Lined canal with circular invert and straight flanks: If the straight sides of a lined canal meet a circular bottom (invert) of radius R and make an angle θ with the horizontal, the cross-sectional area A equals

Introduction / Context:Several lined canal sections use a circular invert to reduce seepage and erosion, with straight side slopes meeting tangentially. A compact formula for area is useful in hydraulic computations (discharge, velocity, and slope checks).

Given Data / Assumptions:

• Bottom is a circular arc of radius R.
• Straight sides meet the invert, making angle θ with the horizontal.
• Geometry forms a circular segment plus two triangular side portions expressed via θ.

Concept / Approach:Area is the sum of the circular segment area and the areas bounded by straight flanks up to intersection levels. For the common proportion used in design charts, the closed-form expression simplifies to A = R^2(θ + tan θ) (θ in radians).

Step-by-Step Solution:Express segment area in terms of R and θ.Express triangular contributions using slope angle θ.Combine to yield A = R^2(θ + tan θ).

Verification / Alternative check:Dimensional check: R^2 multiplies a dimensionless bracket, giving area units. For small θ, tan θ ≈ θ and area behaves as ~R^2(2θ), consistent with a narrow section.

Why Other Options Are Wrong:

• R(θ + …): Wrong dimensions.
• cot θ variants: Do not match this standard geometry.
• Duplicate option with R(θ + tan θ) is dimensionally incorrect.

Common Pitfalls:Using degrees instead of radians in computation; mixing geometric relations of circular segment with trapezoidal assumptions.

Final Answer:R^2(θ + tan θ)