Rectangle ABCD circumscribes the circle of radius $\displaystyle 10$ cm. Prove that ABCD is a square. Hence, find the perimeter of ABCD.

\[\mathrm{AP}=\mathrm{AS} \] \(\displaystyle \mathrm{BP}=\mathrm{BQ}\) \(\displaystyle \mathrm{CR}=\mathrm{CQ}\) DR = DS Adding the above four equations, \[\begin{aligned} & \mathrm{AP}+\mathrm{BP}+\mathrm{CR}+\mathrm{DR}=\mathrm{AS}+\mathrm{BQ}+\mathrm{CQ}+\mathrm{DS} \\ & \Rightarrow \mathrm{AB}+\mathrm{CD}=\mathrm{AD}+\mathrm{CB} \quad \text{--- (i)} \end{aligned} \] Since ABCD is a rectangle \(\displaystyle \therefore \mathrm{AB}=\mathrm{CD}\) and \(\displaystyle \mathrm{BC}=\mathrm{AD}\) ⇒ from (i), \(\displaystyle 2 \mathrm{AB}=2 \mathrm{AD}\) or \(\displaystyle \mathrm{AB}=\mathrm{AD}\) Hence ABCD is a square Clearly side of square = diameter of circle \(\displaystyle =20 \mathrm{~cm}\) ∴ Perimeter of square \(\displaystyle =4 \times 20 \mathrm{~cm}=80 \mathrm{~cm}\) Let the length of shortest side be \(\displaystyle x\) m ∴ length of longest side \(\displaystyle =(x+4) \mathrm{m}\) and length of third side \(\displaystyle =(x+2) \mathrm{m}\) Now, \(\displaystyle (x+4)^{2}=x^{2}+(x+2)^{2}\) \(\displaystyle \Rightarrow x^{2}-4 x-12=0\) \(\displaystyle \Rightarrow(x-6)(x+2)=0\) \(\displaystyle \Rightarrow x=6\) ∴ sides are $\displaystyle 6$ m, $\displaystyle 8$ m and $\displaystyle 10$ m Area \(\displaystyle =\frac{1}{2} \times 6 \times 8=24 \mathrm{~m}^{2}\) Perimeter \(\displaystyle =6+8+10=24 \mathrm{~m}\) Difference = $\displaystyle 0$\(\displaystyle \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}\) \(\displaystyle \Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}\) Simplifying, we get \(\displaystyle 2 x^{2}-19 x+42=0\) \(\displaystyle \Rightarrow(x-6)(2 x-7)=0\) \(\displaystyle \Rightarrow x=6\) or \(\displaystyle x=\frac{7}{2}\) As, \(\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5}\) \[\begin{aligned} & \Rightarrow \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\ & \Rightarrow \angle \mathrm{C}=\angle \mathrm{R} \end{aligned} \]\[\begin{aligned} & \angle \mathrm{ADC}=\angle \mathrm{PSR} \\ & \text { and } \angle \mathrm{C}=\angle \mathrm{R} \\ & \therefore \triangle \mathrm{ADC} \sim \triangle \mathrm{PSR} \\ & \text { (ii) } \frac{\mathrm{AD}}{\mathrm{PS}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{3}{5} \\ & \Rightarrow \frac{4}{\mathrm{PS}}=\frac{3}{5} \\ & \Rightarrow \mathrm{PS}=\frac{20}{3} \mathrm{~cm} \end{aligned} \] \[\text { (iii) } \begin{aligned} \frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{PQR})} & =\frac{\dfrac{1}{2} \times \mathrm{BC} \times \mathrm{AD}}{\dfrac{1}{2} \times \mathrm{QR} \times \mathrm{PS}} \\ & =\frac{3}{5} \times \frac{3}{5}=\frac{9}{25} \end{aligned} \] \[\therefore \operatorname{ar}(\triangle \mathrm{ABC}): \operatorname{ar}(\triangle \mathrm{PQR})=9: 25 \] Join AF intersecting line \(\displaystyle m\) at G In \(\displaystyle \triangle \mathrm{ACF}, \mathrm{BG} \| \mathrm{CF}\) \[\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AG}}{\mathrm{GF}} \] In \(\displaystyle \Delta \mathrm{FDA}, \mathrm{GE} \| \mathrm{AD}\) \[\Rightarrow \frac{\mathrm{EF}}{\mathrm{DE}}=\frac{\mathrm{GF}}{\mathrm{AG}} \text { or } \frac{\mathrm{DE}}{\mathrm{EF}}=\frac{\mathrm{AG}}{\mathrm{GF}} \] From, (i) and (ii), we get \(\displaystyle \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{EF}}\)