Question

Altitude of a triangle increases at 2 cm/min. Its area increases at the rate 5 cm²/min. Find the rate of change of length of base when the altitude is 10 cm and the area is 100 cm².

Answer 1

we know,

area of triangle = 1/2 × altitude × base

given, area of triangle, A = 100 cm²

altitude , a = 10cm

then, base , b = ?

so, A = 1/2 ab

⇒100 cm² = 1/2 × 10cm × b

⇒100 cm² = 5cm × b

⇒b = 20cm

now, area of triangle , A = 1/2 ab

differentiating both sides with respect to time,

i.e., dA/dt = 1/2 [a (db/dt) + b (da/dt)]

given, dA/dt = 5cm²/min , da/dt = 2cm/min

so, 5 = 1/2 [10 × (db/dt) + 20 × 2]

⇒10 = 10 × (db/dt) + 40

⇒10 - 40 = 10 × (db/dt)

⇒-30 = 10 × (db/dt)

⇒db/dt = -3 cm/min

hence, base is decreasing at the rate of 3 cm/min.