The altitude of a triangle is five-thirds the length of the corresponding base. If the altitude is increased by 4cm and the base is decreased by 2cm, the area of the triangle remains the same. Find the altitude and the base of the triangle.

Let the base = b cm and the altitude = h cm.

Given:

h = (5/3) × b

Area of a triangle:

Area = (1/2) × base × height = (1/2) × b × h

Condition: If altitude increases by 4 cm and base decreases by 2 cm, area remains the same:

(1/2) × (b - 2) × (h + 4) = (1/2) × b × h

Multiply both sides by 2:

(b - 2)(h + 4) = b × h

Substitute h = (5/3) b:

(b - 2)((5/3)b + 4) = b × (5/3)b

Expand LHS:

(b - 2) × (5b/3) + (b - 2) × 4 = 5b²/3

Simplify:

(5b² - 10b)/3 + 4b - 8 = 5b²/3

Combine terms:

(5b² - 10b + 12b - 24)/3 = 5b²/3

Simplify further:

2b - 24 = 0

Solving for b:

2b = 24 ⇒ b = 12

Altitude:

h = (5/3) × 12 = 20

Answer:

Base = 12 cm, Altitude = 20 cm