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Area of triangles

A triangle’s slanted sides never line up with rows of squares. How can its area be found exactly?

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Why this exists

Area began as counting unit squares (Area as covering), and for a rectangle the count is quick: rows times columns (Area of rectangles and squares). A parallelogram followed by cutting off a slanted end and sliding it across, which turns it into a rectangle with the same base and height (Area of parallelograms and trapezoids). A triangle does not give way so easily. Its slanted sides cut the squares they cross into odd pieces, and no single cut and slide makes it a rectangle.

Yet triangles are everywhere a flat surface has to be measured: the sail on a boat, the side of a wooden doorstop, the pieces of a wooden puzzle. Every flat shape with straight sides can also be cut into triangles, so a rule for the triangle is a rule for all of them. The way in is a surprise: instead of cutting a triangle up, put two of them together.

Definition

The base of a triangle is whichever side is chosen to measure from. Any of the three sides may be the base; it does not have to be the side drawn at the bottom.

The height that goes with that base is the distance from the opposite corner to the line the base lies on, measured at right angles to that line (Perpendicular lines). It is a straight distance to the base’s line, never the length of a slanted side.

The area of the triangle is then

A = \frac{1}{2}bh

where b is the length of the base and h is the height that goes with it, both in the same unit. The area comes out in square units of that unit.

On the second day two four-sided shapes appear (Quadrilaterals). A diagonal is a segment joining two opposite corners. A kite has two pairs of equal sides, each pair meeting at a corner, and a rhombus has all four sides equal (Rhombuses, rectangles and squares). In both, the two diagonals cross at right angles.

Intuition

Take a triangle and an exact copy of it. Turn the copy upside down and set it against one of the slanted sides, as in Figure 1. The two fit together into a parallelogram, and that parallelogram has the same base b and the same height h as the triangle. Its area is bh, and the triangle is one of its two identical halves, so the triangle’s area is half of bh.

b h copy
Figure 1Two copies of any triangle make a parallelogram with the same base and height.

Where the height sits depends on the shape of the triangle (Classifying triangles), as Figure 2 shows. In an acute triangle it falls inside. In a right triangle one leg stands at right angles to the other, so each leg is the height for the other as base. In an obtuse triangle with a short side as base, the top corner hangs past the end of the base, so the base’s line is extended and the height falls outside the triangle. It is still the distance from the corner straight to the line of the base.

b h b h b h
Figure 2The height goes from the top corner to the line of the base, at right angles. It may lie inside, along a side, or outside.

One more picture: slide the top corner along a line parallel to the base. The triangle leans and its slanted sides change length, but the base and height stay the same, so the area does too.

Results

Area of a triangle. For any triangle, with any side as the base b and h the height to that side,

A = \frac{1}{2}bh

The three sides give three different pairs of base and height, but always the same area.

Area of a kite or rhombus. If a four-sided shape has diagonals of lengths d_1 and d_2 that cross at right angles, as in every kite and every rhombus,

A = \frac{1}{2}d_1 d_2
Why it is true

The triangle. Call the triangle ABC, with base AB of length b and height h from C to the line AB. Make an exact copy and turn it half a turn about the midpoint of side BC. Then B and C swap places and A lands at a new point A'. A half turn sends every line to a line parallel to it, so CA' is parallel to AB, and BA' is parallel to AC. The four-sided shape ABA'C has both pairs of opposite sides parallel, so it is a parallelogram. Its base is AB, of length b, and its height is the distance between the parallel lines AB and CA', which is the distance from C to line AB: the same h. By the parallelogram formula its area is bh. It is made of two identical triangles with no gap and no overlap, so each has area \displaystyle \frac{1}{2}bh.

Nothing in this used the shape of the triangle. When the triangle is obtuse and its height falls outside, the parallelogram leans further and its height falls outside too, and the parallelogram formula holds just the same. Nothing depended on which side was named AB either, so any side may be the base.

