Introduction:
To find the length of the canvas required to make a conical tent, we need to consider the dimensions of the tent and account for any additional material needed for folds and stitching. In this case, the tent has a diameter of 20m and a slant height of 42m. We will calculate the length of the canvas required by applying the formula for the lateral surface area of a cone.

Calculating the slant height:
The slant height is the distance from the apex of the cone to any point on the circumference of the base. It can be calculated using the Pythagorean theorem:
b² = h² + r²
Where b is the slant height, h is the height of the cone, and r is the radius of the base. In this case, the radius is half the diameter, so r = 10m.

Using the given slant height of 42m, we can calculate the height of the cone:
42² = h² + 10²
h² = 42² - 10²
h² = 1764 - 100
h² = 1664
h ≈ 40.8m

Calculating the lateral surface area:
The lateral surface area of a cone can be calculated using the formula:
L = πr * b
Where L is the lateral surface area, r is the radius of the base, and b is the slant height.

In this case, the radius is 10m and the slant height is 42m. Plugging these values into the formula, we get:
L = π * 10 * 42
L ≈ 1324.8m²

Accounting for folds and stitching:
To allow for folds and stitching, we need to add an additional percentage to the calculated lateral surface area. In this case, we are allowing 10% for folds and stitching.

Adding 10% to the lateral surface area:
10% of 1324.8m² = 0.1 * 1324.8m² = 132.48m²

Adding the additional area to the lateral surface area:
1324.8m² + 132.48m² = 1457.28m²

Calculating the length of the canvas:
The length of the canvas required would be equal to the lateral surface area of the cone, accounting for folds and stitching.

Since the lateral surface area is equal to the length of the canvas, the length required would be approximately 1457.28m.

Conclusion:
To make a conical tent with a diameter of 20m and a slant height of 42m, allowing 10% for folds and stitching, a length of approximately 1457.28m of canvas would be required.