The diagonal of a rectangle is 41 cm and its area is 20 sq cm. Based on these two values, what must be the perimeter of the rectangle (approximately, in cm)?

Introduction:
This question relates diagonal, area, and perimeter of a rectangle. Let the sides be a and b. We know a^2 + b^2 = d^2 from the diagonal, and ab = area from the given area. The perimeter is 2(a+b), so we need a+b. A powerful identity connects these: (a+b)^2 = a^2 + b^2 + 2ab. Using this, we can compute a+b directly without solving for a and b individually, and then compute the perimeter. Because the perimeter here is not an integer from exact radicals, the result is best stated approximately.

Given Data / Assumptions:

• Diagonal d = 41 cm• Area ab = 20 sq cm• Let sides be a and b (in cm)

Concept / Approach:
Use the identity (a+b)^2 = a^2 + b^2 + 2ab. Here, a^2 + b^2 = d^2 = 41^2. Also ab = 20, so 2ab = 40. Then find a+b by taking square root. Finally perimeter P = 2(a+b).

Step-by-Step Solution:
Step 1: Use diagonal relation.a^2 + b^2 = 41^2 = 1681Step 2: Use area relation.ab = 20So 2ab = 40Step 3: Compute (a+b)^2 using the identity.(a+b)^2 = (a^2 + b^2) + 2ab = 1681 + 40 = 1721Step 4: Find a+b.a + b = sqrt(1721) ≈ 41.48Step 5: Compute perimeter.Perimeter = 2(a+b) ≈ 2 * 41.48 = 82.96 cmSo perimeter is approximately 83 cm

Verification / Alternative check:
The diagonal 41 cm is extremely large compared to area 20 sq cm, so the rectangle must be very long and very narrow (one side large, the other tiny). That makes a+b close to 41, and perimeter close to 82, which aligns with our computed 82.96. This logical check supports the approximation 83 cm.

Why Other Options Are Wrong:
80 cm is too low; 2(a+b) must be close to 83 because a+b is about 41.5.82 cm underestimates the computed 82.96 significantly for a nearest-integer style answer.85 cm and 86 cm are too high compared to the calculated perimeter.

Common Pitfalls:
• Trying to find a and b directly and getting stuck in a quadratic.• Using a+b = diagonal, which is incorrect.• Forgetting to multiply by 2 at the end to get the perimeter.

Final Answer:
The perimeter of the rectangle is approximately 83 cm.