The kite and the rhombus. Let the diagonals cross at right angles at O, and cut along the diagonal of length d_1. The two pieces are triangles with that diagonal as a shared base. The other diagonal is at right angles to it, so its two parts on either side of O are the two heights. Call them p and q, so p + q = d_2. Adding the two triangles and using the distributive property:

\begin{aligned} \tfrac{1}{2}d_1 p + \tfrac{1}{2}d_1 q &= \tfrac{1}{2}d_1(p + q) \\ &= \tfrac{1}{2}d_1 d_2 \end{aligned}

Worked examples

The lesson runs over two days. Examples 1 to 4 belong to the first: the formula, and where the height sits in right and obtuse triangles. Examples 5 to 7 belong to the second: any side as the base, working back from an area, and kites and rhombuses from their diagonals.

Example 1

A jib sail

Fleur’s sailboat carries a triangular jib sail (Figure 3). Its bottom edge is 4 meters long, and its top corner is 7 meters above the line of the bottom edge. The longest edge measures about 7.6 meters. How much sailcloth is in the sail?

4 m 7 m 7.6 m
Figure 3A jib sail. The slanted edge is not the height.
Solution

Take the bottom edge as the base, so b = 4. The height that goes with it is the distance from the top corner straight down to the line of the base, so h = 7. The 7.6 meter edge is slanted, so it is not a height and plays no part. Substituting into the formula:

A = \frac{1}{2} \times 4 \times 7 = \frac{1}{2} \times 28 = 14

The sail holds 14 square meters of cloth. As a check, a 4 by 7 meter rectangle drawn around the sail has area 28 square meters, and the dashed height splits both rectangle and sail into two parts, each sail part being half of its rectangle part. The sail is half of 28.

Example 2

A wooden doorstop

Tarek cuts a wooden doorstop. Each of its two side faces is a right triangle whose legs are 6\tfrac{1}{2} inches along the floor and 1\tfrac{1}{2} inches at the tall end. He varnishes both side faces. What area does he varnish?

Solution

The two legs meet at a right angle, so either leg is the height for the other as base: b = 6\tfrac{1}{2} and h = 1\tfrac{1}{2}. Write the mixed numbers as fractions and multiply:

A = \frac{1}{2} \times \frac{13}{2} \times \frac{3}{2} = \frac{39}{8} = 4\tfrac{7}{8}

Each face is 4\tfrac{7}{8} square inches. Two faces are \displaystyle 2 \times \frac{39}{8} = \frac{39}{4} = 9\tfrac{3}{4} square inches. Two copies of the triangle make exactly the area of a 6\tfrac{1}{2} by 1\tfrac{1}{2} rectangle, which is the reason the formula has its half.

Example 3

An obtuse puzzle piece

Marit is cutting the pieces of a wooden puzzle. One piece is shaped like triangle ABC in Figure 4. The base AB is 9 cm. The corner C hangs out past B: the line from C at right angles to line AB meets it at D, 3 cm beyond B, and CD = 4 cm. Find the area of the piece.

9 cm 3 cm 4 cm A B C D
Figure 4The piece’s top corner hangs past the end of its base, so the height lies outside it.
Solution

The height for base AB is CD, even though it lies outside the piece, because it is the distance from C to the line the base lies on. So b = 9 and h = 4:

A = \frac{1}{2} \times 9 \times 4 = 18

The piece is 18 square centimeters. Subtraction gives a check. Triangle ADC is a right triangle with legs AD = 9 + 3 = 12 and CD = 4, so its area is \displaystyle \frac{1}{2} \times 12 \times 4 = 24. It is made of the piece and the small right triangle BDC, of area \displaystyle \frac{1}{2} \times 3 \times 4 = 6. The piece is 24 - 6 = 18 square centimeters, as the formula said. The 3 cm was needed only for the check.

Example 4

Two sails with the same base and height

Corwin is choosing between two sail designs. Both have a bottom edge of 3.2 meters and a top corner 5.5 meters above the line of the bottom edge. In the first the top corner is above the middle of the bottom edge; in the second it is above a point well behind the back end, so the sail leans. Which holds more cloth?

