Perimeter and Area: Free Response

Perimeter is the distance all the way round the boundary, so add the four sides. Two of them measure $16$ cm and two measure $7$ cm:

$P = 16 + 7 + 16 + 7 = 2(16 + 7) = 46 \text{ cm}.$

Area is a count of unit squares. Lay $1$ cm squares along the bottom edge and $16$ of them fill one row, because that row is $16$ cm long. Stack rows up the $7$ cm side and there are $7$ of them, each one $1$ cm tall. Every square belongs to exactly one row, with no gaps and no overlaps, so the count is a multiplication rather than a long tally:

$A = 7 \times 16 = 112 \text{ cm}^2.$

The two answers carry different units on purpose. The perimeter is a length, so it is reported in centimetres; the area is a count of squares, so it is reported in square centimetres.

Take the perimeter first:

$P = 2(32 + 14) = 2 \times 46 = 92 \text{ cm},$

which is exactly twice the $46$ cm of the first card. That is no accident: all four sides doubled, so their total doubled.

Now the area:

$A = 32 \times 14 = 448 \text{ cm}^2.$

Set that beside $112 \text{ cm}^2$. Since $448 = 4 \times 112$, the area is four times as large, not twice.

The rows of unit squares say why. Doubling the bottom edge puts twice as many squares in each row, $32$ instead of $16$. Doubling the side gives twice as many rows, $14$ instead of $7$. The count is rows times squares per row, so it picks up a factor of two from each direction at once:

$2 \times 2 = 4.$

Perimeter runs along one direction at a time, so it collects a single factor of two. Area covers two directions together, so it collects two of them.

A perimeter of $46$ cm says only that the length and the width add to $23$ cm, since $P = 2(l + w)$. Plenty of pairs of lengths add to $23$, and they do not have equal products. Alongside $16$ and $7$, take $19$ and $4$:

$P = 2(19 + 4) = 46 \text{ cm}, \qquad A = 19 \times 4 = 76 \text{ cm}^2.$

Same perimeter as the card, smaller area, so the perimeter did not fix the area.

The other direction fails too. An area of $112 \text{ cm}^2$ says only that the length and the width multiply to $112$. Take $28$ and $4$:

$A = 28 \times 4 = 112 \text{ cm}^2, \qquad P = 2(28 + 4) = 64 \text{ cm}.$

Same area as the card, and a boundary $18$ cm longer.

So the claim fails both ways round. A perimeter pins down what the two sides add to; an area pins down what they multiply to. A sum does not determine a product and a product does not determine a sum, which is exactly why a shape needs both measurements reported and why one can never be read off the other.

The perimeter is the distance round the edge, so it is the four sides added. A parallelogram has two sides of $17$ cm and two of $14$ cm:

$P = 2(17 + 14) = 2 \times 31 = 62 \text{ cm}.$

Every side gets walked, so the slanted side certainly counts here.

The area asks a different question, and it does not use the slanted side at all. It is the base multiplied by the perpendicular height, the straight across distance from the base to the side opposite it:

$A = b \times h = 17 \times 6 = 102 \text{ cm}^2.$

Using the slanted side instead would give $17 \times 14 = 238$, and part B says why that number is not the area of anything on this table.

Note the units. The perimeter is a length, so it is in centimetres; the area counts unit squares, so it is in square centimetres.

Drop a vertical line from the top left corner of the panel straight down to the base. That line is the perpendicular height, $6$ cm, and it cuts a right triangle off the left end.

Now slide that triangle across to the right end. Opposite sides of a parallelogram are equal and parallel, so it fits against the slanted right side exactly, with no gap and no overlap. What is left is a rectangle: its bottom is still the base, $17$ cm, and its height is the vertical line you drew, $6$ cm. So the area is

$A = 17 \times 6 = 102 \text{ cm}^2.$

Moving a piece of a shape to a new place never changes how much space the shape covers, so the panel and the rectangle cover the same amount.

