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I recently taught a group of geometry students about properties of rhombuses, and gave them a set of homework exercises created by a previous instructor which included the following problem.

If a rhombus has sides of length $25$ mm and a $45^{\circ}$ angle, determine the lengths of the diagonals.

Here is an image for reference.

enter image description here

I ultimately removed this problem because I haven't found a simple solution at a high school geometry level (i.e. some algebra, no trig). I've solved the problem two ways, which I will post as an answer, one way involved simple trig and the other way was geometrical but messy (involved solving a system of nonlinear equations and a fourth degree polynomial).

I'm curious if anyone can solve this problem a different way. I'd like to know if there is nice solution involving only geometry, but I would also be interested in seeing any novel approaches anyone comes up with.

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  • $\begingroup$ Well the parallelogram law gives a formula relating the diagonals to the sides. Im trying to think of any geometry theorems that can make another relationship $\endgroup$ – Jonathan Davidson Jun 30 '17 at 22:19
  • $\begingroup$ Take a look at math.stackexchange.com/questions/173016/… and my two answers. Not the same as this question, but still... $\endgroup$ – Will Jagy Jun 30 '17 at 22:46
  • $\begingroup$ Law of cosines should do it. $\endgroup$ – Daniel Schepler Jun 30 '17 at 23:38
  • $\begingroup$ My solution might be an easy geometric way to solve this problem. $\endgroup$ – Seyed Jul 1 '17 at 23:58
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It can be done fairly straightforwardly with coordinate geometry.

If we place the origin in at the left vertex and the $x$-axis along the bottom edge, then:

  • The top-left vertex is at $(25/\sqrt{2}, 25/\sqrt{2})$
  • The top-right vertex is at $(25 + 25/\sqrt{2}, 25/\sqrt{2})$
  • The bottom-right vertex is at $(25, 0)$

The first follows because we memorize the ratio of sides of a 45-45-90 triangle (i.e. $1:1:\sqrt{2}$), and the rest just by translating to the right.

The lengths of the two diagonals can then be computed with the distance formula.

This can be converted to an "only geometry" proof by drawing in a few right triangles.

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    $\begingroup$ The Greeks wouldnt be pleased, but it works. $\endgroup$ – krirkrirk Jun 30 '17 at 22:32
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As I mentioned in the question, here are the two ways that I have found to solve this problem.

Method 1: Trigonometry

If we draw in the diagonals, we get the following image, and we know that each diagonal bisects the other as well as the angles of the rhombus:enter image description here

That means that we can say \begin{align*} a &= 25\sin(22.5^{\circ})\text{ mm}\\ b &= 25\cos(22.5^{\circ})\text{ mm}, \end{align*}

meaning that the diagonals are \begin{align*} 2a &= 50\sin(22.5^{\circ})\text{ mm}\approx 19.13 \text{ mm}\\ 2b &= 50\cos(22.5^{\circ})\text{ mm} \approx 46.19 \text{ mm}. \end{align*}

Method 2: (Messy) Geometry

Here I started by drawing a line perpendicular to the top connecting to the lower right angle. Since the upper right angle is $45^{\circ}$ we know that this line makes a $45^{\circ}$ angle as well, meaning we have an isosceles triangle: enter image description here

By the Pythagorean Theorem, we know that

$$2x^{2} = 25^{2} \implies x = \frac{25}{\sqrt{2}} \text{ mm}.$$ Therefore the area of the rhombus is given by

$$\text{Area} = 25 \text{ mm } \times \frac{25}{\sqrt{2}} \text{ mm} = \frac{25^{2}}{\sqrt{2}} \text{ mm}^{2}.$$

Furthermore, going back the the variables introduced in the first image, we can say that

$$a^{2} + b^{2} = 25^{2} \quad \text{and} \quad 4\times \frac{1}{2}ab = \frac{25^{2}}{\sqrt{2}}.$$

Solving for $a$ in the second equation yields

$$ a = \frac{25^{2}}{2b\sqrt{2}},$$

which when substituted into the first equation gives us

$$b^{2} + \frac{25^{4}}{8b^{2}} = 25^{2},$$

which upon multiplying by $b^{2}$ can be written as

$$b^{4} - 25^{2}b^{2} + \frac{25^{4}}{8} = 0.$$

Now, this equation is quadratic in $b^{2}$, so we can just apply the quadratic formula to obtain

$$b^{2} = \frac{25^{2}\pm\sqrt{25^{4} - \frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\sqrt{\frac{25^{4}}{2}}}{2} = \frac{25^{2}\pm\frac{25^{2}}{\sqrt{2}}}{2} = \frac{25^{2}}{2}\left(1 \pm \frac{\sqrt{2}}{2}\right),$$

and so $$b = \frac{25\sqrt{2\pm\sqrt{2}}}{2} \approx 23.096, 9.567.$$

Back substituting then we obtain

$$a = \frac{25^{2}}{2\left(\frac{25\sqrt{2\pm\sqrt{2}}}{2}\right)\sqrt{2}} = \frac{25}{\sqrt{4\pm 2\sqrt{2}}}\approx 9.567, 23.096.$$

