math.wiki
Measurement › Area and perimeter › Area of any polygon by cutting it into triangles and rectangles
Open on the map
Math 6 · 6.G.A.1

Area of any polygon by cutting it into triangles and rectangles

A camp plot or the side of a bookend has straight edges but no formula of its own. How do you measure the surface inside it?

You need
Area of trianglesMath 6
Where you are 15 in Area and perimeter
Measurement12
Area and perimeter15
▴ 9 aboveSplitting a figure into rectanglesArea of triangles
Area of any polygon by cutting it into triangles and rectangles
Area of composite figures and shaded regionsWhy the circle area formula works▾ Show all 15 (1 more)

When you reach for it

Rectangles, triangles, parallelograms and trapezoids each have an area formula (Area of rectangles and squares, Area of triangles, Area of parallelograms and trapezoids). Most real shapes with straight edges are none of these: a camp plot with five corners, the side of a wooden bookend with a slanted brace, a marker sign shaped like an arrow. Such a shape is a polygon, a flat shape closed in by straight sides (Polygons), and any polygon can be broken into pieces that do have formulas.

There are two ways in. Cut and add: slice the shape into rectangles and triangles and add their areas. Surround and subtract: draw the smallest rectangle that holds the shape, then take away the pieces of that rectangle lying outside it. This lesson takes two days: Day 1 (Examples 1 to 3) cuts and adds, and Day 2 (Examples 4 to 6) surrounds and subtracts and asks which way needs the fewest measurements.

The procedure

Cut and add (Day 1)

Area counts the unit squares a shape covers (Area as covering). If the shape is cut into pieces that do not overlap and leave no gaps, every unit square of it lies in exactly one piece, so

\begin{aligned} \text{area of the shape} &= \text{area of piece 1} \\ &\quad + \text{area of piece 2} + \cdots \end{aligned}

Pieces may share an edge, because a line covers no surface. Good cuts run along a line that continues one of the shape’s own edges, or join two corners. A rectangle piece needs its length and width. A triangle piece needs a base and the height at right angles to it, and nothing else: where its top corner sits along the base does not change \displaystyle \frac{1}{2}bh. When the shape is all right angles, cutting into rectangles alone is enough, as in Splitting a figure into rectangles.

A cut makes edges the drawing may not label. Where edges meet at right angles, a missing length is a difference: if the whole bottom is 15 cm and the top part above it is 5 cm, the rest is 15 - 5 = 10 cm.

Different cuts give the same area. The shape covers the same surface however it is sliced, so every correct cut gives the same total. That makes a second cut a free check.

Surround and subtract (Day 2)

Draw a rectangle around the shape, touching it at its farthest points. The rectangle is the shape plus the pieces between the shape and the rectangle’s edge, so

\begin{aligned} \text{area of the shape} &= \text{area of the rectangle} \\ &\quad - \text{areas of the outside pieces} \end{aligned}

This is strongest when the shape’s sides are slanted. The outside pieces are then right triangles whose two short sides, the legs, lie along the rectangle’s edges, so every length needed is measured along a straight edge, and a right triangle’s legs are already a base and its height.

Choosing the cut

Every way that follows the rules gives the same area, so pick the one that needs the fewest measurements and the fewest pieces. A measurement is work, and a wrong one spoils the answer, so fewer is better. Before measuring, list the lengths each way needs and count them.

Worked examples

Example 1

Day 1: a marker sign shaped like an arrow

Ynes paints marker signs for a campground’s footpaths. Each sign is an arrow: a shaft 40 cm long and 12 cm tall, joined to a triangular head. The head’s back edge, standing straight up where it meets the shaft, is 30 cm long, and its tip is 20 cm beyond the end of the shaft. Find the area of one face of the sign.

Solution

A cut along the head’s back edge splits the sign into a rectangle and a triangle with no overlap. The rectangle is 40 \times 12 = 480. The triangle’s base is the 30 cm back edge. Its height is the distance from the tip to that edge at right angles, which is the 20 cm it reaches beyond the shaft, since the shaft meets the back edge at a right angle. So the head is \displaystyle \frac{1}{2} \times 30 \times 20 = 300.

480 + 300 = 780

One face of the sign has area 780\text{ cm}^2.

