Two equal circles are drawn in square in such a way that a side of the square forms diameter of each circle. If the remaining area of the square is 42 cm$^2$. How much will the diameter of the circle measures? [IBPS—RRB Officers Gr. 'B' Exam, 2015]

### Concept & Area Subtraction The problem refers to semicircles formed on the sides of a square that cut into its area. The remaining area is the area of the square minus the area of these two semicircles (which combine to form one full circle). $$Area_{remaining} = Area_{square} - Area_{circle}$$ ### Step-by-Step Solution * Let the side of the square be $a$. The total area of the square is $a^2$. * Two semicircles are drawn with a diameter equal to the side of the square ($a$). Their radius is $r = \frac{a}{2}$. * The combined area of these two semicircles is equal to the area of one full circle: $\pi \times (\frac{a}{2})^2 = \frac{\pi a^2}{4}$. * Substitute $\pi = \frac{22}{7}$: Area $= \frac{22}{7} \times \frac{a^2}{4} = \frac{11a^2}{14}$. * The remaining area of the square is $a^2 - \frac{11a^2}{14} = \frac{3a^2}{14}$. * We are given that the remaining area is 42 cm$^2$. So, $\frac{3a^2}{14} = 42$. * Solving for $a^2$: $a^2 = 42 \times \frac{14}{3} = 14 \times 14 = 196$. * Therefore, $a = 14$ cm. Since the diameter of the circle is equal to the side of the square, the diameter is also 14 cm. ### Exam Strategy & Shortcut Memorize that when a full circle (or two semicircles) of maximum diameter is inscribed in a square, the circle occupies $\frac{11}{14}$ of the square's area, and the remaining space is $\frac{3}{14}$ of the area. Setting $\frac{3}{14} \times a^2 = 42$ quickly yields $a = 14$. ### Common Pitfall Taking "two equal circles" literally as two independent full circles inside the square, which would cause an impossible negative remaining area. Contextually, this phrasing in competitive exams implies semicircles on opposite edges. ### Final Answer Therefore, the correct answer is **14 cm**.