I'm not sure if this is the easiest way of doing this, but it surely work.
Let the base of the triangle be AB, and let CH be the height. Just for reference, we have
Moreover, let CH=y and BC=z
Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system
If we solve the first two equations for y squared, we have
And we can deduce
So that the third equation becomes
(we can't accept the negative root because negative lengths make no sense)
The area of the triangle is
A = (xy)/2
Also,
sqrt(x^2 + y^2) = 19
We solve this for y.
x^2 + y^2 = 361
y^2 = 361 - x^2
y = sqrt(361 - x^2)
Now we substitute this expression for y in the area equation.
A = (1/2)(x)(sqrt(361 - x^2))
A = (1/2)(x)(361 - x^2)^(1/2)
We take the derivative of A with respect to x.
dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]
dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]
dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]
Now we set the derivative equal to zero.
(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0
-2x^2 + 361 = 0
-2x^2 = -361
2x^2 = 361
x^2 = 361/2
x = 19/sqrt(2)
x^2 + y^2 = 361
(19/sqrt(2))^2 + y^2 = 361
361/2 + y^2 = 361
y^2 = 361/2
y = 19/sqrt(2)
We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
Answer:
12
Step-by-step explanation:
"ba" means "b * a", where * is multiplication. So, we want to find the product of a and b.
We know a = 3 and b = 4, so we just plug these in: 3 * 4 = 12.
Thus, ba = 12.
Hope this helps!
Answer:
314 inches³
Step-by-step explanation:
The area of a cone is πr²h/3. We need to find the height and radius of the cone. The height is 12 inches. radius * 2 = diameter.
Radius * 2 = Diameter
/2 /2
Radius = 1/2* Diameter
Radius = 5 inches
We substitute it in:
3.14 * 5² * 12/3:
3.14*25*4
314 inches³