Example 2

Day 1: a bookend cut two ways

Agnieszka cuts the sides of a pair of bookends from wood. One side is shown in Figure 1: every corner is a right angle except the two at the ends of the slanted edge. Find its area by two different cuts.

15\text{ cm}
3\text{ cm}
5\text{ cm}
18\text{ cm}
5\text{ cm}
 A B C D E
Figure 1Example 2. The same bookend cut two ways, into A, B, C and into D, E.
Solution

First the missing lengths. Across the bottom it is 15 cm and across the top 5 cm, so the slant runs 15 - 5 = 10 cm across. Up the left it is 18 cm; the slant starts at the top of the 3 cm right end and stops below the 5 cm upright edge, at a height of 18 - 5 = 13 cm. So it rises 13 - 3 = 10 cm.

Cut 1 (left copy). Strip A is 15 cm by 3 cm. Upright B is 5 cm wide and runs from the top of A to the top, 18 - 3 = 15 cm. Triangle C has legs 10 cm across and 10 cm up, at a right angle, so one leg is the base and the other the height.

\begin{aligned} &15 \times 3 + 5 \times 15 + \frac{1}{2} \times 10 \times 10 \\ &= 45 + 75 + 50 = 170 \end{aligned}

Cut 2 (right copy). Strip D is 5 cm by 18 cm, area 90. Piece E has two parallel upright sides, 13 cm and 3 cm, that are 10 cm apart, so it is a trapezoid: \displaystyle \frac{1}{2}(13 + 3) \times 10 = 80.

90 + 80 = 170

The side has area 170\text{ cm}^2. The cuts agree, as they must: both count the same wood.

Example 3

Day 1: a camp plot with no right angles

Torin’s camp plot has four corners and no right angles. He stretches a tape along one diagonal, 16 m, and measures how far each of the other two corners is from it, at right angles: 5.5 m and 7.5 m (Figure 2). Find the plot’s area.

16\text{ m}
5.5\text{ m}
7.5\text{ m}
Figure 2Example 3. The diagonal cuts the plot into two triangles that share it as their base.
Solution

The diagonal cuts the plot into two triangles that share it, so each has base 16 m. The height of each is the distance from its top corner to the base’s line at right angles, which is exactly what Torin measured.

\begin{aligned} &\frac{1}{2} \times 16 \times 5.5 + \frac{1}{2} \times 16 \times 7.5 \\ &= 44 + 60 = 104 \end{aligned}

The plot has area 104\text{ m}^2. The two triangles share the factor \displaystyle \frac{1}{2} \times 16, so by the The distributive property the sum is also \displaystyle \frac{1}{2} \times 16 \times (5.5 + 7.5), which is 8 \times 13 = 104: half the diagonal times the total of the two heights. Neither triangle has a right angle, and that does not matter; only base and height enter \displaystyle \frac{1}{2}bh.

Example 4

Day 2: a bookend face with its corners sawn off

The front face of one of Agnieszka’s bookends starts as a rectangle 16 cm wide and 10 cm tall. She saws off all four corners, each a right triangle with legs 3 cm along the 16 cm edges and 2 cm along the 10 cm edges. Find the area of the face.

Solution

The rectangle surrounds the face, and the outside pieces are the four corners, each \displaystyle \frac{1}{2} \times 3 \times 2 = 3.

16 \times 10 - 4 \times 3 = 160 - 12 = 148

Check by adding. Upright cuts 3 cm in from each end leave a middle band 16 - 3 - 3 = 10 cm wide and 10 cm tall, area 100. Each end piece has parallel upright sides of 10 cm and 10 - 2 - 2 = 6 cm, 3 cm apart, so it is a trapezoid of area \displaystyle \frac{1}{2}(10 + 6) \times 3 = 24. Then 100 + 2 \times 24 = 148. The face is 148\text{ cm}^2. Subtracting took one kind of piece, used four times; adding took two kinds and two missing lengths.

Example 5

Day 2: a tilted plot in a clearing

Grigor’s camp plot has four slanted sides, and each corner touches one side of a rectangular clearing 20 m by 14 m (Figure 3). Find the area of the plot.

