this is such an amazing problem and I did not expect it to have the outcome that it had the problem is that we have a semicircle right here and a quarter Circle right here and then this red line with the length of 5 cm which goes from this vertex and is tangent to this semicircle we want to find the area of this rectangle so how can we solve it as a first step let us assign a radius R to this semicircle now many of the people think that this radius is the same as this quarter circle's radius but it is nowhere given that they are the same and in the later steps we will also discover that we don't require them to be equal therefore this is one of my favorite problems so now let us label this radius as X this piece will also be equal to r as it is the radius of this semicircle and this height will be equal to X which will be the same as the radius of this quarter Circle we can find the area of this rectangle in terms of R and X the base is x + 2 R and the height is X so the area will be base x height or X plus 2 R * X which will be x² + 2xr now here comes the magic let us draw a segment from this point of tangency to the center of the semicircle and this will be the same as the radius of the semicircle which is R every radius that intersects the tangent line at the point of tangency is at right angle so now we have a nice right angle triangle and therefore yes you have guessed it right we will use Pythagoras Theorem we get r s + 5² = x + r² the left hand side will be r² + 25 and the right hand side will be x² + 2xr + r² these r s will get cancelled out and we are left with x² + 2 XR = 25 but we know that this quantity is the same as the area of this rectangle so the area of this rectangle will be 25 don't forget to put the cm square so good