Question

A rhombus, $ABCD$ has sides of length $10$. A circle with centre $A$, passes through $C$ (the opposite vertex). Likewise, a circle with centre $B$ passes through $D$. If the two circles are tangent to each other, what is the area of rhombus?

Answer 1

The tangent point of the circles must lie on the extension of line $AB$, since their centers lie on $AB$.

By symmetry, the diagonals of a rhombus bisect each other and meet at right angles.

Consider the dot product of

$AC.BD=(AB+AD).(AD-AB)$

$=AD.AD-AB.AB$

$=0$, since $\left|AB\right|=\left|AD\right|$

$\therefore AC$ is perpendicular to $BD$

Let $R=AC=$ radius of larger circle, and $r=BD=$ radius of smaller circle.

Then, considering the four right triangles, the area of rhombus $ABCD=4\times \frac{R}{2}\times \frac{r}{2}\frac{1}{2}=\frac{Rr}{2}$.

Considering one of the right triangles,${\left(\frac{R}{2}\right)}^2+{\left(\frac{r}{2}\right)}^2={10}^2$, from which $R^2+r^2=400$.

Since the circles meet at a tangent, on $AB$, we have $R-r=10$.

Hence ${(R-r)}^2=R^2+r^2-2Rr=100$, and so $2Rr=300$.

Therefore the area of the rhombus $=\frac{Rr}{2}=75$ square units.