Math by First Principles

Area from Rectangles to Triangles

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Learning goals

  • Distinguish perimeter from area and choose the appropriate measurement units
  • Justify rectangle area by tiling with unit squares and rational subdivisions
  • Derive parallelogram and triangle areas using congruent pieces and area addition
  • Identify the perpendicular height matching any chosen side of a triangle
  • Calculate areas and recover missing base or height measurements

What area is allowed to do

The perimeter of a polygon is the total length of its boundary. The area is the measure of the region it encloses. We write PP for perimeter and AA for area. Lengths use units such as centimeters; areas use square units such as square centimeters, written cm2\text{cm}^2.

A rectangle 44 units long and 33 units wide has 1212 unit squares inside. Its boundary length is 1414 units. Counting squares and adding edge lengths answer different questions.

We accept three area postulates, the starting assumptions for our arguments.

These assumptions explain why cutting and reconstructing a region preserves area: each moved piece keeps its area, and the same piece areas add before and after. They say nothing about preserving the outer boundary length.

Check your understanding

Which measurement describes the region inside a triangle?

Rectangles from square tiles

A rectangle’s area comes from counting equal square pieces. A 55 by 22 rectangle contains two rows of five unit squares. Addition gives 1010 square units. In general, positive integer side lengths bb and hh give hh rows of bb squares, so their product counts the tiles.

Fractions need smaller square tiles. Consider a rectangle 53\tfrac53 units long and 11 unit high. Divide a unit square into three rows and three columns. Its nine small squares are congruent, so each has area 19\tfrac19 square unit. The rectangle contains five columns and three rows of those small squares. Its area is 159\tfrac{15}{9}, or 53\tfrac53, square units.

A five-column, three-row rectangle of small squares beside a unit square divided into three columns and three rows. Each small square has side one third of a unit.A five-column, three-row rectangle of small squares beside a unit square divided into three columns and three rows. Each small square has side one third of a unit.5/3 units11 unitEach small side is 1/3 unit
The unit square fixes the area of each small tile before we count the rectangle.

Rectangle area for rational side lengths#

For another numerical case, take a rectangle 74\tfrac74 units by 32\tfrac32 units. A unit square splits into 1616 congruent squares of side 14\tfrac14 unit. The rectangle has 77 columns and 66 rows of them, so its area in square units is

A=42⋅116,A=218.\begin{aligned} A&=42\cdot\frac1{16},\\ A&=\frac{21}{8}. \end{aligned}

This agrees with multiplying the two side lengths. The counting does not depend on these particular fractions. For any two positive rational lengths, choose a positive integer NN so both can be written with denominator NN. Write the lengths as mN\tfrac mN and nN\tfrac nN.

A unit square contains N2N^2 congruent squares of side 1N\tfrac1N, each with area 1N2\tfrac1{N^2}. Our rectangle has mm columns and nn rows, so addition gives

A=mnN2,A=mN⋅nN.\begin{aligned} A&=\frac{mn}{N^2},\\ A&=\frac mN\cdot\frac nN. \end{aligned}

Thus a rectangle with positive rational base bb and perpendicular height hh has area

A=bh.A=bh.

We accept the extension of this rectangle formula to positive irrational side lengths here. For example, a rectangle with sides 22 and 3\sqrt3 units has area 232\sqrt3 square units. A finite rational grid does not tile such a side exactly. The extension is an additional accepted foundation, not something the finite tiling proof has established. No limiting argument is being assumed.

Worked example 1 Count rational tiles and measure the boundary

Problem. A rectangle has side lengths 73\tfrac73 cm and 53\tfrac53 cm. Find its area by counting squares of side 13\tfrac13 cm, and find its perimeter.

A rectangle tiled by 7 columns and 5 rows of congruent squares, each of side 1/3 cm. The sides are 7/3 cm and 5/3 cm.A rectangle tiled by 7 columns and 5 rows of congruent squares, each of side 1/3 cm. The sides are 7/3 cm and 5/3 cm.7/3 cm5/3 cm
Seven columns and five rows make 35 congruent tiles of area one ninth of a square centimeter.

