Spherical excess in geodetic surveying – angular sum difference on the Earth’s surface The sum of the interior angles of a polygon laid on the Earth’s surface exceeds that of the corresponding plane figure by about 1 second for approximately how much area?

Introduction / Context:
On a sphere, the sum of the interior angles of a polygon is greater than on a plane. This excess—called spherical excess—becomes relevant in geodetic surveying, triangulation, and large-area mapping. The question asks for the approximate area on Earth that corresponds to a spherical excess of 1 second of arc in angle sum.

Given Data / Assumptions:

• Spherical excess E (in radians) ≈ Area / R^2 for small regions.
• R is Earth’s mean radius, about 6370 km.
• We want E expressed as 1 second of arc (1'), and the corresponding area in sq. km.

Concept / Approach:

Convert 1 second to radians, then solve Area = E * R^2. One second = 1/206265 of a radian. Using R ≈ 6370 km gives R^2 ≈ 4.06 * 10^7 km^2. Multiplying yields Area ≈ (1/206265) * 4.06 * 10^7 ≈ 197 sq. km, commonly rounded to about 200 sq. km in surveying handbooks. Thus, a spherical polygon of roughly 200 sq. km has an angle-sum excess of about 1' compared to the plane case.

Step-by-Step Solution:

Verification / Alternative check:

Many surveying texts quote 1' ≈ 197–200 sq. km for Earth. Using more precise R marginally changes the value but not the chosen option.

Why Other Options Are Wrong:

100 or 150 sq. km: Underestimates the area corresponding to 1' of spherical excess.

None of these: Incorrect because ≈ 200 sq. km is a standard approximation.

Common Pitfalls:

Confusing seconds with degrees; using diameter instead of radius; forgetting to express E in radians before multiplying by R^2.

Final Answer:

200 sq. km