A Math Forum Project


Geometry Project of the Month

A Single Formula for Area - May 1996

Write a single formula for area that will work for a rectangle, a parallelogram, a trapezoid, a triangle, and a square. Explain how it works.

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Submissions

From: shelly@rgs.edu.sg
Sophia Yong
Grade: 9
School: Raffles Girls School (Secondary)

Answer: Formula: 1/2 x (breadth1+ breadth2) x height,
         where breadth1 and breadth2 are the sides parallel to each
         other and perpendicular to the height, which may be a
         side, or a line constructed to fit the description.

In a rectangle and square, breadth1 and breadth2 are equal, so 
1/2 x (breadth1 + breadth2) = breadth1 [or breadth2].
If lines parallel to the breadth and lines parallel to the height
are drawn across the rectangle or square, so that the square is
divided into small squares 1 unit^2 in area, the area of the
rectangle or square can then be calculated by counting the number 
of small squares in the shape. The number of small squares along 
the breadth would be equal to the length of the breadth, and the 
number of small squares along the height would be equal to the
length of the height; the total number of small squares in the
figure can thus be calculated by multiplying the breadth by the 
height. Thus the area of the square or rect. can be calculated 
by breadth x height, which is equal to 1/2 x (breadth1 + breadth2)
x height.

A parallelogram can be cut along its height, from the part where
the height and one of the breadths meet, and the cut off
part flipped over and moved over to the other side to form a
rectangle. As no part of the breadth or height is added or cut 
off, the breadth and height of the parallelogram are equal to
the breadth and height of the newly formed rectangle. The area of 
the newly formed rectangle = breadth x height (previously proved),
which is equal to 1/2 x (breadth1 + breadth2) x height [of
parallelogram] as the breadths in any parallelogram are equal.
 
Two similar trapezoids placed next to each other, with their 
edges touching, and one trapezoid flipped over, would form
a parallelogram with breadth = (breadth1 + breadth2) of the 
trapezoid. The area of a parallelogram can also be found by
breadth x height, because it's breadth1 = breadth2, so the 
area of one trapezoid is = 1/2 area of parallelogram, which
is equal to 1/2 x breadth x height (of parallelogram), which
is equal to 1/2 x (breadth1 + breadth2) x height (of trapezoid),
as the height in the parallelogram and trapezoid are the same.

There is only one breadth in a triangle, so 1/2 (breadth1 + 
breadth2) is the same as breadth/2. The height of the triangle 
would be a line constructed (if not already present) which is 
perpendicular to the breadth and touches the tip of the triangle. 
By cutting the triangle along its height, flipping the cut off 
part over, and joining it to the other part, a rectangle with 
breadth 1/2 of the triangle would be formed. The area of the 
rectangle would be breadth (of rectangle) x height (same as that 
of triangle), which would be equal to 1/2 x breadth (of triangle) 
x height,  which is the same as 1/2 x (breadth1 + breadth2) x 
height [of triangle], breadth2 being 0.