Solution. A unit square contains nine of these small squares, so each has area 19 cm2\tfrac19\text{ cm}^2. Seven columns and five rows contain 3535 small squares.

A=359 cm2.A=\frac{35}{9}\text{ cm}^2.

The perimeter counts two copies of each side length.

P=2(73+53),P=8 cm.\begin{aligned} P&=2\left(\frac73+\frac53\right),\\ P&=8\text{ cm}. \end{aligned}

Check your understanding

A unit square is divided into four rows and four columns of congruent squares. What is the area of one small square?

Slide a parallelogram into a rectangle

Choose a side of a parallelogram as its base, of length bb. Its corresponding height, of length hh, is the perpendicular distance between the lines containing that base and the opposite side. A slanted side usually has a different length.

For a parallelogram with base 99 units and height 55 units, cut down the perpendicular from its upper left corner. Slide the left triangular piece 99 units to the right. It completes a rectangle 99 units wide and 55 units high, without gaps or overlaps. Both regions therefore have area 4545 square units.

Two complete cut-and-slide stages at the same scale. The initial parallelogram has base 9 and height 5. Its left corner triangle T is cut along a perpendicular, then translated by the base length into the opposite corner. The result is a 9 by 5 rectangle with the moved triangle red and dashed.Two complete cut-and-slide stages at the same scale. The initial parallelogram has base 9 and height 5. Its left corner triangle T is cut along a perpendicular, then translated by the base length into the opposite corner. The result is a 9 by 5 rectangle with the moved triangle red and dashed.Cut perpendicular to the base95TTranslate T by the base length95T
The moved triangle is labeled T in both stages; its dashed outline also distinguishes it in print.

The parallelogram cut and slide#

The same move works for a base of 77 units and height 33 units when the sideways shift is no more than 77 units. The cut-off right triangle has the same height and sideways width as the empty corner on the right. A translation by 77 units makes it fit that corner exactly. The resulting rectangle has area 2121 square units.

For general positive bb and hh, when the sideways shift is at most bb, translation by bb fits the cut-off triangle into the opposite corner. Parallel sides give matching perpendicular heights and matching sideways widths. The remaining piece and the moved triangle have disjoint interiors and fill a rectangle with sides bb and hh. The postulates give equal total areas.

If the parallelogram leans left, reflect this construction. A rectangle itself needs no cut.

If the sideways shift is greater than the base, one vertical cut might miss the opposite side segment. For instance, a parallelogram with base 33, height 66, and sideways shift 66 can be cut into three equal-height horizontal strips. Each is a parallelogram with base 33, height 22, and sideways shift 22. Rearrange each strip as above. Their three areas of 66 square units add to 1818 square units.

A strongly leaning parallelogram divided into three equal-height strips. Each strip has base 3, height 2, and sideways shift 2. The total sideways shift is 6 and total height is 6.A strongly leaning parallelogram divided into three equal-height strips. Each strip has base 3, height 2, and sideways shift 2. The total sideways shift is 6 and total height is 6.3Each stripheight 23 strips; total height 6
Shorter strips make the cut possible even for a large sideways shift.

In general, choose a finite number nn of equal-height strips large enough that the shift of each strip is at most bb. Parallel-line proportionality makes each shift the original shift divided by nn. Each strip has height hn\tfrac hn and base bb. Adding their rectangle areas gives

A=n(bhn).A=n\left(b\frac hn\right).

Therefore every nondegenerate parallelogram has area

A=bh.A=bh.

This uses the rectangle formula, including its accepted irrational extension when needed, and only finitely many cuts.

Worked example 2 Use the height for area and the side for perimeter

Problem. A parallelogram has base 1313 cm, perpendicular height 66 cm, and adjacent side 1010 cm. Find its area and perimeter.

