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Geometry · G.GPE.B.7

Perimeter and area from coordinates

A polygon is known only by the coordinates of its corners. How long is its boundary, and how much space does it enclose?

You need
The distance formulaMath 8Area of a polygon from coordinatesMath 6
Where you are 9 in Coordinate geometry
Geometry11
Coordinate geometry9
▴ 4 aboveHorizontal and vertical distanceThe distance formula
Perimeter and area from coordinates
The midpoint formulaCoordinate proofs with variables

When you reach for it

A polygon drawn on a grid with its corners at known coordinates is a polygon you can measure without a ruler. Two questions come up again and again: how far it is around, its Perimeter, and how much of the plane it covers, its area.

When every side runs along the grid, both are counting. Once a side is slanted, counting squares stops working: the side cuts through squares, and its length is not a whole number of anything. This lesson combines two tools you already have, the distance formula for lengths and boxing and cutting for areas, into one method that measures any polygon from its corner coordinates alone.

Reach for it whenever a figure is given by points rather than by measurements: a plot of land on a site plan, a shape on a design grid, or a triangle in a geometry problem where you need its area to find something else.

The procedure

Throughout, the vertices of a polygon are listed in order around it, so that each vertex is joined to the next and the last is joined back to the first. That order is what says which pairs of points are sides.

Perimeter: one distance per side

The perimeter is the sum of the side lengths. For a side from (x_1, y_1) to (x_2, y_2):

If the side is horizontal or vertical, its length is a difference of coordinates, |x_2 - x_1| or |y_2 - y_1|. The absolute value is there because a length is never negative, whichever end you start from.

If the side is slanted, it is the hypotenuse of a right triangle whose legs run along the grid: a horizontal leg of length |x_2 - x_1| and a vertical leg of length |y_2 - y_1|. The legs meet at a right angle because the axes do, so the Pythagorean theorem gives

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

The absolute values are dropped because squaring makes them unnecessary.

(x_1,\,y_1)
(x_2,\,y_2)
|x_2 - x_1|
|y_2 - y_1|
d
Figure 1Every slanted side is the hypotenuse of a right triangle whose legs run along the grid.

Keep each length exact. Simplify each square root so that like roots show themselves: \displaystyle \sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}, whereas \displaystyle \sqrt{2} + \sqrt{3} cannot be combined at all. Round only at the very end, if a decimal is wanted, so that the rounding errors of several sides do not pile up.

Area: three routes, one principle

The principle behind all three is that area adds: if a region is cut into pieces that do not overlap and that fill it exactly, its area is the sum of the areas of the pieces. A shape’s area also does not change when it is moved, turned or reflected, since a congruent copy covers exactly as much of the plane.

A rectangle, even a tilted one. A rectangle’s area is length times width (for the reason you already know), and that stays true when its sides are slanted: turn the rectangle until its sides lie along the grid, and neither its side lengths nor its area has changed. So find two adjacent sides with the distance formula and multiply. First make sure it really is a rectangle. Four corners alone do not promise right angles, and perpendicular slopes are the test: two slanted sides meet at a right angle exactly when their slopes multiply to -1, and a horizontal side meets a vertical one at a right angle.

A triangle with a horizontal or vertical side. Take that side as the base. The area of a triangle is \displaystyle \frac{1}{2} \times \text{base} \times \text{height}, where the height is measured at right angles to the base. A line at right angles to a horizontal line is vertical, so the height is simply the vertical distance from the third vertex to the line the base lies on: a difference of y-coordinates. For a vertical base, swap the roles and use x-coordinates.

Anything else: cut it or box it. Cut the polygon along horizontal or vertical lines through its vertices into pieces of the two kinds above, or into trapezoids whose parallel sides run along the grid, and add their areas. Or draw the smallest grid-aligned rectangle around it and subtract the right triangles left in the corners, the method of area from coordinates. Both are the same principle, area adds; one adds pieces inside, the other takes pieces away from outside. Doing a hard one both ways is a good check.

Areas come in square units. If one grid unit stands for some real length, every length in the answer is multiplied by that length, and every area by its square, since an area is a product of two lengths.

Worked examples

Example 1

A triangle with a horizontal base

Find the perimeter and area of the triangle with vertices A(1, 1), B(7, 1) and C(4, 5).

Solution

A and B have the same y-coordinate, so AB is horizontal and AB = 7 - 1 = 6. The other two sides are slanted, so use the distance formula:

AC = \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5
BC = \sqrt{(4 - 7)^2 + (5 - 1)^2} = \sqrt{9 + 16} = 5

The perimeter is 6 + 5 + 5 = 16 units.