From: GTSamf@aol.com Melissa Miller and Kevin Mayer Grades 9 and 10 Bishop Lynch High School, Dallas, Texas: Length of top (base1) plus length of bottom (base2) divided by 2 times the vertical height. We demonstrate that this formula works for a square when base1 and base2 are both equal to 3. The same formula also works for a rectangle when base1 and base2 are both equal to 5. The formula also works for a trapezoid when base1 is equal to five and base2 is equal to 7. The formula also works for a triangle when base1 is equal to zero. The same formula works for all of the figures.
From: GTSamf@aol.com Josh Walshe, Steve Maher, and Jennifer Ruehl Grade 11 Bishop Lynch High School, Dallas, Texas: Our solution is that the formula for the area of a trapezoid (Area=.5(h)(base1 + base2)) works for all five figures. First, we demonstrate that for a triangle we consider one of the bases to be zero. For the parallelogram, the rectangle, and the square, we consider the two bases equal, so we get Area=.5(h)(2)(base1). Finally, for the trapezoid, we consider base1 From: GTSamf@aol.com Alfred Lopez, Joel Lenzini, Boeh Tal-Placido, and Emily Herrin Grades 10 and 9 Bishop Lynch High School, Dallas, Texas: Area formula:((base1 + base2)/2)(h) Base of triangle = 0 Sides parallel to ground are bases. Bases are parallel to one another.
From: GTSamf@aol.com Beth Digby Grade 10 Bishop Lynch High School, Dallas, Texas: The formula is .5(h)(base1 + base2). This will work for a rectangle, a parallelogram, a trapezoid, a triangle, and a square because you're finding the median distance between the two bases, the multiplying it by the height, which comes from the formula A=bh. We demonstrate this by constructing a chart: Individual formula Single formula Rectangle(b1=5,b2=5,h=4) .5(5 + 5)(4)=20 A=bh (5)(4)=20 Parallelogram(b1=5,b2=5,h=4) .5(5 + 5)(4)=20 A=bh (5)(4)=20 Trapezoid(b1=4,b2=6,h=4) .5(4+6)(4)=20 A=.5(b1 +b2)(h) (5)(4)=20 Triangle(b1=0,b2=10,h=4) .5(0 + 10)(4)=20 A=.5bh 5(10)(4)=20 Square(b1=4,b2=4,h=4) .5(4+4)(4)=16 A=s^2 4^2=16 As seen in the chart, the answers with the single formula are the same as the answers from the individual formulas.
From: ssusd2@owens.ridgecrest.ca.us Cassie Gorish Grade 8 School: Murray Middle School The area of a square, parallelogram, rhombus, and rectangle is base times height. The area of a triangle is half base times height. But the trapezoid has a unique area equation: (base1 + base2) X height 2 Now if you add two numbers together and then divide them by two, it is an average. So the area of a trapezoid is average base times height. If applied to any of the above figures, the equation also works. The base1 (the top) of a triangle is , so the average is half the base. So the equation that applies to all of the above figures is average base times height.
From: arthur@iolani.honolulu.hi.us Jason Yeung Grade: 9 School: Iolani School My formula for this is: (the sum of the two bases) * height / 2 This formula is the same formula for the area of the trapezoid, therefore, it will work for trapezoid, square, parallelogram, and rectangle. For square, the two bases and the height are the same as its height. For rectangle, the bases are either the length or the width, and the height is the left over one (either length or width). For a parallelogram, the base is the same base and the height is the same height. Triangle is a little bit tricky. Since there is only one base, you have to assume the other base is zero. This makes sense because without a base, it is zero. Remember that the height is the perpendicular distance. Another formula that is good to know is Heron's formula. The formula is the square root of (s-a)(s-b)(s-c)(s-d), where s is the semi-perimeter and a, b, c, and d are the sides. Remember that the figure must be able to be enclosed in a circle.
From: Eglsnest@aol.com Claire Gross Grade 8 Garland Street Middle School, Bangor, Maine To find the area of a rectangle, parallelogram, square, trapezoid, or triangle, I came up with the formula generally used for trapezoids: A=1/2(b1+b2)h. The following chart explains how the individual formulas for each figure can be derived from this one formula: Figure Special Properties Formula rectangle b1=b2, b is A=1/2(b1+b2)h or substituted for A=1/2(b+b)h parallelogram both A=1/2(2b)h A=bh square b1=b2=h, s is A=1/2(b1+b2)h substituted for A=1/2(s+s)s all A=1/2(2s)s A=s*s A=s^2 trapezoid A=1/2(b1+b2)h (This is the usual formula for trapezoids.) triangle b2=0, b is A=1/2(b1+b2)h substituted for b1 A=1/2(b+0)h A=1/2bh