6\text{ m}
14\text{ m}
4\text{ m}
10\text{ m}
5\text{ m}
15\text{ m}
9\text{ m}
5\text{ m}
Figure 3Example 5. Four right triangles lie between the plot and the edge of the clearing.
Solution

Cutting the plot itself would need a diagonal and heights measured across rough ground. Surrounding it is easier, since the clearing is already a rectangle around it: 20 \times 14 = 280. Between the plot and the clearing’s edge lie four right triangles, each with its legs along two sides of the clearing, meeting at a corner of the clearing.

Corner of the clearingLegsArea
bottom left6 and 5\displaystyle \frac{1}{2} \times 6 \times 5 = 15
bottom right14 and 4\displaystyle \frac{1}{2} \times 14 \times 4 = 28
top right10 and 5\displaystyle \frac{1}{2} \times 10 \times 5 = 25
top left15 and 9\displaystyle \frac{1}{2} \times 15 \times 9 = 67.5

The four together are 15 + 28 + 25 + 67.5 = 135.5, so

280 - 135.5 = 144.5

The plot has area 144.5\text{ m}^2. Check. It covers about half the clearing, and 144.5 is a little over half of 280.

Example 6

Day 2: choosing the measurements

Maelle’s camp plot (Figure 4) is a rectangle 14 m along the front and 9 m deep, with a triangle on its back edge whose tip lies 5 m beyond that edge, nearer the left end. Before she measures, she compares two ways. Which needs fewer lengths? Find the area.

14\text{ m}
9\text{ m}
9\text{ m}
5\text{ m}
Figure 4Example 6. Three lengths are enough; where the tip sits along the back does not matter.
Solution

Cut across along the back edge of the rectangle. The rectangle needs its length and depth, 14 and 9. The triangle’s base is the same 14 m edge, and its height is the 5 m, so it needs one new length. That is three measurements in all, and the tip’s position along the back is not one of them, since \displaystyle \frac{1}{2}bh uses only base and height.

Surround and subtract. The surrounding rectangle is 14 m by 9 + 5 = 14 m. Each corner triangle beside the tip has the 5 m as one leg, but its other leg is the distance from the tip to that end of the back edge, so the tip’s position has to be measured too: four measurements.

Cutting across wins:

\begin{aligned} &14 \times 9 + \frac{1}{2} \times 14 \times 5 \\ &= 126 + 35 = 161 \end{aligned}

The plot is 161\text{ m}^2. Check. If the tip is 4 m from the left end, the corner triangles have legs 4 and 5, and 10 and 5, so 14 \times 14 - 10 - 25 = 196 - 35 = 161.

Boundaries

  • No overlaps, no gaps. The pieces added must cover the shape exactly once, and the outside pieces taken away must cover exactly the space between the shape and the rectangle.
  • A height is at right angles to its base. A slanted side is not a height. A triangle piece is usable only when its height is known or can be worked out.
  • Lengths come from stated facts. Reading a missing length as a difference needs right angles, or parallel edges, that are marked or stated; a drawing may not be to scale.
  • A diagonal must stay inside. If a corner of the shape points inward, one of the diagonals runs outside it, and the two triangles on that diagonal are not pieces of the shape. Use a diagonal that lies inside.
  • One unit throughout. Bring centimeters and meters to one unit before multiplying.
  • Straight sides only. These pieces have straight edges. A shape with curved edges needs circle pieces, as in Area of composite figures and shaded regions.

Where this goes next

Area of composite figures and shaded regions uses the same two moves on shapes with half circles and quarter circles. On the coordinate plane the grid supplies every length, so surrounding a tilted shape with a rectangle and subtracting right triangles, as in Example 5, becomes the usual way in. Cutting from the center into equal triangles gives the area of a regular polygon, and adding the areas of flat faces gives surface area.

Exercises

Exercises 1 to 7 go with Day 1 and 8 to 16 with Day 2. A corner described as square is a right angle.