Schematic parallelogram with base 13 cm, perpendicular height 6 cm, and slanted side 10 cm. Not drawn to scale.Schematic parallelogram with base 13 cm, perpendicular height 6 cm, and slanted side 10 cm. Not drawn to scale.13 cm6 cm10 cm
The height meets the base at a right angle; the adjacent side belongs to the boundary.

Solution. The area uses the base and its perpendicular height.

A=13⋅6,A=78 cm2.\begin{aligned} A&=13\cdot6,\\ A&=78\text{ cm}^2. \end{aligned}

The boundary has two sides of each given side length.

P=2(13+10),P=46 cm.\begin{aligned} P&=2(13+10),\\ P&=46\text{ cm}. \end{aligned}

Check your understanding

A parallelogram is cut into pieces that fill a rectangle without gaps or interior overlaps. The rectangle has the same base and perpendicular height. Why do the areas agree?

Pair two triangles

A triangle is half of a suitable parallelogram. Take a triangle with base 66 units and height 44 units. Turn a congruent copy halfway around the midpoint of one of its other sides. The two copies fill a parallelogram with the same base and height. Its area is 2424 square units, so each triangle has area 1212 square units.

Two congruent triangles ABC and DCB form parallelogram ABDC with base AB of length 6 and height 4. BC is the shared diagonal. A half turn about the midpoint of BC carries one triangle to the other.Two congruent triangles ABC and DCB form parallelogram ABDC with base AB of length 6 and height 4. BC is the shared diagonal. A half turn about the midpoint of BC carries one triangle to the other.ABCD46M
A half turn pairs the triangles along BC. The two copies have equal areas.

Why a triangle has half the base times the height#

For a triangle with base 55 and height 33, the half-turn construction makes a parallelogram with area 1515 square units. Congruence gives equal areas for its two triangular pieces. Addition gives 2A=152A=15, so one has area 152\tfrac{15}{2} square units.

Now take any triangle ABCABC whose vertices are not collinear, with chosen base ABAB of length bb and perpendicular height hh from CC to line ABAB. Turn the triangle halfway around the midpoint of BCBC. The images of BB and CC exchange places; call the image of AA point DD.

The image of ABAB is DCDC, parallel and equal to ABAB. The image of ACAC is DBDB, parallel and equal to ACAC. Thus the union is parallelogram ABDCABDC. The copies lie on opposite sides of BCBC, meet only along that side, and fill the parallelogram. Its base is bb and its height is hh.

Let AA be the original triangle’s area. The postulates and the parallelogram formula give

2A=bh,A=12bh.\begin{aligned} 2A&=bh,\\ A&=\frac12bh. \end{aligned}

The half turn works whether the triangle is acute, right, or obtuse. No rectangle or trapezoid formula for a larger enclosing shape is needed.

Worked example 3 Find an area with an exterior height

Problem. Triangle ABCABC has an 1111 cm base ABAB. The perpendicular from CC meets line ABAB at HH beyond BB, and its length is 66 cm. Find the triangle’s area.

Schematic triangle with base 11 cm, perpendicular height 6 cm, with the foot beyond the right endpoint. Not drawn to scale.Schematic triangle with base 11 cm, perpendicular height 6 cm, with the foot beyond the right endpoint. Not drawn to scale.11 cm6 cmABCH
The supporting line is extended to H; the base remains segment AB.

Solution. The height is the perpendicular distance to the base’s line, so the exterior segment CHCH is the correct height. The extension BHBH is not part of the base.

A=12⋅11⋅6,A=33 cm2.\begin{aligned} A&=\frac12\cdot11\cdot6,\\ A&=33\text{ cm}^2. \end{aligned}

Check your understanding

A triangle has a 15 cm base. Its opposite vertex is 8 cm from the base line, measured perpendicularly, with the foot outside the base. What is the area?

