Area of segments?

The area of a segment can be found out using the following formula:

$\large \color{#3D99F6} A = \dfrac{1}{2} r^2 (\theta - \text{sin} \theta )$

where r is radius and $\theta$ is in radians. In our case $\theta$ = $\dfrac{\pi}{2}$

Hence we can solve for the following equation:

$\large \color{#D61F06} \dfrac{1}{2} r^2 (\theta - \text{sin} \theta ) = 1 \\ \large \color{#EC7300} \dfrac{1}{2} r^2 ( \dfrac{\pi}{2} - 1 ) = 1 \\ \large \color{#20A900} r^2 ( \dfrac{\pi}{2} - 1 ) = 2 \\ \large \color{#69047E} r^2 = \dfrac{4}{\pi - 2} \\ \large \color{#624F41} r = \sqrt{\dfrac{4}{\pi - 2}} \ \text{cm}$

The area of blue region and red region combined is: $\large \frac{\pi r^2}{4}.$

The area of the red triangle is : $\large\frac{ base \times height}{2}= \frac{r \times r}{2}.$

Thus, the area of the blue region is: $\large \frac{\pi r^2}{4} - \frac{r^2}{2}=1\implies r^2(\frac{\pi}{4}-\frac{1}{2}) = 1\implies r^2=\frac{4}{\pi-2}.$

Therefore, $\boxed{ r = \sqrt{\frac{4}{\pi-2}}}.$