The slanted side cannot be one of the rectangle's dimensions because it is not a side of the rectangle at all: the cut and the slide consumed both slanted edges, turning them into the two upright ends. It is also longer than the height, since it travels sideways as well as upward, so multiplying by it counts space the panel does not cover:

$17 \times 14 = 238,$

more than twice the true area.

Work out the second panel's area from the same rule as the first. Its base is $17$ cm and its perpendicular height is $6$ cm, so

$A = 17 \times 6 = 102 \text{ cm}^2,$

the same as the first panel. The slanted side never enters the rule, so changing it from $14$ cm to $15.5$ cm changes nothing at all about the area.

The perimeters are another matter. The first panel gives $2(17 + 14) = 62$ cm, and the second gives

$P = 2(17 + 15.5) = 2 \times 32.5 = 65 \text{ cm}.$

Priya is right that the second panel is built from longer sides, and wrong about what that buys. Leaning a parallelogram further lengthens its boundary while the space inside stays put.

The cut and slide picture explains it. Both panels stand on the same base and reach the same perpendicular height, so the same cut turns each of them into the same rectangle, $17$ cm by $6$ cm. Leaning further only carries the material further sideways; it does not add any. If the area did depend on the slanted side, two panels that rearrange into the identical rectangle would have to cover different amounts of space, which cannot happen.

Rotate the copy by half a turn and set it against the original along one of the sides that is not the base. The two shapes fit with no gap and no overlap, and the two $14$ cm bases finish as the opposite pair of sides of the parallelogram. The perpendicular distance between that pair is the distance from the triangle's base to its far vertex, which is the triangle's own height, $13$ cm. So the parallelogram stands on a base of $14$ cm with a perpendicular height of $13$ cm, and its area is base times height:

$A_{\text{parallelogram}} = 14 \times 13 = 182 \text{ cm}^2.$

That parallelogram is built from two triangles, and they are copies of one another, so they cover equal amounts of space. One triangle is therefore exactly half of the pair:

$A = \frac{1}{2} \times 14 \times 13 = \frac{1}{2} \times 182 = 91 \text{ cm}^2.$

This is where the factor of one half in the triangle rule comes from. It is not a decoration on the formula: it is the record of the triangle being one of two equal pieces of a parallelogram that shares its base and its height. The unit is square centimetres, because the answer counts unit squares.

Cut from one end of the short parallel side to the far end of the long one. The two triangles that result have the parallel sides as their bases, one of $19$ m and one of $11$ m, and the perpendicular distance between those parallel sides, $5$ m, is the height of both.

Take them one at a time:

$A_1 = \frac{1}{2} \times 19 \times 5 = 47.5 \text{ m}^2, \qquad A_2 = \frac{1}{2} \times 11 \times 5 = 27.5 \text{ m}^2.$

Area is additive, so the trapezoid is their sum:

$A = 47.5 + 27.5 = 75 \text{ m}^2.$

The trapezoid rule gives the same total:

$A = \frac{11 + 19}{2} \times 5 = 15 \times 5 = 75 \text{ m}^2.$

The agreement is not luck. Because both triangles stand on the same height, their areas are $\frac{1}{2} \times 19 \times 5$ and $\frac{1}{2} \times 11 \times 5$, and the common factor $\frac{1}{2} \times 5$ can be taken out of the sum:

$\frac{1}{2} \times 5 \times (19 + 11) = \frac{1}{2}(a + b)h,$

writing $a$ and $b$ for the parallel sides and $h$ for the height. So adding the two triangles is the trapezoid rule, written out the long way. The average of the parallel sides, $15$ m, is then the width of a rectangle $5$ m tall that covers the same space.

The two rectangles are quick:

$11 \times 5 = 55 \text{ m}^2, \qquad 19 \times 5 = 95 \text{ m}^2.$

The trapezoid covers $75 \text{ m}^2$, more than the narrow rectangle and less than the wide one, which it had to be: the trapezoid is wider than $11$ m everywhere except along its short parallel side, and narrower than $19$ m everywhere except along its long one.

