---
title: "Most efficient sections for area 6 m²"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:efficient-sections-area-6-triangle-rect-trap"
source: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-sections-area-6-triangle-rect-trap
site: BCT Engineering
---

# Most efficient sections for area 6 m²

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2070 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-sections-area-6-triangle-rect-trap
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

The area of cross-section of flow in a channel is 6 m². Calculate the dimensions of the most efficient section if the channel is (a) triangular, (b) rectangular and (c) trapezoidal (2:1). Which has the least perimeter?

## Short answer

For the most efficient triangular section (side slope 1:1) with area 6 m²: depth 2.45 m, top width 4.90 m, perimeter 6.93 m. Rectangular (b=2y): depth 1.73 m, width 3.46 m, perimeter 6.93 m. Trapezoidal 2H:1V: depth 1.56 m, bed width 0.74 m, perimeter 7.70 m. Triangular and rectangular have the least perimeter (6.93 m).

## Answer

A most efficient section has the least wetted perimeter for a given area. Area $A=6$ m² in all cases.

#### (a) Triangular section
Best side slope is $z=1$ (45°): $A=y^2$, $P=2\sqrt2\,y$.
$$
y=\sqrt6=2.449\ \text{m},\qquad \text{top width}=2y=4.899\ \text{m},\qquad P=2\sqrt2(2.449)=6.928\ \text{m}
$$

#### (b) Rectangular section
$b=2y$, $A=2y^2$:
$$
y=\sqrt{\frac62}=1.732\ \text{m},\qquad b=2y=3.464\ \text{m},\qquad P=4y=6.928\ \text{m}
$$

#### (c) Trapezoidal section, side slope 2:1 (taken as 2H:1V, $z=2$)
$$
A=y^2(2\sqrt{1+z^2}-z)=y^2(2\sqrt5-2)=2.4721\,y^2
$$
$$
y=\sqrt{\frac{6}{2.4721}}=1.558\ \text{m},\quad b=2y(\sqrt5-2)=0.736\ \text{m}
$$
$$
P=b+2y\sqrt5=0.736+2(1.558)(2.2361)=7.703\ \text{m}\quad(\text{also }P=2A/y)
$$

#### Comparison
| Section | Depth (m) | Bed width (m) | Wetted perimeter (m) |
|---|---|---|---|
| (a) Triangular, 45° sides | 2.449 | 0 | 6.928 |
| (b) Rectangular | 1.732 | 3.464 | 6.928 |
| (c) Trapezoidal 2H:1V | 1.558 | 0.736 | 7.703 |

**Answer: the triangular and the rectangular sections both have the least perimeter (6.93 m); the 2H:1V trapezoid has a larger perimeter (7.70 m).**

Note: if "2:1" is read as 2V:1H ($z=0.5$), the trapezoid has $y=1.859$ m, $b=2.298$ m and $P=6.455$ m, which is the least of the three. The perimeter is smallest when the side slope is close to the half-hexagon value $z=0.577$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The most efficient triangular section has side slope:
   A. z = 0.5
   B. z = 2
   C. z = 1 (45 degrees), giving A = y^2 and P = 2 sqrt2 y
   D. z = 0.577
   Answer: C. A triangle has least perimeter for given area at z = 1. (Key point: Efficient triangle)

2. For A = 6 m2 the most efficient rectangle has:
   A. b = y = 2.45 m
   B. b = 3y
   C. b = 2y = 3.46 m, y = 1.73 m, P = 6.93 m
   D. b = 2y = 4.9 m, y = 2.45 m
   Answer: C. A = 2y^2 = 6 gives y = 1.732 m, b = 3.464 m, P = 4y = 6.93 m. (Key point: Efficient rectangle)

3. For the trapezoid with z = 2, which relation gives the depth?
   A. A = y^2(2 sqrt(1+z^2) - z) = 2.4721 y^2, y = 1.56 m
   B. A = 2y^2, y = 1.73 m
   C. A = y^2 z, y = 1.73 m
   D. A = 3y^2, y = 1.41 m
   Answer: A. The economical trapezoid has A = 2.4721 y^2, so y = 1.558 m with P = 7.70 m. (Key point: Trapezoid 2H:1V)

4. Which statement matches the answer?
   A. The 2H:1V trapezoid has the least perimeter
   B. Triangle and rectangle tie at 6.93 m; if 2:1 meant 2V:1H (z = 0.5) the trapezoid would be the least at 6.46 m, since z is near 0.577
   C. The rectangle is always the least
   D. The triangle has the greatest perimeter
   Answer: B. With z = 2 the trapezoid has P = 7.70 m, but with z = 0.5 it is close to the half-hexagon slope and has P = 6.455 m. (Key point: Least perimeter and reading of slope)

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