An equilateral triangle has side length 8 cm. Find the area of the region that lies between its circumcircle and its incircle (use π = 22/7 where needed).

Introduction:
This problem asks you to find the area of an annular region formed between the incircle and circumcircle of an equilateral triangle. It tests knowledge of standard formulas for the inradius and circumradius of an equilateral triangle, as well as the area formula for a circular ring (annulus).

Given Data / Assumptions:

• Triangle is equilateral with side length a = 8 cm.
• We consider its circumcircle (radius R) and incircle (radius r).
• We want the area of the region between these two circles.
• Use π = 22/7 for numerical evaluation.

Concept / Approach:
For an equilateral triangle of side a:
Circumradius R = a / √3.Inradius r = a / (2√3).The area between the circumcircle and incircle is the difference of their areas: Area = πR² − πr² = π(R² − r²). We first compute R and r in terms of a, then find R² − r², and finally substitute a = 8 cm and approximate using π = 22/7.

Step-by-Step Solution:
Step 1: Use formulas for R and r.R = a / √3 = 8 / √3.r = a / (2√3) = 8 / (2√3) = 4 / √3.Step 2: Compute R² and r².R² = (8 / √3)² = 64 / 3.r² = (4 / √3)² = 16 / 3.R² − r² = (64 / 3) − (16 / 3) = 48 / 3 = 16.Step 3: Area between the two circles.Area = π(R² − r²) = π * 16 = 16π square centimetres.Step 4: Substitute π = 22/7.Area = 16 * (22/7) = 352/7.352 ÷ 7 = 50 with remainder 2, so 352/7 = 50 2/7 sq. cm.

Verification / Alternative check:
We can also compute actual numeric values: R ≈ 8 / 1.732 ≈ 4.618 and r ≈ 4 / 1.732 ≈ 2.309. Then R² ≈ 21.33 and r² ≈ 5.33, so the difference ≈ 16, and 16π ≈ 50.27, which aligns well with 50 2/7 ≈ 50.2857, confirming the calculation when π is approximated by 22/7.

Why Other Options Are Wrong:
50 1/7, 75 1/7 and 75 2/7 correspond to different multiples of π and do not match 16π when computed with π = 22/7. The value 64 sq. cm is unrelated to the circular areas and is closer to the square of the side length. Only 50 2/7 sq. cm is equal to 16 * (22/7) and correctly represents the annular area between circumcircle and incircle.

Common Pitfalls:
Students often mix up the formulas for R and r or forget that the area difference uses R² − r², not R − r. Another error is to use approximate decimal values for π inconsistently while the question asks specifically for π = 22/7. Staying systematic with the formulas avoids these mistakes.

Final Answer:
The area of the region between the circumcircle and incircle is 50 2/7 sq. cm.