Question

the area of the deep dark square​

Answer 1

In the above figure,

let ABCD be the square, O be the centre of the circle and OP and OQ be the radii 'r'

join O and D, we see that OB = r and BD = d

since x - axis acts as a tangent to the circle,

angle OQD = 90° ... (radius _|_ tangent)

In ∆OQD, angle OQD = 90°

OD² = OQ² + DQ² ... (Pythagorean Theorem)

(r + d)² = r² + r² (OP = DQ; OPDQ forms a square)

(r + d)² = 2r²

taking square root on both sides

r + d = r√2

d = (r√2) - r = r(√2 - 1)

squaring both sides

d² = r²(2 + 1 - 2√2)

d² = r²(3 - 2√2) ... (i)

now, area of square ABCD

$= \frac{1}{2} {d}^{2} \\</p><p>= \frac{{r}^{2}(3 - 2 \sqrt{2})}{2}$ ... (from (i))

Hence, area of the darkened square is

$\frac{{r}^{2}(3 - 2 \sqrt{2})}{2}$