Problem
Given, we have a square ▢ABCD with a side of 2cm. We construct the diagonal and draw a circle tangent to all the 3 sides of the triangle △ABD formed. Then, we divide this circle into 4 equal quadrants. In one of the quadrants, we construct another circle tangent to the 2 sides and one part of the circle. How can we find its area?
It is highlighted in black in the figure given below.
My Progress
All is was able to figure out was the radius of the bigger circle which turned out to be $2-\sqrt 2$ which is $\approx 0.59$
How can I find its area?
PS: I would be grateful to anyone who would be able to explain keeping in mind the concepts learned in grade 8th.
Thanks!
Let $R$ and $r$ be the radii of the bigger and smaller circles respectively. The required area is $\pi r^2$.
To follow your approach using "the tangents meeting are equal theory", consider the common tangent line of the two circles:
$$r + R = \sqrt{R^2+R^2} = R\sqrt 2$$
Alternatively, I use the distance from the outside right-angle to the tangent point to calculate both radii:
$$\begin{align*} R + \sqrt{R^2+R^2} = R \left(1+\sqrt2\right) &= BM\\ r + \sqrt{r^2+r^2} = r \left(1+\sqrt2\right) &= R \end{align*}$$