Circle inscribed in a quadrant of radius 7 cm: In quadrant PQR with radius 7 cm, a circle is inscribed touching both radii and the arc. Find the exact area (in sq cm) of the circle.

Introduction / Context:An inscribed circle in a quadrant (a 90° sector) is tangent to the two perpendicular radii and the circular arc. The circle’s center lies on the 45° line from the corner, giving a simple relationship between the big radius R and the small radius r.

Given Data / Assumptions:

• Quadrant radius R = 7 cm.
• Inscribed circle radius r satisfies distance from corner: √(r^2 + r^2) = R − r.
• Use π = 22/7 to match the answer format without π.

Concept / Approach:From √2 * r = R − r ⇒ r(√2 + 1) = R ⇒ r = R(√2 − 1). Then area = π r^2 = π * R^2 * ( (√2 − 1)^2 ) = π * R^2 * (3 − 2√2 ). Substitute R = 7 and π = 22/7 and simplify.

Step-by-Step Solution:

Verification / Alternative check:Numeric check with √2 ≈ 1.414: r ≈ 7(0.414) ≈ 2.898; area ≈ 3.1416 * (2.898)^2 ≈ 26.4, and 462 − 308√2 ≈ 26.4, consistent.

Why Other Options Are Wrong:Other expressions arise from incorrect placement of r or from using (R − r)^2 = r^2 instead of 2r^2.

Common Pitfalls:Forgetting that the center-to-corner distance is along 45°, giving √2r, not r; using sector area instead of the circle’s area.

Final Answer:462-308√2