From: PDALEY@fair1.fairfield.edu Bilal Seyal Grade 9 School: Fairfield High School, Fairfield, CT A formula that I found that works for the area of a square, rectangle, parallelogram, trapezoid, and a triangle is 1/2 * (b1 + b2) * h where b means base and h means height. This formula is also the real formula for a trapezoid. To prove that this formula works for the area of the five polygons, I first found the area of each polygon using the regular formula and then my formula. Lets say one side of a square is 5 inches long. Using the regular formula, a side squared, the area is 25 square inches. Using my formula, it also comes out to 25 square inches. Let's say the base of a rectangle is 4 inches and the height is 5 inches. Using the regular formula, b*h, the area is 20 square inches. Using my formula, 1/2 * (4 + 4) * 5, it also comes out to 20 square inches. Let's say the base of a parallelogram is 6 inches and the height is 4 inches. Using the regular formula, b*h, the area comes out to 24 square inches. Using my formula, the area also comes out to 24 square inches. Since the regular formula of a trapezoid is the same as my formula, it doesn't have to be tested on a trapezoid. Let's say the base of a triangle is 8 inches and the height is 5 inches. Using the regular formula, 1/2 * bh, the area is 20 square inches. My formula asks for two bases, but a triangle has one, but you can consider it having two if you say the length of one of the bases is zero. So my formula also works on a triangle because the area also comes out to 20 square inches. So I conclude that the formula 1/2 * (b1 + b2) * h works on the five polygons.
From: upernova@sasd.k12.pa.us Liz Federowicz Shaler Area High School A Web page by Meghan Von Geis, Sarah Clark, and Liz Federowicz can be found at clark.pom5.96.html.
From: drews@sasd.k12.pa.us Drew Schaub Shaler High School Hi, This time our solution is in homepage form: see schaub.pom5.96.html.
From: jredding@telerama.lm.com Matt Rich Grade 11 School: North Allegheny Senior High School, Pittsburgh, PA rich.pom5.96.html.
From: sawdavid@freoshs.wa.edu.au David Saw Grade: 8 School: South Fremantle Senior High School This is my first entry for the Project of the Month. My answer can be found at saw.pom5.96.html.
From: daviscjd@freoshs.wa.edu.au Cameron Davis Grade: 8 School: South Fremantle Senior High School I am at davis.pom5.96.html.
From: brownsg@freoshs.wa.edu.au Steven Brown Grade: 8 School: South Fremantle Senior High School Hello. This is the first time I have entered the Project of the Month. I have tried a weekly problem once and I find your problems very interesting. You will find my solution on the Web at brown.pom5.96.html.
From: kealleym@freoshs.wa.edu.au Megan Kealley and Rebbekah Bullock Grade: 8 School: South Fremantle Senior High School Hi, we're Megan and Rebbekah. This is the first time we have entered the monthly problem. Thank you for looking at our solution. Our entry can be found at kealley.pom5.96.html.
From: batesca@freoshs.wa.edu.au Chris Bates and Ceri Blond Grade: 8 School: South Fremantle Senior High School Here is our solution to the May POM: bates.pom5.96.html
From: anne_d._sandler@sshs1.ccsd.k12.co.us From: Jennifer Sprangers and Yunny Chen Grades 9 & 10 School: Smoky Hill High School The single formula for area that will work for a rectangle, a parallelogram, a trapezoid, a triangle, and a square is the formula for a trapezoid. This is (b1+b2)h/2 where b1 = base one, b2 = base two, and h = height. This works because the area for a rectangle is normally b*h. Since the bases are congruent in a rectangle, the formula (b1+b2)h/2 would equal the same thing as b*h. The same is true for a square and a parallelogram, because in the formula (b1+b2)h/2 one base and the divided by 2 cancel out and you are left with b*h, which is the normal area formula. The formula works for a triangle because there is only one base, so when the formula says (b1+b2) then you would go (b1+0), multiply by the height, and then divide by 2. Simplified, the formula for the triangle is b*h/2, which is the normal formula. This formula works for a trapezoid because the (b1+b2)/2 equals the midline, and then you multiply by the height to find the area.
