Question 52

Area of the circle inscribed in a square of diagonal 6√2 cm (in sq cm) is

Solution

Length of AC = $$6\sqrt{2}$$ cm

Let side of square = $$x$$ cm = Diameter of circle

In $$\triangle$$ ABC, => $$(AB)^2 + (BC)^2 = (AC)^2$$

=> $$(x)^2 + (x)^2 = (6\sqrt{2})^2$$

=> $$2x^2 = 72$$

=> $$x^2 = \frac{72}{2} = 36$$

=> $$x = \sqrt{36} = 6$$ cm

Thus radius of circle = $$\frac{6}{2} = 3$$ cm

$$\therefore$$ Area of circle = $$\pi r^2$$

= $$\pi \times (3)^2 = 9\pi$$ $$cm^2$$

=> Ans - (A)

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