Correction: Jonathan pointed out that I did not specify the order of the vertices. Thanks, Jonathan! Here is the revised version in which A and C are opposite vertices as are B and D:
Consider parallelogram ABCD, three of whose vertices are A(0,0), B(2,3) and D(3,2).
Find the coordinates of C and the area.
Of course, we expect our Geometry students to celebrate even more by generalizing:
Note: This has been revised for the reasons stated in the correction at the top.
Three of the vertices of a parallelogram ABCD are A(0,0), B(a,b) and D(b,a), where b>a>0.
(a) Show that vertex C has coordinates (b+a,b+a).
(b) Prove that this figure is actually a rhombus.
(c) Show that its area is b2 - a2. Can you find FIVE ways? (ok, that's a stretch but anything is possible on May 5th!).
Sunday, May 4, 2008
A Geometry Tribute to Cinco de Mayo
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9 comments:
Thinking out loud, can I poke a hole in this?
Let the fourth vertex be (a-b, b-a)
OK, that's not a rhombus (unless ab=0). So I think we need to specify that the origin is the second coordinate to make the rest fly.
Jonathan
Jonathan, You're right!
I need to label the vertices in 'order' like OABC. Very careless on my part - thanks!! I'll correct that in the post. With that corrected, give me your thoughts about the question and how you think your geometry students will prove it would be a rhombus. There are several ways, only one of which is to use the distance formula for all 4 sides.
The fourth point has the coordinates (a+b,b+a). It can be
constructed by mirroring the triangle formed by the given
three points along the side opposite to the vertex that
is the origin.
(a)
The figure is formed by mirroring an isosceles triangle
along its baseline. This means that the lengths of opposite
sides are equal as are opposite angles. Therefore the new
figure must be a rhombus.
(b)
Because of symmetry the surface of the
rhombus equals 2* the size of the triangle.
(i) Use the formula for the area of an isosceles triangle
(requires calculation of the sides of the triangle which
need sqrt operation for the height)
(ii) Use the formula for the area of a rhombus (requires
calculation of the diagonals of the rhombus which needs sin, cos)
(iii) Inscribe the triangle into a smallest square (=b^2) and
substract the three trangles that are not port of the triangle.
Multiply the result by two (=mirroring the isocees triangle).
There are many other ways by selecting differnt area segments
of the rhombus .. you get the idea.
I wonder if there is a another unique way to do it?
I revised the problems and the title to accommodate Jonathan's suggestion. The problem is that the RSS feed might reflect the original version - not sure of that.
Florian,
Nice methods. I found a couple of others. One way to show that it's a rhombus is to show that 2 adjacent sides are congruent since it's already a paralleogram.
One can show that the diagonals are (b+a)√2 and (b-a)√2. Also, the midpoint is ((b+a)/2,(b+a)/2). This leads to several methods.
The symmetry you noted is what I really like about this diagram. That and the coordinate connection with the reflection over the line y = x.
Also:
(1) No one uses this much anymore but the determinant approach is always fun to see how it produces a simple algebraic expression:
Consider the determinant whose rows are
0 0 1
a b 1
b a 1
Expanding by minors of the first row, this reduces to the 2x2 determinant:
a b
b a
which provides us with the b^2-a^2 form we wanted for the area of the rhombus. Normally we would use determinant for the area of a triangle but I modified this from the symmetry of the diagram.
(2) Another view is to use use trig as you suggested. The area of the parallelogram is (side1)(side2)sinθ
We can determine sinθ from
sinθ = 2sin(θ/2)cos(θ/2).
The sin and cos of the half angles follow easily from the right triangle formed by the diagonals of the rhombus.
There are also coordinate methods. The beat goes on...
A slight correction to the determinant approach. The order in which you place the coordinates can affect the sign of the result (there's an orientation involved). Either reverse the rows or take absolute values to guarantee the area is positive.
Florian's method of placing the triangle in a square is very nice for students to see. Any triangle can be placed in a rectangle whose sides are parallel to the axes. The area can then be obtained by subtraction.
OK, revised and clean, how do we show that (a,b), (0,0), (b,a), (a+b, a+b) is a rhombus.
1. Distance^2 for each side is
a^2 + b^2
2,3. Show parallelogram, then show two adjacent sides congruent (for Parallelogram, lots of choices. AB||DC, AD||BC (by slopes b/a and a/b). Or diagonals bisect each other: 2*midpoint(AC) = (a+b,a+b), 2*midpoint(BD) = (a+b,a+b)). Then distance for the adjacent sides
4,5 Show parallelogram, then show diagonals perpendicular. Same options for parallelogram, then slopes of diagonals are -1 and 1 (does not work if a=0, b=0, or a=-b)
How do we show that the area is b^2 - a^2 ?
hm
Or if you allow calculus, just integrate over one of the four
trianges of the rhombus. Then
multiply by 4.
For that one needs the difference
of the slope of the diagonal and
the slope of the point (a,b). If
we call this the slope m we get:
f(x) = mx
F(X) = (mx^2)/2 + C
Area = 4*[F(SQRT(a^2+b^2)) -F(0)] = b^2-a^2
If you consider the points in three dimensions so that you have A(0,0,0), B(a,b,0), and D(b,a,0), you can also find the area by taking the cross product of vectors AB and AD.
AB X AD =(a^2-b^2)k.
|AB x AD|=b^2-a^2
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