The altitude (i. e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 95 square centimeters? The base is changing at cm/min.

Question

Terrance Patrick

Mathematics

7 June, 12:37

The altitude (i. e., height) of a triangle is increasing at a rate of 1.5 cm/minute while the area of the triangle is increasing at a rate of 4.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 95 square centimeters? The base is changing at cm/min.

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Answers (1)

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Coffey · Answer 1

Step-by-step explanation:

at time = 0min,

height, h0 = 10.5cm

area, a0 = 95cmsq

base, b0 = a0 x2/h0

=> b0 = 95 x2 / 10.5 = 18.1cm

at time = 1 min,

increase of height, rh = 1.5cm/min

height at 1 min, h1 = h0 x rh

=> h1 = 10.5 * 1.5 = 15.75cm

increase of area, ra = 4.5cmsq/min

area after 1 min, a1 = a0 x ra

=> a1 = 95 x 4.5 = 427.5cm/sq

base at 1 min, b1 = a1x2/h1

=> b1 = 427.5 x 2 / 15.75 = 54.3 cm

rate of increase for base, rb = b1/b2

=> rb = 54.3/18.1 = 3cm/min