Area of square that is inscribed in a circle that is also inscribed in a rhombus

Question

A circle is inscribed in a rhombus whose diagonals are $17 cm$ and $27 cm$. What is the area of square inscribed on the same circle?

Solution:

Centred at the origin, one side can lie on the line $\frac{x}{(13.5)} + \frac{y}{(8.5)} = 1$. The square of the distance from the origin to the line/side $\frac{x}{(13.5)} + \frac{y}{(8.5)} = 1$ is $\frac{1} {\frac{1}{(13.5)^2}} + \frac{4}{(8.5)^2}$ which is $\frac{210681}{4072}$ which is half the area of the inscribed square is which is $\frac{210681}{2036}cm²$.

Question:

Is $\frac{210681}{2036}cm²$ correct? Or is there something wrong with my solution?

Answer 1

Let $r$ be the circle of the radius. Then,

$$r=\frac{a_1a_2}{\sqrt{a_1^2+a_2^2}}$$

with $a_1$ and $a_2$ being the half diameters of the rhombus. Then, the area of the square is $$ Area = 2r^2 = \frac{2\left( \frac{17}2 \cdot\frac{27}2\right)^2}{{\left(\frac{17}2\right)^2+\left(\frac{27}2\right)^2}} = \frac12\cdot\frac{(17\cdot27)^2}{17^2+27^2}$$

Answer 2

Alternatively, you can compute the area of the rhombus as $(17)(27)/2 = 459/2$, but since this area is also equal to the diameter of the inscribed circle multiplied by a side length, it follows that $$d = \frac{459}{2\sqrt{(17/2)^2 + (27/2)^2}} = \frac{459}{\sqrt{1018}}.$$ The area of the square is simply $$\frac{d^2}{2} = \frac{210681}{2036}.$$