Which of the following must be true about the area A of the triangle?

  • Thread starter Thread starter lude1
  • Start date Start date
  • Tags Tags
    Area Triangle
lude1
Messages
34
Reaction score
0

Homework Statement



If the base b of a triangle is increasing at a rate of 3 inches per minute while its height h is decreasing at a rate of 3 inches per minute, which of the following must be true about the area A of the triangle?

a. A is always increasing
b. A is always decreasing
c. A is decreasing only when b < h
d. A is decreasing only when b > h
e. A remains constant

Correct answer is d. A is decreasing only when b > h

Homework Equations





The Attempt at a Solution



I don't understand why the answer is d. If the area of a triangle is (1/2)(base)(height) and the base increases by 3 while the height decreases by 3, wouldn't they just cancel out each other?

Moreover, I don't understand why the area would only be decreasing when the base is bigger than the height and not vice versa. But I guess I won't understand this part until I understand why the area can't be constant.
 
Physics news on Phys.org
lude1 said:

Homework Statement



If the base b of a triangle is increasing at a rate of 3 inches per minute while its height h is decreasing at a rate of 3 inches per minute, which of the following must be true about the area A of the triangle?

a. A is always increasing
b. A is always decreasing
c. A is decreasing only when b < h
d. A is decreasing only when b > h
e. A remains constant

Correct answer is d. A is decreasing only when b > h

Homework Equations





The Attempt at a Solution



I don't understand why the answer is d. If the area of a triangle is (1/2)(base)(height) and the base increases by 3 while the height decreases by 3, wouldn't they just cancel out each other?

Moreover, I don't understand why the area would only be decreasing when the base is bigger than the height and not vice versa. But I guess I won't understand this part until I understand why the area can't be constant.

You need to find the dA/dt. Since A is a function of both h and b, and both can be assumed to be functions of t, dA/dt will involve the partial derivatives with respect to h and the partial derivative with respect to b.
 
To see why the area wouldn't be constant, take a triangle with height a and base b. Then its area would be ab/2, right? Now suppose that a minute passes. So now it has base b+3 and height a-3.

Find the area of the new triangle and figure out when it's larger or smaller than the other one.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top