triangle mathematics?

Question

The altitude of a triangle is increasing at a rate of 2.5cm/min while the area of the triangle is increasing at a rate of 5.00 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8.00 centimeters and the area is 93 square cm?

Answer 1

1/2 * 2.5b = 5
1.25b = 5
b= 5/1.25
b=4
the base of the triangle is changing at a rate of 4cm/min

Answer 2

I have a problem with this question. No info is given at whether the height, base, and area are increasing linearly. If the height and base are increasing linearly, then the area will be increasing qualilaterally (2nd order).
With Non-sufficient info given, all claimed solutions are wrong, unless additional assumptions are added!

Answer 3

First, write an equation for the altitude as a function of time:

alt = (2.5cm/min)(t) + alt0

where alt0 is the initial altitude at t=0 (we don't know that value yet).

Likewise, for the area:

area = (5cm²/min)(t) + area0

Now, we are told that, at some time t1, the altitude is 8 and the area is 93:

(2.5cm/min)(t1) + alt0 = 8cm
(5cm²/min)(t1) + area0 = 93cm²

From this we can determine a relationship between alt0 and area0:

t1 = (8cm - alt0)/(2.5cm/min) = (93cm² - area0)/(5cm²/min)
from which:
area0 = 77cm² + alt0(2cm)

(We'll use that later).

The problem is concerned with the base; we can write a function for that by combining the previous two:

base = 2·area/alt = (2)((5cm²/min)(t) + area0) / ((2.5cm/min)(t) + alt0)

Now for the calculus. The rate at which the base is changing can be determined by taking the derivative:

(d/dt)base
= (2)((5cm²/min)(alt0) - (2.5cm/min)(area0)) / ((2.5cm/min)(t) + alt0)²
= ((10cm²/min)(alt0) - (5cm/min)(area0)) / alt²

Substituting the value we derived earlier for area0:

(d/dt)base = ((10cm²/min)(alt0) - (5cm/min)(77cm² + alt0(2cm))) / alt²
= ((10cm²/min)(alt0) - 385cm^3/min - alt0(10cm²/min)) / alt²
= –(385cm^3/min) / alt²

At time t1, the problem tells us that alt=8cm; so:

base′(t1) = –(385cm^3/min)/(8cm)² = –6.015625 cm/min

So, the base is actually shrinking at the indicated time.

Answer 4

a) i) reason ABCD is a sq.,so AB=BC=CD=DA, reason AB=AP+PB,BC=BQ+high quality administration, so AP+PB=BQ+high quality administration, reason AP=BQ, so PB = high quality administration. ii) reason ABCD is a sq.,so PBQ=QCR=ninety(good perspective) from i) PB = high quality administration, from the question, BQ=CR, so triangle BPQ is congruent to triangle CQR. iii) from ii) ,triangle BPQ is congruent to triangle CQR so perspective BQP=perspective CRQ, perspective BPQ=perspective CQR, so perspective BQP+perspective CQR=perspective BPQ+perspective CRQ, reason perspective BQP+perspective BPQ= a hundred and eighty-anglePBQ=a hundred and eighty-ninety=ninety, so angleBQP+perspective CQR=ninety. reason angleBQP+perspective CQR+perspective RQP=a hundred and eighty so perspective RQP=a hundred and eighty-angleBQP+perspective CQR=ninety.