Show that a rhombus with diagonals $x$ and $y$ has sides of length $\frac12\sqrt{x^2+y^2}$

Question

I have no idea how to answer this question.

If the diagonals of a rhombus are $x$ and $y$, show that the length of its side is $\frac12 \sqrt{x^2+y^2}$.

If somebody could explain ...

Thank you

Answer 1

You do not need any right angles as the relation to show is a special case of the parallelogram identity.

So, here we go:

If $\vec a$ and $\vec b$ are the sides spanning the rhombus you have

• $|\vec a| = |\vec b| = a$ the side length of the rhombus
• $\vec a + \vec b = \vec x$ and $\vec a - \vec b = \vec y$ are the diagonals
• with lengths $|\vec x| = x$, $|\vec y| = y$
• to show: $4a^2 = x^2 + y^2$

Use the scalar product to show this: $x^2 + y^2 = (\vec a + \vec b) \cdot (\vec a + \vec b) + (\vec a - \vec b) \cdot (\vec a - \vec b) = a^2 + b^2 + 2\vec a \cdot \vec b + a^2 + b^2 -2\vec a \cdot \vec b = 4a^2$

Answer 2

In a rhombus, the diagonals are perpendicularly bisected. Thus, the side of a rhombus of length $l$ is the hypotenuse of a right triangle with the semi diagonals as it's sides. Thus we have: $$l^2=(x/2)^2+(y/2)^2$$ $$\Longrightarrow l = \frac{1}{2}\sqrt{x^2+y^2}$$