Two equal circles intersect in such a way that their centres and their two points of intersection form the vertices of a square of side 1 cm. What is the area (in sq.cm) of the region that is common to both circles?

Introduction / Context:
This problem concerns the intersection area (lens-shaped region) of two equal circles. The configuration is special: the centres of the circles and the two intersection points together form the four vertices of a square of side 1 cm. We must find the area common to both circles using this geometric information. This involves understanding the relationship between the distance between the centres, the radius of each circle, and the area of overlap of two equal circles.

Given Data / Assumptions:

• Two circles have equal radii (say radius R).
• The centres of the circles and the two intersection points form a square of side 1 cm.
• All sides of this square measure 1 cm.
• The distance between the centres of the circles equals the length of one side of the square or its diagonal, depending on which vertices correspond to the centres.
• We must find the area common to both circles in square centimetres.

Concept / Approach:
In the described configuration, it turns out that the centres of the circles lie at opposite corners of the square and the intersection points lie at the other two corners. This means that the distance between the centres equals the diagonal of the square, which is side * √2 = 1 * √2 = √2. For two equal circles of radius R that intersect, the distance between centres is usually denoted by d. Here d = √2. Geometry of the square implies that the radius of each circle is R = 1. Once R and d are known, we use the standard formula for the overlap area of two equal circles of radius R with centre distance d.

Step-by-Step Solution:
Let R be the radius of each circle and d be the distance between the centres.Since the centres and intersection points form a square of side 1 cm, the distance between opposite vertices (a diagonal) is 1 * √2 = √2.The centres occupy opposite corners, so d = √2.In such a configuration, geometry shows that the radius R equals 1 cm (so that the intersection points also lie at distance 1 from each centre).The area of intersection of two equal circles of radius R with centre distance d is:Area = 2R^2 * cos^−1(d / (2R)) − (d / 2) * √(4R^2 − d^2).Here R = 1 and d = √2.Compute d / (2R) = √2 / 2 = 1 / √2.cos^−1(1 / √2) = 45° or π / 4 radians.Thus 2R^2 * cos^−1(d / (2R)) = 2 * 1^2 * (π / 4) = π / 2.Now compute the second term: (d / 2) * √(4R^2 − d^2).4R^2 − d^2 = 4 * 1^2 − (√2)^2 = 4 − 2 = 2.So √(4R^2 − d^2) = √2.Therefore (d / 2) * √(4R^2 − d^2) = (√2 / 2) * √2 = (√2 / 2) * √2 = 1.Hence the common area = (π / 2) − 1 square centimetres.

Verification / Alternative check:
We can reason qualitatively as well. Since R = 1, each circle has area π. The overlap region is less than half of either circle, and π / 2 − 1 is indeed less than π / 2. Numerically, π / 2 ≈ 1.57, so π / 2 − 1 ≈ 0.57 square centimetres, which seems reasonable for a modest overlapping lens from unit circles with centre distance √2. This rough magnitude check supports the correctness of the detailed formula-based calculation.

Why Other Options Are Wrong:
π/4 is roughly 0.785, which is larger than the derived overlap and does not match the standard overlap formula for circles at this distance. π/5 is even smaller and does not follow from any standard expression here. The value √2 − 1 is approximately 0.414, which is unrelated to the circle overlap formula. The option π − 1 would give an area close to 2.14, which is much larger than half of a unit circle and impossible for the overlap of two such circles. Only π/2 − 1 fits the exact expression derived from the known formula for the intersection of two equal circles with centre distance √2 and radius 1.

Common Pitfalls:
Many students are unfamiliar with the exact formula for the area of overlap of two circles and may try to approximate or guess the answer. Others might misinterpret the square configuration and incorrectly set the distance between centres equal to 1 instead of √2. Mistakes also arise from forgetting that the radius is 1, not 1/2 or another value. Careful geometric visualization of the square and consistent use of the standard formula are essential to solve this correctly.

Final Answer:
The area of the region common to both circles is π/2 − 1 square centimetres.