Any side can be the base

The base is a choice, not a permanent label for the lowest or longest side. Changing the base changes which altitude supplies the height. The triangle itself and its area stay unchanged.

For triangle ABCABC, base ABAB uses the altitude from CC; base ACAC uses the altitude from BB; base BCBC uses the altitude from AA. In an obtuse triangle, two of these feet lie on extensions of sides. The third lies on a side segment. In a right triangle, choosing either leg makes the other leg the height; choosing the hypotenuse requires the perpendicular from the right-angle vertex.

The same obtuse triangle ABC is shown in three panels. Base AB uses exterior height CH beyond B. Base AC uses interior height BK. Base BC uses exterior height AL beyond B. Every foot is marked perpendicular.The same obtuse triangle ABC is shown in three panels. Base AB uses exterior height CH beyond B. Base AC uses interior height BK. Base BC uses exterior height AL beyond B. Every foot is marked perpendicular.Base AB; height CHABCHBase AC; height BKABCKBase BC; height ALABCL
Each panel uses the same triangle and a different chosen base. Right-angle marks identify the matching height.

Worked example 4 Use three base-and-height pairs

Problem. A triangle has sides 1313 cm, 2020 cm, and 2121 cm. Its perpendicular height to the 2121 cm side is 1212 cm. Find its area, then find the heights to the other two sides.

Schematic triangle with base 21 cm, perpendicular height 12 cm, and a side labeled 13 cm. Not drawn to scale.Schematic triangle with base 21 cm, perpendicular height 12 cm, and a side labeled 13 cm. Not drawn to scale.21 cm12 cm13 cm
The other slanted side is 20 cm. The given height belongs to the 21 cm base.

Solution. First use the supplied matching pair.

A=12⋅21⋅12,A=126 cm2.\begin{aligned} A&=\frac12\cdot21\cdot12,\\ A&=126\text{ cm}^2. \end{aligned}

For the 1313 cm base, call the height h13h_{13}.

126=12⋅13h13,h13=25213 cm.\begin{aligned} 126&=\frac12\cdot13h_{13},\\ h_{13}&=\frac{252}{13}\text{ cm}. \end{aligned}

For the 2020 cm base, call the height h20h_{20}.

126=12⋅20h20,h20=635 cm.\begin{aligned} 126&=\frac12\cdot20h_{20},\\ h_{20}&=\frac{63}{5}\text{ cm}. \end{aligned}

The three matching products all equal 252 cm2252\text{ cm}^2, twice the triangle’s area.

Worked example 5 Recover a height from the enclosed area

Problem. A triangle has area 65 cm265\text{ cm}^2 and a 1010 cm chosen base. Find the corresponding height and check it.

Schematic triangle with base 10 cm, perpendicular height h. Not drawn to scale.Schematic triangle with base 10 cm, perpendicular height h. Not drawn to scale.10 cmh
The height is unknown; this schematic supplies no measuring scale.

Solution. Substitute the known area and base into the triangle formula.

65=12⋅10h,65=5h,h=13 cm.\begin{aligned} 65&=\frac12\cdot10h,\\ 65&=5h,\\ h&=13\text{ cm}. \end{aligned}

Check by substituting the height back into the area calculation.

12⋅10⋅13=65 cm2.\frac12\cdot10\cdot13=65\text{ cm}^2.

Area and one base determine the corresponding height. They do not supply all the side lengths needed for perimeter.

Check your understanding

A triangle has area 48 square units and a chosen base of 6 units. Which height belongs to that base?

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A bit of history (optional)

In Euclid’s Elements, a diagonal of a parallelogram is called its “diameter.” The word can surprise a modern reader expecting a circle. In Book I, Proposition 34, Euclid first uses parallel-line angles to compare the two triangles formed by that diagonal. Their congruence supplies equal sides and angles, then equal areas. The proposition bundles properties that a modern textbook might spread across several pages. There is no numerical area calculation in the argument: matching pieces establishes the equality directly.

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