In a square of side length a, two circular arcs are drawn with centres at opposite vertices, joining the other two vertices and forming two overlapping sectors inside the square. What is the area common to both sectors?

Introduction / Context:
This geometry question deals with the intersection area formed by two circular sectors inside a square. It uses ideas from circles, sectors, and symmetry. The configuration creates a lens shaped region that is common to two quarter circles drawn with centres at opposite corners of the square. Problems like this are important for building spatial reasoning and for understanding how standard formulas for areas of sectors and triangles can be combined to find areas of complex shapes.

Given Data / Assumptions:
• The figure is a square of side length a.
• Opposite vertices of the square are used as centres of two equal circles of radius a.
• Each circular arc is drawn inside the square and passes through the remaining two vertices.
• We are asked to find the common area of the two circular sectors inside the square.
• The distance between opposite vertices of the square is a√2, which is the distance between the two circle centres.

Concept / Approach:
Each circle has radius a and the distance between their centres is a√2. The lens shaped common region of the two circles is symmetric with respect to the line joining their centres. For two equal circles of radius r whose centres are distance d apart, the common area can be expressed in terms of circular sectors minus the area of an isosceles triangle. In this special case, the geometry simplifies because the angle between the radii at the centre is a right angle. We can therefore work with one circle and then double the area of a single sector based component to obtain the total intersection area.

Step-by-Step Solution:
Step 1: Let the square be ABCD with side a. Take A and C as the centres of two circles of radius a. The other vertices B and D lie on both circles. Step 2: In each circle, the angle subtended at the centre by chord BD is 90 degrees, since angle BAD and angle BCD in a square are right angles. Step 3: Therefore, in each circle, the relevant region corresponding to the intersection is based on a sector of angle 90 degrees, that is π/2 radians. Step 4: For one circle, the area of the sector with angle π/2 is (π/2) * a^2. Step 5: The triangle formed by the two radii and the chord BD inside that sector is a right isosceles triangle with legs of length a, so its area is (1/2) * a * a = a^2 / 2. Step 6: The area of the circular segment in one circle is area of sector minus area of triangle = (π/2) * a^2 - a^2 / 2. Step 7: The common lens shaped area is made of two identical segments, one from each circle, so the total common area is 2 * [(π/2) * a^2 - a^2 / 2] = a^2 * (π/2 - 1).

Verification / Alternative check:
We can verify the formula using the general expression for the intersection area of two equal circles of radius r whose centres are distance d apart. In general, the common area is 2 r^2 cos⁻¹(d / (2r)) - (d / 2) * √(4r^2 - d^2). Here r = a and d = a√2. We get d / (2r) = (a√2) / (2a) = √2 / 2, so cos⁻¹(d / (2r)) is 45 degrees or π/4 radians. Substituting, the area becomes 2 a^2 * (π/4) - (a√2 / 2) * √(4a^2 - 2a^2). This simplifies to (a^2 π) / 2 - a^2, which is exactly a^2 * (π/2 - 1). This matches the earlier reasoning and confirms that the chosen formula is correct.

Why Other Options Are Wrong:
a^2 * (π/2 + 1) is too large because it uses a plus sign where a subtraction is required, effectively adding the triangular area instead of subtracting it.
a^2 is incorrect because it ignores the circular geometry and treats the region as equivalent to a simple square region of area a^2.
a^2 * (π - 1) overestimates the area, roughly corresponding to a sector of 180 degrees rather than 90 degrees in each circle.
a^2 * (π/4) underestimates the area by taking only half of the needed sector contribution and ignoring the triangular part. None of these match the exact lens area expression.

Common Pitfalls:
Learners often confuse the geometry and compute the area of a single sector instead of the overlapping region. Another frequent mistake is forgetting to subtract the triangular area from the sector to obtain the segment, which leads to overestimation. Some also forget that there are two identical segments contributing to the total common area. Finally, mixing up degrees and radians or misusing the diagonal length of the square can also lead to errors. Keeping the picture clear and working systematically with one circle at a time helps avoid these issues.

Final Answer:
The common area of the two sectors inside the square is a^2 * (π/2 - 1).