Triangle ABC has side lengths AB = 5 cm, AC = √41 cm and BC = 8 cm. AD is drawn perpendicular to BC from vertex A. What is the area, in square centimetres, of triangle ABD?

Introduction / Context:
This question combines coordinate geometry thinking with classical triangle area formulae. You are given all three side lengths of a triangle and an altitude from one vertex to the opposite side. You must find the area of just one of the smaller triangles formed by this altitude. The problem tests understanding of how altitudes split the base and how area is computed using base and height.

Given Data / Assumptions:

• Triangle ABC has sides AB = 5 cm, AC = √41 cm, and BC = 8 cm.
• AD is the altitude from A to side BC, so AD is perpendicular to BC.
• D lies on BC, and we consider triangle ABD.
• We must find the area of triangle ABD.
• The triangle is non-degenerate and lies in a Euclidean plane.

Concept / Approach:
We can place the triangle on a coordinate system to find the height AD easily. Let BC be on the x-axis with B at (0, 0) and C at (8, 0). Then the coordinates of A = (x, y) must satisfy the distance conditions AB = 5 and AC = √41. Solving these gives the height y, which equals the length of AD. Once we know the height, the area of triangle ABD is given by one-half the product of base BD and the height AD.

Step-by-Step Solution:
Step 1: Place B at (0, 0) and C at (8, 0). Step 2: Let A have coordinates (x, y). Step 3: From AB = 5, we get: x² + y² = 5² = 25. Step 4: From AC = √41, we get: (x - 8)² + y² = 41. Step 5: Subtract the first equation from the second: (x - 8)² - x² = 41 - 25 = 16. Step 6: Expand (x - 8)² = x² - 16x + 64, so x² - 16x + 64 - x² = 16. Step 7: Simplify to get -16x + 64 = 16 ⇒ -16x = -48 ⇒ x = 3. Step 8: Substitute x = 3 into x² + y² = 25: 9 + y² = 25 ⇒ y² = 16 ⇒ y = ±4. Step 9: The altitude length AD is the absolute value of the y-coordinate, so AD = 4 cm. Step 10: D lies on BC at the vertical projection of A, so D has coordinates (3, 0). This means BD = 3 cm and DC = 5 cm. Step 11: Area of triangle ABD is (1 / 2) * base * height = (1 / 2) * BD * AD = (1 / 2) * 3 * 4 = 6 square centimetres.

Verification / Alternative check:
We can verify by computing the total area of triangle ABC using base BC = 8 cm and height AD = 4 cm. The total area is (1 / 2) * 8 * 4 = 16 square centimetres. The segment BD is 3 cm, so area of ABD is (3 / 8) times the total area, because it uses base 3 instead of 8 with the same height. (3 / 8) * 16 = 6 square centimetres, which matches the value we found earlier. This consistency confirms our answer.

Why Other Options Are Wrong:
An area of 12 or 20 square centimetres is too large considering that the entire triangle only has area 16 square centimetres. An area of 10 square centimetres would again exceed the total area proportionally when compared with the base segment BD = 3 cm on a base of length 8 cm. The option 8 square centimetres would represent half the total area, which would require BD = 4 cm, not the 3 cm derived from the coordinate solution. Only 6 square centimetres fits all the geometric relationships exactly.

Common Pitfalls:
Students sometimes incorrectly assume that BD is half of BC or make guesses about its length without using the distance conditions. Another error is to misapply the distance formula or to forget that both equations for AB and AC must be satisfied simultaneously. Some also compute the total area correctly but then divide it into two smaller triangle areas in the wrong ratio. A systematic coordinate approach avoids guesswork and leads to the correct values.

Final Answer:
The area of triangle ABD is 6 square centimetres.