Question

Area of a triangle is increasing at a rate of 4 cm²/sec and its altitude is increasing at a rate of 2 cm/sec. At what rave is the length of the base of the triangle changing, when its altitude is 20 cm and area is 30 cm² ?

Answer 1

we know,

area of triangle = 1/2 × altitude × base

given, area of triangle, A = 30 cm²

altitude , a = 20cm

then, base , b = ?

so, A = 1/2 ab

⇒30 cm² = 1/2 × 20cm × b

⇒60 cm² = 20cm × b

⇒b = 3cm

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 = 4cm²/s , da/dt = 2cm/s

so, 4 = 1/2 [20 × (db/dt) + 3 × 2 ]

⇒8 = 20 × (db/dt) + 6

⇒8 - 6 = 20 × (db/dt)

⇒2 = 20 × (db/dt)

⇒db/dt = 1/10 cm/s = 0.1 cm/s

hence, base is increasing at the rate of 0.1 cm/s