Solution

Both sails have b = 3.2 and h = 5.5, and the formula uses only these two numbers, so their areas are equal. The commutative and associative properties of multiplication let the three factors be multiplied in any order, so halve 3.2 first:

A = \frac{1}{2} \times 3.2 \times 5.5 = 1.6 \times 5.5 = 8.8

Each sail holds 8.8 square meters. The leaning sail is an obtuse triangle with its height outside, like Example 3. Sliding the top corner along a line parallel to the base changes the shape and the slanted edges, but not the area.

Example 5

Using a different side as the base

Marit’s next puzzle piece is a right triangle with sides 6 cm, 8 cm and 10 cm, the right angle between the 6 cm and 8 cm sides. Find its area, then the height that goes with the 10 cm side as the base (Figure 5).

10 cm 6 cm 8 cm h
Figure 5The same triangle with its longest side as the base.
Solution

The legs are a base and height for each other:

A = \frac{1}{2} \times 6 \times 8 = 24

The area is 24 square centimeters, whichever side is called the base. With b = 10 the formula still gives 24, so

\frac{1}{2} \times 10 \times h = 24

Half of 10 is 5, so 5h = 24. Dividing both sides by 5 keeps them equal (Inverse operations and the properties of equality), so h = 24 \div 5 = 4.8 cm. Check: \displaystyle \frac{1}{2} \times 10 \times 4.8 = 5 \times 4.8 = 24. The height is shorter than both legs, as it must be, since it is the shortest path from the right-angle corner to the long side.

Example 6

A kite-shaped piece

Marit’s puzzle also has a kite-shaped piece TLUR (Figure 6). Its diagonals are TU = 16 cm and LR = 10 cm, crossing at right angles at O, with LO = OR = 5 cm. Find its area by cutting it into two triangles, then with the diagonal formula.

T U L R O
Figure 6Cutting along the diagonal TU makes two triangles, with heights LO and OR.
Solution

Cut along TU. Triangles TLU and TRU share the base TU = 16. Because LR is at right angles to TU, the height from L to line TU is LO = 5, and the height from R is OR = 5. Each triangle has area

\frac{1}{2} \times 16 \times 5 = 40

so the kite is 40 + 40 = 80 square centimeters. The diagonal formula gives \displaystyle \frac{1}{2} \times 16 \times 10 = 80 in one step, because it is the two triangles added together.

Example 7

A rhombus and its box

Siddharth cuts six identical rhombus pieces for a puzzle, each with diagonals 12 cm and 7.5 cm. How much wood do the six pieces cover?

Solution

The diagonals of a rhombus cross at right angles, so

A = \frac{1}{2} \times 12 \times 7.5 = 6 \times 7.5 = 45

Each piece is 45 square centimeters, and six are 6 \times 45 = 270 square centimeters.

There is a second way to see the 45. Draw the rectangle whose sides pass through the four corners, parallel to the diagonals. It is 12 by 7.5, area 90. The diagonals cut it into four small rectangles, and each side of the rhombus cuts one of them exactly in half. So the rhombus is half its box: half the product of the diagonals.

Boundaries

  • The height must meet the line of the base at a right angle. A slanted side is a height only in a right triangle, where the two legs are base and height for each other.
  • Base and height must belong together. The height is measured to the line of the chosen base; pairing a base with the height to a different side gives a wrong area.
  • Both lengths must be in the same unit. A base in meters and a height in centimeters must be converted before multiplying.
  • The diagonal formula needs diagonals that cross at right angles. A rectangle 8 cm by 1 cm has area 8 \times 1 = 8 square centimeters. Each diagonal runs the whole 8 cm across and climbs 1 cm as well, so each is longer than 8 cm, and half their product is more than \displaystyle \frac{1}{2} \times 8 \times 8 = 32, far more than the true area. Its diagonals do not cross at right angles, so the formula does not apply.
  • The formulas measure flat surfaces. A sail that billows in the wind has a little more cloth than the flat triangle between its corners.