But it does better than merely landing in between. It lands exactly halfway:

$\frac{55 + 95}{2} = \frac{150}{2} = 75 \text{ m}^2.$

The reason is inside the formula. Multiplying the trapezoid rule out splits it into those very rectangles:

$\frac{1}{2}(a + b)h = \frac{1}{2}(ah + bh),$

and $ah$ and $bh$ are the areas of the rectangle on the short parallel side and the rectangle on the long one. Averaging the two parallel sides and averaging the two rectangles are therefore the same operation done in a different order. That is what the trapezoid rule is averaging.

One warning falls out of the same line. Adding the parallel sides and multiplying by the height without halving gives $30 \times 5 = 150 \text{ m}^2$, which is the two rectangles added rather than averaged. That number is a genuine area, but of the wrong figure: it is the parallelogram made by the trapezoid and a flipped copy of it, so it counts the trapezoid twice.

Both routes need the two lengths the figure does not mark, and both of those come from subtracting along the figure. The top edge is $17 - 5 = 12$ m, and the right hand edge is $8 - 3 = 5$ m.

Decompose first. A vertical cut $12$ m from the left splits the floor into a full depth rectangle on the left and a shallower one on the right:

$A = (12 \times 8) + (5 \times 5) = 96 + 25 = 121 \text{ m}^2.$

A horizontal cut is just as good, and it is worth seeing that it lands in the same place. Cutting across $5$ m up from the front leaves a full width band below and a narrower one above:

$A = (17 \times 5) + (12 \times 3) = 85 + 36 = 121 \text{ m}^2.$

Now subtract instead. Put the corner back and the floor would be the full $17$ m by $8$ m rectangle, from which the missing corner of $5$ m by $3$ m has to come off:

$A = (17 \times 8) - (5 \times 3) = 136 - 15 = 121 \text{ m}^2.$

The two agree, as they must. The region of floor is the same either way, and how you choose to cut it up cannot change how much of it there is. The unit is square metres, because the answer counts unit squares.

The edging follows the boundary, so the length needed is the perimeter. Walk right round the floor, taking the sides in turn: the front, up the right hand edge, in along the bottom of the stairwell, up its side, back along the top, and down the left hand edge.

$P = 17 + 5 + 5 + 3 + 12 + 8 = 50 \text{ m}.$

Now the two costs. Tiles are sold by the square metre, so they are priced off the area:

$121 \times 8 = 968 \text{ dollars}.$

Edging is sold by the metre, so it is priced off the perimeter:

$50 \times 6 = 300 \text{ dollars}.$

The bill is the two added:

$968 + 300 = 1268 \text{ dollars}.$

The two prices come from different measurements, and swapping them is the classic slip. Pricing tiles off the perimeter, or edging off the area, produces a number that answers no question at all.

Compare the two boundaries. The full rectangle would need

$P = 2(17 + 8) = 50 \text{ m},$

and the notched floor needs $50$ m as well. Meanwhile the area fell from $136 \text{ m}^2$ to $121 \text{ m}^2$. So the owner is right that the floor got smaller and wrong that the edging did.

Here is why it happens for a corner cut of any size. Taking out a corner piece $5$ m by $3$ m removes $5$ m from the top edge and $3$ m from the right hand edge, so the old boundary loses $5 + 3 = 8$ m. In their place the cut supplies two new edges: $5$ m in along the bottom of the notch and $3$ m up its side. What is removed and what is added have the same two lengths, so they cancel:

$50 - (5 + 3) + (5 + 3) = 50 \text{ m}.$

Nothing in that argument used the sizes $5$ and $3$ beyond their cancelling, so it holds for any corner notch that leaves part of each wall it cuts still standing: the notch's own two edges are then exactly as long as the pieces of wall they stand in for. A notch running the full length of a wall is a different shape of question, because it leaves no wall to stand in for and the floor is simply a smaller rectangle. Area and boundary are answering different questions again, and this is a case where one moves and the other does not.