Day 1: cut and add

  1. Ynes’s camp plot is a rectangle 18 m along the lane and 11 m deep, with a triangle on its back edge. The triangle’s base is the whole back edge, and its tip is 6 m beyond it. Find the area.
  2. The end wall of a trail shelter is 9 m wide, its two side walls are 5 m tall, and the roof rises in a straight line from the top of each side wall to a peak 8 m above the ground. Find the area of the end wall.
  3. The side of an L-shaped bookend has six square corners. Its bottom edge is 14 cm, its left edge 17 cm, its top edge 3 cm and its right edge 2 cm. Find the two unlabeled edges, and the area by two different cuts.
  4. The front of a trailhead noticeboard is a rectangle 2.5 m wide and 1.5 m tall with a triangular gable on top, whose base is the top of the rectangle and whose height is 1 m. Find the area of the front.
  5. A four-sided camp plot has a diagonal of 22 m. The other two corners are 6 m and 9 m from the diagonal, measured at right angles. Find the area.
  6. Grigor’s tent pad is a rectangle 15 m by 8 m with a triangle on each 8 m end. Each triangle’s base is that end, and its tip is 4 m beyond it. Find the area.
  7. A stepped terrace is three strips, each 4 m deep, stacked with their left ends lined up: the bottom strip 15 m long, the middle 10 m and the top 5 m. Find the area by adding the strips, then by cutting into upright columns instead.

Day 2: surround and subtract

  1. A bookend face is a square of side 15 cm with one corner sawn off, a right triangle with legs 6 cm and 4 cm. Find the area of the face.
  2. A wooden book trough, seen from the end, is a rectangle 24 cm wide and 10 cm tall with a rectangular slot 16 cm wide and 7 cm deep cut down from the middle of the top edge. Find the area of the end, and the width of each side arm, then check by adding.
  3. A four-sided plot has one corner on each side of a rectangular clearing 16 m wide and 12 m tall. Along the bottom its corner is 5 m from the left end; up the right side, 3 m from the bottom; along the top, 4 m from the right end; down the left side, 7 m from the top. Find the area of the plot.
  4. A triangular brace is drawn on a card 10 cm wide and 8 cm tall. One corner is the bottom left corner of the card, one is on the right edge 3 cm above the bottom, and one is on the top edge 6 cm from the left. Find the area of the brace.
  5. A trail shelter’s end wall is 12 m wide with side walls 5 m tall and a peak 9 m high in the middle. It has a doorway 3 m wide and 4 m tall. Find the area to be painted, using a surrounding rectangle, and check by adding.

Day 2: choosing the cut

  1. Maelle’s next plot is a rectangle 20 m wide and 12 m deep with a triangle on its back edge, tip 7 m beyond the edge and 6 m from its left end. Which lengths are needed, at fewest? Find the area.
  2. A trail marker is a chevron: a rectangle 30 cm wide and 20 cm tall with a triangle cut into its left end (base the whole 20 cm end, tip 8 cm in) and a matching triangle added to its right end (tip 8 cm out). Find its area with as little arithmetic as possible.
  3. A square plot 18 m on a side has a right triangle fenced off in one corner for a fire pit, with legs 5 m and 4 m along the sides. Find the area of the rest of the plot, and say which way needs fewer pieces.
  4. A four-sided camp plot sits in a rectangular clearing with one corner on each side. Compare the number of lengths needed to find its area by surrounding (as in Example 5) and by cutting along a diagonal (as in Example 3).