Since we chose $b$ to be the longer diagonal then we can say

\begin{align*} 2a &= \frac{25}{\sqrt{4+ 2\sqrt{2}}}\text{ mm} \approx 19.13 \text{ mm}\\ 2b &= \frac{50\sqrt{2+\sqrt{2}}}{2}\text{ mm} \approx 46.19 \text{ mm}. \end{align*}

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  • $\begingroup$ I liked the approach in CandiedOrange’s answer, which unfortunately has been deleted. It’s similar to your second solution, but I find it more direct: drop an altitude from the upper-right vertex and after using the isosceles right triangle to find the length of this altitude, you can compute the long diagonal directly via the Pythagorean theorem. I think I’ll add to to my answer. $\endgroup$ – amd Jul 1 '17 at 21:26
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Because the rhombus lie in the euclidean space $\Bbb R^2$ we can use vectors:

rhombus

Here clearly $\|u\|=\|v\|$ and $\alpha+\beta=\pi$. Then suppose that $v$ lie in the $X$-axis, then $v=(\|v\|,0)$. This imply, by trigonometry, that $u=(\|v\|\cos\alpha,\|v\|\sin\alpha)$.

And because $\|r\|=\sqrt{r_1^2+r_2^2}$ for some vector $r=(r_1,r_2)$ in $\Bbb R^2$ then

$$\|u+v\|=\|v\| \sqrt{(1+\cos\alpha)^2+(\sin\alpha)^2}=\|v\|\sqrt{2+2\cos\alpha}$$

$$\|v-u\|=\|v\|\sqrt{(1-\cos\alpha)^2+(\sin\alpha)^2}=\|v\|\sqrt{2-2\cos\alpha}$$

In our case $\|v\|=25\rm mm.$ and $\alpha=\pi/4$.

P.S.: I've learned basic euclidean space vector manipulation in high school, at least in the last years.

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Alternative hint using complex numbers: set the origin of the complex plane at the leftmost vertex $a=0$ of the rhombus, and the real axis along the bottom side. Then the other vertex on the real axis is $b=25\,$, and its opposite vertex is give by a rotation of $\pi/4$ counterclockwise around the origin, so it is $d = b \cdot e^{i \pi / 4}=25(1+i)/\sqrt{2}\,$. The lengths of the diagonals are $|b-d|$ and $|b+d|\,$, which reduces the problem to a routine calculation in complex numbers.

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Answer using geometry: $$d = 25 \sqrt{2-\sqrt{2}},$$ $$D = 25 \sqrt{2+\sqrt{2}}.$$

We just need to use Pythagorean Theorem for some rectangle, as we can see in the diagram below: Diagram 1

We have $ l = \frac{25}{\sqrt{2}}$, triangle (CDE), and from the larger triangle (AEC): $$ D^2 = l^2 + (l+25)^2.$$

Diagram 2

For the minor diagonal $d$, it is similar but instead: $$d^2 =l^2 +(25-l)^2,$$ from triangle (BDE).

So we can find it easily!

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We know that the area of the rhombus is equal to $25^2\sin\frac\pi4$. If you don’t want to introduce trig functions, you can derive this value the way you did, by dropping a perpendicular. Letting $a$ be the length of either diagonal, by Heron’s formula this area is equal to $$2\sqrt{s(s-a)(s-25)^2}$$ where $s=\frac12(a+50)$. Equating these and squaring both sides produces a quadratic equation in $a^2$, which students at that level ought to be able to solve. The nice thing about this approach is that it gives you both diagonal lengths at once.


CandiedOrange gave a simple approach to a solution in an answer that’s since been deleted, but I like it enough to reproduce here.

enter image description here

Drop an altitude from the upper-right vertex. You can use the isosceles triangle argument from your second solution to find the lengths $DE=BE$, and from there a direct application of the Pythagorean theorem to $\triangle{AED}$ gets you $AD$. The other diagonal can be recovered by observing that the area of the rhombus is $4\cdot\frac12\cdot(\frac12AD\cdot\frac12BC)=\frac12AD\cdot BC=AB\cdot DE.$ Alternatively, drop another altitude from $B$ or $C$ and repeat the calculation done for the first diagonal.

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This might be an easy geometric solution to this problem. We can think of a regular octagon which is made up of 8 halves rhombuses. enter image description here

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