Where this goes next

Every straight-sided shape can be cut into triangles, so this formula is the tool behind Area of any polygon by cutting it into triangles and rectangles, and behind Area of a polygon from coordinates for shapes plotted on a grid. Area of composite figures and shaded regions adds and subtracts pieces in just the way Example 3 subtracted one triangle from another. A regular polygon splits from its center into equal triangles, which gives Area of a regular polygon, and the slanted faces of a pyramid are triangles whose areas add up to Surface area of pyramids and cones. Later, one of the classic proofs of the The Pythagorean theorem fits triangles together and compares areas, and when only the three side lengths are known, Heron’s formula finds the area without a height.

Exercises

Exercises 1 to 11 go with the first part of the lesson (Examples 1 to 4), Exercises 12 to 20 with the second (Examples 5 to 7), and the reasoning group draws on both.

Part 1: base and height

  1. A triangle has base 10 cm and height 7 cm. Find its area.
  2. A triangle has base 15 m and height 8 m. Find its area.
  3. A triangle has base 9 inches and height 5 inches. Find its area.
  4. A triangle has base 6.4 m and height 2.5 m. Find its area.
  5. A triangle has base \displaystyle \frac{3}{4} foot and height \displaystyle \frac{2}{3} foot. Find its area.
  6. A right triangle has sides 9 cm, 12 cm and 15 cm, with the right angle between the 9 cm and 12 cm sides. Find its area.
  7. Marit makes a doorstop whose two side faces are right triangles with legs 8 inches and 2\tfrac{1}{4} inches. Find the total area of the two side faces.
  8. A puzzle piece is an obtuse triangle. With its 7 cm side as the base, the height falls outside the piece and measures 6 cm. Find the area of the piece.
  9. A mainsail is a right triangle. Its bottom edge is 2.8 m, its edge along the mast is 6 m and at right angles to the bottom edge, and its third edge is about 6.6 m. Find its area.
  10. Two triangles share a base 12 cm long, and both top corners lie on a line parallel to the base, 5 cm from it. One triangle is acute and the other obtuse. Find each area.
  11. A parallelogram and a triangle each have base 8 m and height 5 m. Find both areas. How many copies of the triangle cover the same area as the parallelogram?

Part 2: any side, working back, and diagonals

  1. A triangle has area 36 square centimeters and base 9 cm. Find its height.
  2. A triangular sail has area 27 square meters and height 6 m. Find its base.
  3. A right triangle has legs 15 cm and 20 cm, and its longest side is 25 cm. Find its area and the height that goes with the 25 cm side.
  4. A triangle has a 14 cm side, and the height that goes with it is 6 cm. Find the height that goes with its 12 cm side.
  5. A kite has diagonals 20 cm and 9 cm. Find its area.
  6. A rhombus has diagonals 6.4 m and 5 m. Find its area.
  7. A square has two diagonals, each 10 cm long. Use the diagonals to find its area.
  8. A kite has area 60 square inches and one diagonal of 15 inches. Find the other diagonal.
  9. A rectangular puzzle tray 40 cm by 30 cm holds a rhombus piece whose four corners touch the midpoints of the tray’s sides; four triangle pieces fill the rest. Find the area of the rhombus and of the four triangles together.