Reasoning

  1. Explain why cutting a shape two different ways must give the same area, and what goes wrong if two of the pieces overlap.
  2. A shape is a rectangle with a triangle on its back edge, as in Example 6, but with a back edge of b and a tip h beyond it. The tip is p from the left end. Show that the two corner triangles of the surrounding rectangle always total \displaystyle \frac{1}{2}bh, whatever p is.
  3. Explain why a polygon with no corner pointing inward can be cut into triangles by drawing every diagonal from one corner. How many triangles does a six-sided polygon give?
  4. Is surrounding and subtracting always the shorter way? Use the examples of this lesson to answer.
Answers
  1. 252\text{ m}^2. Rectangle 18 \times 11 = 198, triangle \displaystyle \frac{1}{2} \times 18 \times 6 = 54.
  2. 58.5\text{ m}^2. A cut across at the top of the side walls leaves a rectangle 9 \times 5 = 45 and a triangle of base 9 and height 8 - 5 = 3, area 13.5.
  3. The inner edges are 11 cm across (14 - 3) and 15 cm up (17 - 2); the area is 73\text{ cm}^2. Cut across: 14 \times 2 + 3 \times 15 = 28 + 45 = 73. Cut up: 3 \times 17 + 11 \times 2 = 51 + 22 = 73.
  4. 5\text{ m}^2. 2.5 \times 1.5 = 3.75 and \displaystyle \frac{1}{2} \times 2.5 \times 1 = 1.25.
  5. 165\text{ m}^2. \displaystyle \frac{1}{2} \times 22 \times 6 + \frac{1}{2} \times 22 \times 9 = 66 + 99.
  6. 152\text{ m}^2. 15 \times 8 = 120, and each triangle \displaystyle \frac{1}{2} \times 8 \times 4 = 16.
  7. 120\text{ m}^2. Strips: 15 \times 4 + 10 \times 4 + 5 \times 4 = 60 + 40 + 20. Columns, each 5 m wide, 12, 8 and 4 m tall: 60 + 40 + 20 = 120.
  8. 213\text{ cm}^2. \displaystyle 15^2 - \frac{1}{2} \times 6 \times 4 = 225 - 12.
  9. 128\text{ cm}^2, arms 4 cm wide. 24 \times 10 - 16 \times 7 = 240 - 112 = 128, and (24 - 16) \div 2 = 4. Adding: two arms 4 \times 10 and a base 16 \times 3 under the slot, 80 + 48 = 128.
  10. 103\text{ m}^2. The clearing is 16 \times 12 = 192. Corner triangles: legs 5 and 5 (12.5), 11 and 3 (16.5), 9 and 4 (18), 12 and 7 (42), total 89. 192 - 89 = 103.
  11. 31\text{ cm}^2. 10 \times 8 = 80. Outside: legs 10 and 3 (15), 5 and 4 (10), 6 and 8 (24), total 49. 80 - 49 = 31.
  12. 72\text{ m}^2. Surround: 12 \times 9 = 108, less two corner triangles with legs 6 and 4, 12 each: 84. Less the door 3 \times 4 = 12: 72. Adding: 12 \times 5 = 60 and \displaystyle \frac{1}{2} \times 12 \times 4 = 24, so 60 + 24 = 84, and 84 - 12 = 72.
  13. Three lengths, 20, 12 and 7; the 6 is not needed. Area 240 + 70 = 310\text{ m}^2, from 20 \times 12 and \displaystyle \frac{1}{2} \times 20 \times 7.
  14. 600\text{ cm}^2. The notch removes \displaystyle \frac{1}{2} \times 20 \times 8 = 80 and the point adds the same 80, so the area is the rectangle’s, 30 \times 20 = 600.
  15. 314\text{ m}^2. Subtract, one piece: \displaystyle 18^2 - \frac{1}{2} \times 5 \times 4 = 324 - 10. Cutting the rest needs three pieces, for example 18 \times 14 = 252, 13 \times 4 = 52 and the triangle left beside the fence, also with legs 5 and 4, 10: 314.
  16. Surrounding needs six: the clearing’s two sides and one distance on each side for the four corners. The diagonal way needs three: the diagonal and the two heights to it. The diagonal way needs fewer lengths, but they lie across the plot and the heights need a right angle found; the surrounding lengths all lie along straight edges.
  17. Each cut covers the same surface with pieces that do not overlap, so every unit square is counted exactly once either way, and the totals match. If two pieces overlap, the shared part is counted twice and the total is too big by the area of the overlap.
  18. The corner triangles have legs p and h, and b - p and h. Their total is \displaystyle \frac{1}{2}ph + \frac{1}{2}(b - p)h = \frac{1}{2}h(p + b - p) = \frac{1}{2}bh by the distributive property, so p drops out. In words: both triangles have height h, and their bases p and b - p lie end to end along the top, making the whole b.
  19. Six sides give 4 triangles. With no corner pointing inward, every diagonal from one corner lies inside, and the diagonals do not cross, so they split the shape into triangles that do not overlap and leave no gaps. From one corner a diagonal goes to every corner except itself and its two neighbors, 6 - 3 = 3 diagonals, and 3 cuts make 3 + 1 = 4 pieces.
  20. No. It was shorter for Grigor’s tilted plot (Example 5) and the sawn bookend (Example 4). For the arrow sign (Example 1) the surrounding rectangle leaves two strips and two triangles outside, four pieces, against two pieces when cutting; for Maelle’s plot (Example 6) it needs one more measurement.
unattributed · CC-BY-SA-4.0