The area is the full rectangle less the alcove:

$A = 136 - (3 \times 4) = 136 - 12 = 124 \text{ m}^2.$

Now follow the boundary along the front. The alcove takes $3$ m out of the middle of that wall, leaving $17 - 3 = 14$ m of it in two pieces, and it supplies three new edges: $4$ m in, $3$ m across the back of the alcove, and $4$ m back out. So the front now contributes

$14 + 4 + 3 + 4 = 25 \text{ m}$

where it used to contribute $17$ m. The other three sides are untouched, so

$P = 50 + 8 = 58 \text{ m}.$

The difference lies in where the new edges come from. At a corner, the two new edges stand in for two stretches of wall of exactly their own lengths, so the boundary breaks even. In the middle of a wall, the $3$ m across the back of the alcove stands in for the $3$ m of wall taken out, but the two sides of the alcove, $4$ m each, stand in for nothing: they are boundary that did not exist before. So the boundary grows by twice the depth, $2 \times 4 = 8$ m, however wide the alcove is made.

Name the unknown. Let $w$ be the width of the plot in metres. The rectangle rule says the area is length times width, so

$19w = 133.$

The left side multiplies $w$ by $19$, so dividing both sides by $19$ undoes it:

$w = \frac{133}{19} = 7.$

The plot is $7$ m wide. Check it against the rule it came from: $19 \times 7 = 133 \text{ m}^2$, the area given. The answer is a length, so its unit is metres, not square metres.

Let $h$ be the perpendicular height in metres. The triangle rule gives

$\frac{1}{2} \times 19 \times h = 133.$

The left side does two things to $h$: it multiplies by $19$ and it halves. Undo the halving first, by doubling both sides:

$19h = 266.$

Then divide both sides by $19$:

$h = \frac{266}{19} = 14.$

The height is $14$ m, and the check holds: $\frac{1}{2} \times 19 \times 14 = \frac{1}{2} \times 266 = 133 \text{ m}^2$.

The rectangular plot stands on the same $19$ m base and covers the same area, and its width came out at $7$ m, so the triangle's height is exactly twice it. That is the factor of one half seen from the other side. A triangle is half of a parallelogram on the same base and height, so at equal heights it covers only half as much space, and the only way to make the shortfall up is to double the height.

Let $b$ be the unknown parallel side in metres. The trapezoid rule gives

$\frac{1}{2}(6 + b) \times 19 = 133.$

Clear the halving first, by doubling both sides:

$(6 + b) \times 19 = 266.$

Divide both sides by $19$:

$6 + b = 14,$

and subtract $6$ from both sides:

$b = 8.$

The other parallel side is $8$ m, so the average of the two parallel sides is

$\frac{6 + 8}{2} = 7 \text{ m}.$

That average is what the rule multiplies by the height: $7 \times 19 = 133 \text{ m}^2$, the area given. So this plot covers exactly the space of a rectangle $7$ m wide and $19$ m tall, which is the rectangular plot from part A. A trapezoid always covers what a rectangle of its average width and its own height covers.

In all three plots the length Reza knows is $19$ m, so his rule performs the same division three times:

$133 \div 19 = 7.$

What that $7$ means is different every time.

For the rectangle it is the width, and it is right. The area rule there is length times width and nothing else, so one division undoes the whole rule. The check confirms it: $19 \times 7 = 133 \text{ m}^2$.

For the triangle it is wrong, and the check is what shows it. A triangle of base $19$ m and height $7$ m covers

$\frac{1}{2} \times 19 \times 7 = 66.5 \text{ m}^2,$

only half the area asked for. What the division undid was the multiplication by the base; what it left standing was the halving, so the value has to be doubled before it is a height.

For the trapezoid it is not a side at all. That rule involves the height and both parallel sides, so dividing by the height returns the average of the two sides. The average is a real and useful length, since $7 \times 19 = 133 \text{ m}^2$ checks out, but it is not what was asked for: the known parallel side still has to be taken off before the unknown one appears, as part C does.

The pattern is clear once you look at what each rule does to the unknown. Dividing an area by a known length undoes the rule exactly when the rule is a plain product of two lengths and nothing else, which is the rectangle and the parallelogram, where the area is base times perpendicular height. As soon as the rule carries anything more, a factor of one half for the triangle or a sum of two sides for the trapezoid, that extra operation has to be undone as well, peeling the operations off from the outside in.