Macedonian Mathematical Competitions

In the problem it is given that $d_1 = 7.2\,\text{cm}$ and $P = 34.56\,\text{cm}^2$. From the formula for area $P = \frac{d_1 \cdot d_2}{2}$ we have $d_2 = \frac{2 \cdot P}{d_1} = \frac{2 \cdot 34.56}{7.2} = 9.6\,\text{cm}$. For the side length of the rhomb we have $a = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2} = \sqrt{3.6^2 + 4.8^2} = 6\,\text{cm}$. So the perimeter of the rhomb is $L = 4 \cdot a = 4 \cdot 6 = 24\,\text{cm}$. The radius of the inscribed circle is half of its altitude. If we denote with $x$ the length of the orthogonal projection of one side of the rhomb on its neighboring side we have $h^2 = a^2 - x^2$ and $h^2 = d_1^2 - (a-x)^2$. From these two equalities we have $a^2 - x^2 = d_1^2 - (a-x)^2$ or $a^2 - d_1^2 + 2ax = 0$. From this we have $x = \frac{2a^2 - d_1^2}{2a} = \frac{2 \cdot 6^2 - 7.2^2}{2 \cdot 6} = 1.68\,\text{cm}$. Then $h = \sqrt{a^2 - x^2} = \sqrt{6^2 - 1.68^2} = 5.76\,\text{cm}$ and the radius of the inscribed circle is $r = \frac{1}{2}h = \frac{1}{2} \cdot 5.76 = 2.88\,\text{cm}$.