From: jordanna@cerf.net Jordanna Schutz Grade: going into 11th School: La Jolla Country Day Answer: (base1 + base2)/2 * height = common A formula for trapezoids First I will deal with squares, rectangles, parallelograms, and trapezoids. I will save triangles for after because....TRIANGLES ARE SPECIAL CREATURES! By definition a square is a rectangle which is a parallelogram which is a trapezoid (but not the reverse). Thus the equation for the area of a trapezoid ( (base1+base2)/2 * height ) should work for the other listed shapes, which are also trapezoids. But I'll show why... Area of a square is the length of a side squared. The area of a rectangle is length of one side times the length of an adjacent side. It is pretty clear that this formula works for a square where the adj. side will merely be the same length as the first. Area of a parallelogram is base times height. The base of a rectangle can be defined as the long side and the height will equal an adj. side because all angles are 90 degrees. The area of a trapezoid is the sum of the bases, divided by 2, multiplied by the height. The sum of the bases of a rectangle divided by 2 is the length of a single base. So the formula for a trapezoid works for squares, rectangles, and parallelograms. Now for triangles we have to make a little stretch. The formula for the area of a triangle is base*height/2. If the trapezoid area formula is applied to a triangle then the first step is to find the sum of the bases and divide it by two. But a triangle has only one base! Well, the opposite base is a point with a length of 0. So the sum of the base plus 0, divided by two is the same as the base/2. The next step in the trapezoid formula is to multiply the number we just got by the height. Thus base/2*height. This is the same as base*height/2. So assuming that the second base of a triangle is 0 then the trapezoid formula works for squares, rectangles, parallelograms, trapezoids, and triangles!
From: tor@ii.uib.no Tor Erling Bjorstad Grade: 9th School: Nattland Skole, Norway Answer: You measure the width at the 'top' of the figure, then the bottom width, average them, and multiply by the height of the figure: This simplifies for the square, rectangle and parallelogram to base * height, while the trapezoid is ((a+b)/2)*h (just like 'the' normal formula) The triangle can be simplified to base+0/2*height, or base*height/2 if you wish.
From: bnhs13a@mail.erols.com Jeffrey Chang Grade: 8 There is a formula that finds the area of a rectangle, a parallelogram, a trapezoid, a triangle, and a square, given two bases, be they existent or non-existent bases. Area, k, equals the average of the two bases, divided by two and multiplied by the height. Bases are defined here as any two parallel sides of the figure in question. A triangle would have a point from which a base could be constructed parallel to a true base of the triangle (from a point may be constructed an immediately parallel line to a given line). This formula works because the trapezoid is merely a triangle, a special triangle with a side instead of a point, a parallel side to another side. The parallelogram is merely the average of two equivalent bases multiplied by height, while the trapezoid has the same formula for area, where the bases are merely not the same, as in the parallelogram. Both rectangle and square are types of parallelogram, so that the formula would easily work. The only trouble with incorporating this formula into mathematics is that the definition of base is bothersome in the case of the triangle, and has two meanings in the parallelogram, two pairs of bases, both of which are correct.
From: jhaemmer@whitecap.psesd.wednet.edu Ben Wilhelm I've got a solution for the Problem of the Month. My name is Ben Wilhelm, I'm in 9th grade, and my school is the Northwest School, Seattle, WA. I'm doing this alone for fun, not part of a team and/or assignment. The formula is this: ( L_B + L_C ) / 2 * H = A Using English instead of the more obscure symbols and letters, it's: ( Length sub base + Length sub ceiling ) / 2 * Height = Area And plain English: The average of the base and the ceiling times the height is the area. The actual proper term for "ceiling" is also base, but using "lower base" and "upper base" would get very confusing, so I'll use ceiling. Here's the explanation: Everyone knows that the area of a rectangle is the base times the