For the area, use the horizontal side AB as the base. The height is the vertical distance from C to the line y = 1, which is 5 - 1 = 4. So the area is

\frac{1}{2} \times 6 \times 4 = 12 \text{ square units}

The two slanted sides came out equal, so the triangle is isosceles, and the height lands at the midpoint (4, 1) of the base, as it must.

Example 2

A perimeter in square roots

Find the exact perimeter of the triangle with vertices P(-2, -1), Q(4, 1) and R(0, 5), and give it to two decimal places.

6
2
4
4
2
6
P
Q
R
Figure 2The triangle of Examples 2 and 7, with P at (-2, -1), Q at (4, 1) and R at (0, 5). The box edges carry the run and rise of every side.
Solution

No side is horizontal or vertical, so each needs the distance formula. The figure shows the run and rise of each side as pieces of the box edges.

\begin{aligned} PQ &= \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10} \\ QR &= \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \\ RP &= \sqrt{(-2)^2 + (-6)^2} = \sqrt{40} = 2\sqrt{10} \end{aligned}

Simplifying shows that two of the lengths are like roots and combine, \displaystyle 2\sqrt{10} + 2\sqrt{10} = 4\sqrt{10}. The third is a different root and stays separate. The exact perimeter is

4\sqrt{10} + 4\sqrt{2} \approx 18.31 \text{ units}

The rounding is done once, on the final sum: \displaystyle 4\sqrt{10} \approx 12.649 and \displaystyle 4\sqrt{2} \approx 5.657.

Example 3

A tilted rectangle

Show that A(1, 0), B(7, 3), C(5, 7), D(-1, 4) are the vertices of a rectangle, and find its perimeter and area.

A
B
C
D
\displaystyle 3\sqrt{5}
\displaystyle 2\sqrt{5}
Figure 3The rectangle of Example 3. Its sides are slanted, but it is still a rectangle, so its area is still length times width.
Solution

Work out the slope of each side, going round in order:

SideRun, riseSlope
AB6, \ 3\displaystyle \frac{1}{2}
BC-2, \ 4-2
CD-6, \ -3\displaystyle \frac{1}{2}
DA2, \ -4-2

Opposite sides have equal slopes, so they are parallel and ABCD is a parallelogram. Adjacent sides have slopes \displaystyle \frac{1}{2} and -2, whose product is -1, so every corner is a right angle. It is a rectangle.

Now two adjacent sides give everything:

\begin{aligned} AB &= \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \\ BC &= \sqrt{(-2)^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \end{aligned}

The perimeter is \displaystyle 2(3\sqrt{5} + 2\sqrt{5}) = 10\sqrt{5} \approx 22.36 units, and the area is

3\sqrt{5} \times 2\sqrt{5} = 6 \times 5 = 30

square units. The square roots multiplied out to a whole number, which happens because both sides are multiples of the same root.

Check by boxing it. The box runs from x = -1 to x = 7 and from y = 0 to y = 7, so it is 8 by 7, area 56. The four corner triangles have legs 6 and 3, 2 and 4, 6 and 3, 2 and 4, so their areas are 9, 4, 9 and 4, which add to 26. And 56 - 26 = 30.

Example 4

Cutting a triangle with a vertical line

Find the area and the exact perimeter of the triangle with vertices A(0, 1), B(4, 7) and C(8, 3).

5
4
4
A
B
C
M
Figure 4The vertical cut through B in Example 4 is a base shared by both halves, and each half’s height is a horizontal distance.
Solution

No side is horizontal or vertical, so cut. The vertex with the middle x-coordinate is B, and the vertical line x = 4 through it cuts the triangle into two triangles. To find where the cut meets AC: the slope of AC is \displaystyle \frac{3 - 1}{8 - 0} = \frac{1}{4}, so moving 4 to the right from A rises 1, and the cut meets AC at M(4, 2).

The cut BM is vertical, with length 7 - 2 = 5, and it is a side of both pieces. Use it as the base of each. A vertical base has a horizontal height: the distance from A to the line x = 4 is 4, and so is the distance from C. So

\begin{aligned} \text{area} &= \tfrac{1}{2}(5)(4) + \tfrac{1}{2}(5)(4) \\ &= 10 + 10 = 20 \end{aligned}

square units.

