have perpendicular heights xx and yy:

A=12d1x+12d1y,A=12d1(x+y),A=12d1d2.\begin{aligned} A&=\frac12 d_1x+\frac12 d_1y,\\ A&=\frac12 d_1(x+y),\\ A&=\frac12 d_1d_2. \end{aligned}

The numerical split into 22 and 77 did not matter. This argument applies to every convex quadrilateral with perpendicular diagonals, so it includes every kite and rhombus; it does not require both diagonals to bisect each other.

Worked example 4 A diagonal split and an area

Problem. In convex kite ABCDABCD, the equal-side pairs meet at AA and CC. Diagonals ACAC and BDBD meet at OO. The full diagonal ACAC measures 1616 cm, its part AOAO measures 55 cm, and the full diagonal BDBD measures 1111 cm. Find OCOC and the kite’s area.

Perpendicular diagonals partition the areaSchematic convex kite ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 11 cm. AC is labeled 16 cm. AO is 5 cm. Single side ticks pair AB with AD; double ticks pair BC with CD.ABCDO11 cm16 cm5 cmNot to scale
The vertical diagonal is not marked as bisected.

Solution. For the missing part, subtract along ACAC:

OC=16−5,OC=11 cm.\begin{aligned} OC&=16-5,\\ OC&=11\text{ cm}. \end{aligned}

For the area, use the two full diagonals. Chapter 8’s kite theorem guarantees their perpendicularity.

A=12⋅16⋅11,A=88 cm2.\begin{aligned} A&=\frac12\cdot16\cdot11,\\ A&=88\text{ cm}^2. \end{aligned}

Check your understanding

A convex quadrilateral has diagonals of 1010 cm and 1717 cm. Which extra fact justifies an area of 8585 square cm?

Equal diagonals alone do not supply perpendicular heights. A rectangle generally has equal diagonals without perpendicular diagonals, so equal lengths alone do not justify taking half their product.

Decompose when a special formula does not apply

A diagonal divides any convex quadrilateral into two triangles with nonoverlapping interiors. Use that diagonal as a base and find each vertex’s perpendicular distance to its line; these heights need not be parts of the other diagonal.

Worked example 5 Two different heights to one diagonal

Problem. In convex quadrilateral ABCDABCD, diagonal ACAC is 1818 cm. The perpendicular distances from BB and DD to line ACAC are 33 cm and 88 cm, respectively. Find the quadrilateral’s area.

One diagonal gives two trianglesSchematic convex quadrilateral ABCD, not to scale. AC is a horizontal diagonal labeled 18 cm. Vertex B lies above AC and vertex D below it. Dashed perpendicular heights from B and D to AC are labeled 3 cm and 8 cm. Right-angle squares mark their feet. BD is not drawn or asserted perpendicular to AC.ABCD18 cm3 cm8 cmNot to scale
Each marked height belongs to a different triangle.

Solution. Let A1A_1 be the area of triangle ABCABC and A2A_2 the area of triangle ACDACD. They lie on opposite sides of ACAC, so add their areas.

A1=12⋅18⋅3,A1=27 cm2.\begin{aligned} A_1&=\frac12\cdot18\cdot3,\\ A_1&=27\text{ cm}^2. \end{aligned}A2=12⋅18⋅8,A2=72 cm2.\begin{aligned} A_2&=\frac12\cdot18\cdot8,\\ A_2&=72\text{ cm}^2. \end{aligned}A=27+72,A=99 cm2.\begin{aligned} A&=27+72, \\ A&=99\text{ cm}^2. \end{aligned}

We did not need a pair of parallel sides or perpendicular diagonals. The two marked distances were the required triangle heights.

Check your understanding

A diagonal of a convex quadrilateral is 99 cm. The other vertices have perpendicular distances 44 cm and 66 cm to its line. What is the total area?

Match the formula to the evidence

Given informationArea methodRequired condition
Two base lengths and their separationA=12(b1+b2)hA=\tfrac12(b_1+b_2)hChosen bases parallel; hh perpendicular to them
Both full diagonal lengthsA=12d1d2A=\tfrac12 d_1d_2Convex quadrilateral with perpendicular diagonals
A diagonal and two distances to its lineAdd the two triangle areasConvex quadrilateral; both distances perpendicular to the chosen diagonal

When a problem asks for a missing dimension, use the same condition check before forming its area equation. An equation solved correctly cannot repair an unjustified formula.

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Practice

Multiple Choice Questions (MCQ)

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A bit of history (optional)

The Indian mathematician and astronomer Brahmagupta wrote the Brahmasphutasiddhanta in the seventh century. Alongside astronomical calculations, it contained rules for lengths and areas, including a remarkable quadrilateral rule. His mathematical instructions were written in verse, a compact form that could be remembered and passed on.

That quadrilateral rule later prompted a disagreement among historians. It works for quadrilaterals whose four vertices lie on one circle, but Brahmagupta did not explicitly state that restriction. Some readers regard this as an omission; others think the intended setting was already understood. The surviving words leave room for both readings. MacTutor’s biography describes the debate.

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