height. In the case of a rectangle, the base is equal to the ceiling. Since the average of any number of equal numbers is that number, this formula works for rectangles. A parallelogram is roughly the same problem. It is simply a skewed rectangle. This can be made obvious by: +------------+-----+ / | / / | / / |/ +------------------+ Simply move the right triangle to the left side, and you get: This is a rectangle, and we have already proven it works for rectangles. +-----+-----------+ | / | | / | |/ | +-----------------+ This equation is the definition of area for a trapezoid, but I'll go into it anyway. If you take a trapezoid, like so, and cut a triangle off the right side: +-----------+ / |\ / | \ / | \ +-----------------+-----+ then flip the triangle vertically and move it over to the left side, you get: +-----+-----------+ | / | | / | |/ | +-----------------+ Look familiar? It should. It's also the end figure for the parallelogram. The length of this rectangle is the average of the base and the ceiling. This also works for a non-isosceles. For a triangle it works somewhat like a trapezoid, except the ceiling has a length of 0. It's just a point, with no length. A square is just a special-case rectangle. Very nice problem, by the way. That's it!
From: peterson@eng16.rochny.uspra.abb.com Dave Peterson I am a software engineer and homeschooling math teacher. My son (11) thought this was too easy for a monthly project; although I'm not a student, I thought I'd submit an answer just for fun. The rectangle, parallelogram, trapezoid, and square are in fact all (special cases of) trapezoids (since all have two parallel sides), so the trapezoid formula applies to all of them. The triangle can be thought of as a (very!) special case of trapezoid, with a "top" of length zero. So the common formula is, to use my favorite version, area = (altitude) * (average of top and bottom) where the top and bottom are defined to be the two parallel sides of a trapezoid, or one side and the opposite vertex of a triangle, and the altitude is defined to be the distance from the top to the bottom. As for how this works, the simplest explanation for the formula I know of, which happens to make the triangle case clear as well, is to divide a trapezoid into two triangles along a diagonal: b2 B +----------------------+ C | \____ \ | \____ \ | \____ \ | \____ \ | \____ \ | \ \ A +-----------------------------+ D b1 (Nothing in the drawing is assumed to be equal or right angles, and the zigzag is supposed to be a straight line.) The area of ABC is h * b1 / 2, where h is the altitude, or distance from line AD to line BC, and b1 is the length of AD. The area of BCD is h * b2 / 2; for the triangle case, this reduces to zero, since b2 is zero. The total area of the trapezoid is b1 + b2 Area = h * -------- 2 Then I tried to find a similar formula for the area of a general quadrilateral, which would involve an "average" altitude (the distance of the midpoint of the "top" side from the opposite side) as well as the "average" width. I discovered something not far from that: the area is the average of the products of two opposite sides times the average altitude to the opposite side; expressed more visually, it is the average of the areas of a rectangle on the "bottom" passing through the center of the top, and a rectangle on the "top" passing through the center of the bottom. Specifically, if AD and BC are not necessarily parallel in the figure above, and we construct midpoints X and Y of the sides AD and BC respectively, and find h1 and h2, the distances of point Y from line AD and of point X from BC respectively, then the area is: Y B +-----------+---+------+ C | b2 | |_| \ | | | \ | h1| | \ | | |h2 \ | _| | \ | | | | \ A +-----------+---+-------------+ D b1 X b1 * h1 + b2 * h2 Area = ----------------- 2 This is a reasonable formula, since it requires knowing only four distances (two sides and two "altitudes"). I proved it by adding together the area of each of the four triangles formed by diagonals, then dividing by two because the triangles cover the quadrilateral twice. For a trapezoid, the two altitudes are the same, yielding the same formula as above. For a triangle, b2 becomes zero and B=C=Y, so we have the usual triangle formula. So this formula goes one step beyond what was asked for.
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8 December 1996