The perimeter needs all three slanted sides:

\begin{aligned} AB &= \sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13} \\ BC &= \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2} \\ CA &= \sqrt{(-8)^2 + (-2)^2} = \sqrt{68} = 2\sqrt{17} \end{aligned}

These are three different roots, so nothing combines: the perimeter is \displaystyle 2\sqrt{13} + 4\sqrt{2} + 2\sqrt{17} \approx 21.11 units.

Check the area by boxing: the box is 8 by 6, area 48, and the corner triangles have areas \displaystyle \frac{1}{2}(4)(6) = 12, \displaystyle \frac{1}{2}(4)(4) = 8 and \displaystyle \frac{1}{2}(8)(2) = 8. Then 48 - 28 = 20.

Example 5

A field on a site plan

On a site plan, each grid unit stands for 10 meters. A field has corners at A(0, 0), B(6, 0), C(6, 3), D(3, 7) and E(0, 4). How much fencing goes round it, and what is its area in square meters?

4
7
3
3
3
A
B
C
D
E
Figure 5The field of Example 5, in grid units. The vertical cut through D leaves two trapezoids, each with two vertical parallel sides.
Solution

Work in grid units first and convert at the end.

Three sides run along the grid: AB = 6, BC = 3 and EA = 4. The other two are slanted:

\begin{aligned} CD &= \sqrt{(-3)^2 + 4^2} = 5 \\ DE &= \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2} \end{aligned}

So the perimeter is \displaystyle 6 + 3 + 5 + 3\sqrt{2} + 4 = 18 + 3\sqrt{2} grid units.

For the area, cut along the vertical line x = 3 through D. It meets AB at (3, 0) and splits the field into two trapezoids, each with two vertical parallel sides and a horizontal width of 3. The area of a trapezoid is half the sum of the parallel sides times the distance between them:

\begin{aligned} \text{left} &= \frac{1}{2}(4 + 7)(3) = 16.5 \\ \text{right} &= \frac{1}{2}(7 + 3)(3) = 15 \end{aligned}

The field covers 16.5 + 15 = 31.5 square grid units.

Now convert. Each grid length is 10 meters, so the fencing is \displaystyle 10(18 + 3\sqrt{2}) = 180 + 30\sqrt{2} \approx 222.43 meters. Each square grid unit is a square 10 meters on a side, which is 100 square meters, so the area is 31.5 \times 100 = 3150 square meters. Lengths were multiplied by 10 and the area by 10^2.

Example 6

Placing a vertex to make an area

A(1, 2) and B(9, 2) are two vertices of a triangle. Find every point C on the y-axis that makes the area of triangle ABC equal to 20.

Solution

AB is horizontal with length 9 - 1 = 8, so take it as the base. The area is then \displaystyle \frac{1}{2} \times 8 \times h = 4h, where h is the vertical distance from C to the line y = 2. For an area of 20 the height must be h = 5.

A point on the y-axis is (0, y), and its vertical distance from y = 2 is |y - 2|. So

|y - 2| = 5

so y - 2 = 5 or y - 2 = -5, that is, y = 7 or y = -3.

There are two answers, C(0, 7) and C(0, -3), one on each side of the base.

Notice that C is not directly above or below any point of the base. The foot of the height, (0, 2), is outside the segment AB, and the area formula does not care: the height is measured to the line the base lies on.

Example 7

Using area to find a height

For the triangle PQR of Example 2, find the distance from R to the line through P and Q.

Solution

That distance is the height of the triangle when PQ is the base, so if both the area and PQ are known, the height follows from \displaystyle \text{area} = \frac{1}{2} \times PQ \times h.

The area comes from the box in the figure for Example 2. The box runs from x = -2 to x = 4 and from y = -1 to y = 5, so it is 6 by 6, area 36. The corner triangles have legs 6 and 2, 4 and 4, and 2 and 6, so their areas are 6, 8 and 6. The triangle’s area is 36 - 20 = 16.

From Example 2, \displaystyle PQ = 2\sqrt{10}. So

16 = \frac{1}{2} \times 2\sqrt{10} \times h = \sqrt{10}\,h
h = \frac{16}{\sqrt{10}} = \frac{16\sqrt{10}}{10} = \frac{8\sqrt{10}}{5} \approx 5.06

The height to a slanted side is hard to measure directly, because its foot is not at a grid point. Computing the area another way first, and then dividing, finds it without ever locating the foot. This is one route to the distance from a point to a line.