Reasoning

  1. Explain, using two copies of a triangle, why the triangle formula has a \displaystyle \frac{1}{2} in it.
  2. A triangle has a 10 cm side and a 7 cm side. Can its area be found from these two lengths alone? Explain.
  3. The base of a triangle is doubled and its height is halved. What happens to its area? Explain.
  4. A four-sided shape has diagonals 8 cm and 6 cm that cross at right angles, but neither crosses the other at its midpoint. Show that its area is still 24 square centimeters.
Answers
  1. 35 square centimeters. \displaystyle \frac{1}{2} \times 10 \times 7 = 35.
  2. 60 square meters. \displaystyle \frac{1}{2} \times 15 \times 8 = 60.
  3. 22\tfrac{1}{2} square inches. \displaystyle \frac{1}{2} \times 9 \times 5 = \frac{45}{2}.
  4. 8 square meters. Half of 6.4 is 3.2, and 3.2 \times 2.5 = 8.
  5. \displaystyle \frac{1}{4} square foot. \displaystyle \frac{1}{2} \times \frac{3}{4} \times \frac{2}{3} = \frac{6}{24} = \frac{1}{4}.
  6. 54 square centimeters. The legs are base and height: \displaystyle \frac{1}{2} \times 9 \times 12 = 54. The 15 cm side is not used.
  7. 18 square inches. One face is \displaystyle \frac{1}{2} \times 8 \times \frac{9}{4} = 9, and two faces are 18.
  8. 21 square centimeters. \displaystyle \frac{1}{2} \times 7 \times 6 = 21; a height outside the piece counts the same.
  9. 8.4 square meters. The mast edge is at right angles to the bottom edge, so it is the height: \displaystyle \frac{1}{2} \times 2.8 \times 6 = 1.4 \times 6 = 8.4. The 6.6 m edge is not used.
  10. Both are 30 square centimeters. They have the same base and the same height: \displaystyle \frac{1}{2} \times 12 \times 5 = 30.
  11. The parallelogram is 40 square meters and the triangle 20 square meters, so two copies. 8 \times 5 = 40 and \displaystyle \frac{1}{2} \times 8 \times 5 = 20.
  12. 8 cm. \displaystyle \frac{1}{2} \times 9 \times h = 36 means 4.5h = 36, so h = 36 \div 4.5 = 8. Check: \displaystyle \frac{1}{2} \times 9 \times 8 = 36.
  13. 9 m. \displaystyle \frac{1}{2} \times b \times 6 = 27 means 3b = 27, so b = 9.
  14. Area 150 square centimeters, height 12 cm. The legs give \displaystyle \frac{1}{2} \times 15 \times 20 = 150. Then \displaystyle \frac{1}{2} \times 25 \times h = 150, so 25h = 300 and h = 300 \div 25 = 12.
  15. 7 cm. The area is \displaystyle \frac{1}{2} \times 14 \times 6 = 42. With the 12 cm side as base, \displaystyle \frac{1}{2} \times 12 \times h = 42, so 6h = 42 and h = 7.
  16. 90 square centimeters. \displaystyle \frac{1}{2} \times 20 \times 9 = 90.
  17. 16 square meters. \displaystyle \frac{1}{2} \times 6.4 \times 5 = 3.2 \times 5 = 16.
  18. 50 square centimeters. A square is a rhombus, so its diagonals cross at right angles: \displaystyle \frac{1}{2} \times 10 \times 10 = 50.
  19. 8 inches. \displaystyle \frac{1}{2} \times 15 \times d = 60 means 7.5d = 60, so d = 60 \div 7.5 = 8.
  20. The rhombus is 600 square centimeters and the four triangles 600 square centimeters together. The rhombus’s diagonals join midpoints of opposite sides, so they are 40 and 30 and cross at right angles: \displaystyle \frac{1}{2} \times 40 \times 30 = 600. The tray is 40 \times 30 = 1{,}200, leaving 1{,}200 - 600 = 600. As a check, each triangle has legs 20 and 15, so the four make \displaystyle 4 \times \frac{1}{2} \times 20 \times 15 = 600.
  21. Turn a copy of the triangle upside down and fit it against one side. The two make a parallelogram with the triangle’s base b and height h, whose area is bh. The triangle is one of two identical halves, so its area is \displaystyle \frac{1}{2}bh.
  22. No. With the 10 cm side as base, the height depends on the angle between the two sides. It is largest, 7 cm, when the 7 cm side stands at right angles to the base, giving 35 square centimeters; at any other angle the height is less than 7 cm. The area can be anything above 0 up to 35 square centimeters.
  23. It stays the same. \displaystyle \frac{1}{2} \times (2b) \times \frac{h}{2} = \frac{1}{2}bh, because by the commutative and associative properties the 2 and the \displaystyle \frac{1}{2} can be multiplied first, and \displaystyle 2 \times \frac{1}{2} = 1.
  24. Cut along the 8 cm diagonal into two triangles with base 8. The other diagonal is at right angles to it, so its two parts, p and q, are the heights, and p + q = 6. The area is \displaystyle \frac{1}{2} \times 8 \times p + \frac{1}{2} \times 8 \times q = 4p + 4q = 4(p + q) = 4 \times 6 = 24 square centimeters.
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