In triangle ABC, sides are BC = 5 cm, AC = 12 cm, AB = 13 cm (a 5–12–13 right triangle). Find the length of the altitude drawn from B to AC.

Introduction / Context:
The sides 5, 12, and 13 form a classic right triangle with hypotenuse 13. Recognizing the right angle at C (since 5^2 + 12^2 = 13^2) allows us to equate the area computed via the two legs to the area computed using the altitude from B to the leg AC.

Given Data / Assumptions:

• AB = 13 cm, AC = 12 cm, BC = 5 cm
• Right angle at C (5–12–13 triangle)
• Altitude from B to AC has length h

Concept / Approach:
Compute area as (1/2) * AC * BC. Also, area = (1/2) * (altitude from B to AC) * AC. Equate the two expressions to solve for the altitude h.

Step-by-Step Solution:
Area via legs: (1/2) * 12 * 5 = 30 cm^2Area via altitude: (1/2) * h * 12 = 6hSet equal: 6h = 30 ⇒ h = 5 cm

Verification / Alternative check:
The altitude from the vertex on the hypotenuse to the opposite leg equals the other leg in a right triangle when comparing areas as above; here, h matches BC numerically, which is consistent with the computation.

Why Other Options Are Wrong:
4, 6, and 7 cm produce areas 24, 36, and 42 cm^2 via (1/2)*h*12, none of which match the true area 30 cm^2 derived from leg multiplication.

Common Pitfalls:
Dropping the 1/2 factor or using AB (hypotenuse) incorrectly in the area formula. The area based on legs is the most straightforward path here.

Final Answer:
5 cm