Boundaries

The vertices must be taken in order. The method joins each vertex to the next. Listed in a different order, the same four points can trace a path that crosses itself, and then the sum of the “sides” is not the perimeter of anything you meant. The points (0, 0), (4, 0), (4, 3), (0, 3) in that order give a rectangle with perimeter 14; taken as (0, 0), (4, 3), (4, 0), (0, 3) the path has length 5 + 3 + 5 + 3 = 16.

The polygon must not cross itself. Cutting and boxing both assume one region with one inside. For a figure whose sides cross, “the area” has no single meaning until you say which region you mean.

Length times width needs right angles. Equal opposite sides are not enough. The quadrilateral (0, 0), (5, 0), (8, 4), (3, 4) has four sides of length 5, but it is a rhombus with no right angle, and its area is 5 \times 4 = 20, not 25. Check the slopes before multiplying.

The slope test for perpendicular sides works for slanted sides only. A vertical side has no slope, so a corner between a horizontal and a vertical side is a right angle without any product to compute.

The height of a triangle is measured to the line through the base, not to the segment. For an obtuse triangle such as (0, 0), (4, 0), (7, 3) the foot of the height lands outside the base, at (7, 0), and the area is still \displaystyle \frac{1}{2}(4)(3) = 6.

Exact first, decimals last. Unlike square roots do not combine, \displaystyle \sqrt{2} + \sqrt{3} is not \displaystyle \sqrt{5}, and rounding each side before adding can change the last digit of the total.

The two axes must use the same unit. The distance formula and every area here assume a grid unit means the same length across as up. On a plan whose horizontal and vertical scales differ, convert the coordinates to real lengths before measuring anything slanted.

Straight sides only. A boundary that curves, such as an arc of a circle, is not a polygon, and its length and the area it encloses need other tools.

Where this goes next

The same measurements answer questions about shape. Comparing side lengths and slopes is how you decide which kind of quadrilateral four points form, and Example 3 was a small case of it. Example 7’s move, area first and then a height, is one way to find the distance from a point to a line; that lesson also gives a direct route. Three side lengths from the distance formula are also enough on their own: Heron’s formula turns them into the area, though with coordinates in hand the box is usually quicker.

Cutting by vertical lines can be pushed all the way. Drop a vertical line from every vertex to the x-axis, and every side becomes the top of a trapezoid; adding the trapezoids under the top of a polygon and subtracting those under its bottom gives its area from the coordinates in one formula, often called the shoelace formula. And with letters for coordinates instead of numbers, the calculations of this lesson become coordinate proofs about every figure of a kind at once.

Exercises

Perimeter

  1. Find the perimeter of the triangle with vertices (0, 0), (8, 0) and (0, 6).
  2. Find the perimeter of the quadrilateral with vertices (1, 1), (4, 5), (9, 5) and (6, 1).
  3. Find the exact perimeter of the triangle with vertices (-1, 2), (3, 4) and (5, -2), and give it to two decimal places.
  4. Find the exact perimeter of the quadrilateral with vertices (0, 0), (3, 3), (6, 0) and (3, -3).

Rectangles and triangles

  1. Show that (2, 1), (5, 4), (3, 6), (0, 3) are the vertices of a rectangle, and find its area and exact perimeter.
  2. Find the area of the triangle with vertices (-3, -2), (5, -2) and (1, 4).
  3. Find the area of the triangle with vertices (2, -1), (2, 7) and (-3, 4).
  4. Find the area and the exact perimeter of the triangle with vertices (1, 1), (7, 3) and (3, 6).
  5. Find the area of the quadrilateral in Exercise 2.

Cutting into pieces

  1. Find the area and the exact perimeter of the quadrilateral with vertices (0, 0), (6, 0), (6, 4) and (2, 6).
  2. Find the area and the exact perimeter of the pentagon with vertices (-2, 0), (4, 0), (4, 3), (1, 6) and (-2, 3).
  3. Find the area and the exact perimeter of the quadrilateral with vertices (0, 1), (3, 5), (9, 5) and (10, 1).

Real units

  1. On a map each grid unit stands for 5 meters. A triangular plot has corners at (0, 0), (8, 0) and (8, 6). Find its perimeter in meters and its area in square meters.
  2. On a drawing each grid unit stands for 2 cm. A rectangle has vertices (0, 0), (4, 2), (3, 4) and (-1, 2). Find its area in square centimeters and its perimeter in centimeters, to two decimal places.

Working backwards

  1. Triangle ABC has A(-1, 1), B(7, 1) and C(3, k) with k > 1. Its area is 24. Find k, and then the exact perimeter.
  2. Rectangle ABCD has A(0, 0) and B(6, 3), and its area is 30. Find the two possible positions of D, and C in each case.
  3. Find the distance from (1, 5) to the line through (0, 0) and (6, 2), by finding the area of the triangle with these three vertices first.

Reasoning

  1. Explain why the area of a rectangle whose sides are slanted on the grid is still its length times its width, even though its sides do not run along the grid lines.
  2. A(0, 0) and B(8, 0) are fixed, and C moves along the line y = 6. Explain why the area of triangle ABC never changes, and show by comparing C(0, 6) with C(4, 6) that the perimeter does.
  3. Explain why the perimeter of a polygon must be found by taking its vertices in order around it, using the points (0, 0), (4, 0), (4, 3), (0, 3) as an example.
  4. Let a, b and c be positive numbers. Prove that the parallelogram with vertices (0, 0), (a, 0), (a + b, c) and (b, c) has area ac, by boxing it and subtracting the corners.
Answers
  1. 24 units. The legs are 8 and 6; the slanted side is \displaystyle \sqrt{8^2 + 6^2} = \sqrt{100} = 10, and 8 + 6 + 10 = 24.
  2. 20 units. The sides from (1, 1) to (4, 5) and from (9, 5) to (6, 1) each have run 3 and rise 4, so length 5; the horizontal sides are 9 - 4 = 5 and 6 - 1 = 5.
  3. \displaystyle 2\sqrt{5} + 2\sqrt{10} + 2\sqrt{13} \approx 18.01 units. The sides are \displaystyle \sqrt{4^2 + 2^2} = \sqrt{20}, \displaystyle \sqrt{2^2 + 6^2} = \sqrt{40} and \displaystyle \sqrt{6^2 + 4^2} = \sqrt{52}, three unlike roots.
  4. \displaystyle 12\sqrt{2} \approx 16.97 units. Every side has run 3 and rise 3, so length \displaystyle \sqrt{18} = 3\sqrt{2}, and four of them make \displaystyle 12\sqrt{2}.
  5. Area 12, perimeter \displaystyle 10\sqrt{2}. The sides in order have slopes 1, -1, 1, -1, so opposite sides are parallel and adjacent ones perpendicular, since 1 \times (-1) = -1. The lengths are \displaystyle \sqrt{3^2 + 3^2} = 3\sqrt{2} and \displaystyle \sqrt{2^2 + 2^2} = 2\sqrt{2}, so the area is \displaystyle 3\sqrt{2} \times 2\sqrt{2} = 12 and the perimeter \displaystyle 2(5\sqrt{2}).
  6. 24 square units. The base from (-3, -2) to (5, -2) is horizontal, length 8; the height is 4 - (-2) = 6, and \displaystyle \frac{1}{2}(8)(6) = 24.
  7. 20 square units. The base from (2, -1) to (2, 7) is vertical, length 8; the height is the horizontal distance 2 - (-3) = 5, and \displaystyle \frac{1}{2}(8)(5) = 20.
  8. Area 13, perimeter \displaystyle 5 + 2\sqrt{10} + \sqrt{29} \approx 16.71. The box is 6 by 5, area 30; the corner triangles have areas \displaystyle \frac{1}{2}(6)(2) = 6, \displaystyle \frac{1}{2}(4)(3) = 6 and \displaystyle \frac{1}{2}(2)(5) = 5, and 30 - 17 = 13. The sides are \displaystyle \sqrt{40} = 2\sqrt{10}, \displaystyle \sqrt{16 + 9} = 5 and \displaystyle \sqrt{4 + 25} = \sqrt{29}.
  9. 20 square units. It is a parallelogram with horizontal base 5 from (1, 1) to (6, 1) and height 5 - 1 = 4. (It is in fact a rhombus, all sides 5, but not a square: its area is not 25.)
  10. Area 26, perimeter \displaystyle 10 + 2\sqrt{5} + 2\sqrt{10}. Cut at x = 2: a triangle with vertical side 6 and width 2, area 6, and a trapezoid with vertical parallel sides 6 and 4, width 4, area \displaystyle \frac{1}{2}(6 + 4)(4) = 20. The sides are 6, 4, \displaystyle \sqrt{16 + 4} = 2\sqrt{5} and \displaystyle \sqrt{4 + 36} = 2\sqrt{10}.
  11. Area 27, perimeter \displaystyle 12 + 6\sqrt{2} \approx 20.49. Cut along y = 3: a 6 by 3 rectangle, area 18, under a triangle with base 6 and height 3, area 9. The sides are 6, 3, \displaystyle 3\sqrt{2}, \displaystyle 3\sqrt{2} and 3.
  12. Area 32, perimeter \displaystyle 21 + \sqrt{17} \approx 25.12. It is a trapezoid with horizontal parallel sides 10 and 6 and height 4: \displaystyle \frac{1}{2}(10 + 6)(4) = 32. The slanted sides are \displaystyle \sqrt{3^2 + 4^2} = 5 and \displaystyle \sqrt{1^2 + 4^2} = \sqrt{17}.
  13. Perimeter 120 m, area 600 square meters. In grid units the sides are 8, 6 and 10, so the perimeter is 24 units, 24 \times 5 = 120 m; the area is \displaystyle \frac{1}{2}(8)(6) = 24 square units, and each is 5^2 = 25 square meters.
  14. Area 40 square centimeters, perimeter \displaystyle 12\sqrt{5} \approx 26.83 cm. Slopes \displaystyle \frac{1}{2} and -2 multiply to -1, so it is a rectangle with sides \displaystyle \sqrt{20} = 2\sqrt{5} and \displaystyle \sqrt{5}: area 10 square units, each 2^2 = 4 square centimeters; perimeter \displaystyle 6\sqrt{5} units, each 2 cm.
  15. k = 7, perimeter \displaystyle 8 + 4\sqrt{13} \approx 22.42. The base AB = 8 is horizontal, so \displaystyle \frac{1}{2}(8)(k - 1) = 24 gives k - 1 = 6. Then \displaystyle CA = CB = \sqrt{4^2 + 6^2} = 2\sqrt{13}.
  16. D(-2, 4) with C(4, 7), or D(2, -4) with C(8, -1). \displaystyle AB = \sqrt{45} = 3\sqrt{5}, so \displaystyle AD = 30 \div 3\sqrt{5} = 2\sqrt{5}. AD is perpendicular to AB, so its slope is -2: it runs 1 across and 2 down, or 1 back and 2 up, for each \displaystyle \sqrt{5} of length, giving D = (-2, 4) or (2, -4). Then C = B + (D - A).
  17. \displaystyle \frac{7\sqrt{10}}{5} \approx 4.43 units. Box the triangle: 6 by 5, area 30, minus corners 6, 7.5 and 2.5, leaves area 14. The base from (0, 0) to (6, 2) is \displaystyle 2\sqrt{10}, so \displaystyle h = \frac{2 \times 14}{2\sqrt{10}} = \frac{14}{\sqrt{10}}.
  18. Turning the rectangle about a corner until its sides lie along the grid moves it without stretching it, so its side lengths and its area stay the same. In that position it is an ordinary grid-aligned rectangle, whose area is length times width. The lengths are the ones the distance formula gave, so the product is the area of the tilted rectangle too.
  19. Take AB, length 8, as the base. Wherever C is on y = 6, its vertical distance from the base line y = 0 is 6, so the area is always \displaystyle \frac{1}{2}(8)(6) = 24. The perimeter is not fixed: for C(0, 6) it is 8 + 6 + 10 = 24, while for C(4, 6) the two slanted sides are each \displaystyle \sqrt{16 + 36} = 2\sqrt{13}, and the perimeter is \displaystyle 8 + 4\sqrt{13} \approx 22.42.
  20. A polygon’s sides join each vertex to the next one around it. Taken in the order (0, 0), (4, 0), (4, 3), (0, 3) the four points give a 4 by 3 rectangle, perimeter 14. Taken as (0, 0), (4, 3), (4, 0), (0, 3) the “sides” include both diagonals, of length 5 each, and the path crosses itself, with length 5 + 3 + 5 + 3 = 16. The coordinates alone do not say which points are joined; the order does.
  21. The box runs from x = 0 to x = a + b and from y = 0 to y = c, so its area is (a + b)c. The two corner pieces are right triangles: one with vertices (a, 0), (a + b, 0), (a + b, c), and one with vertices (0, 0), (b, c), (0, c). Each has legs b and c, so each has area \displaystyle \frac{1}{2}bc. The parallelogram’s area is (a + b)c - bc = ac, its base a times its height c.
kg1128 · updated 2026-10-04 · CC-BY-SA-4.0