You are constructing a concrete landing pad for a helicopter. You have 4704 cubic feet of concrete.
1 answer:
a) We have to calculate the height of the pad.
We know that the volume will be 4704 cubic feet.
We also know that the base of the pad is a square of side length 56 feet.
As the volume is the area of the base times the height, we can find the height by dividing the volume by the base area:
The height is 1.5 feet.
b) Now, we have a circular pad of which we know the height (1.5 feet) and the volume (4705 ft³). The area of the base will have to be the same as the are of the square: 3136 ft².
The area of this circular pad can be expressed as πD²/4, so we can calculate the diameter D as:
The diameter is approximately 63.21 feet.
Answer:
a) The pad will be 1.5 feet thick
b) The diameter will be 63.21 feet.
You might be interested in
Polar coordinates give the distance from the origin and the angle from the positive x axis. Cartesian coordinates give the distance from the x and y axes.
You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.
Let's convert (4, 16°) to Cartesian coordinates.
Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375
sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>
So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).
Let's convert (-2, 177°) to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)
In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).
Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951
sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248
So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).
Now we must use the distance formula with our two points.
You can draw a right triangle with these values. (see attached)
If you know the r value and theta of that triangle below, you can use trig to find x and y.
Let's convert (4, 16°) to Cartesian coordinates.
Note that since our angle is acute, (in Quadrant I) our sine and cosine will both be positive, as you should be able to derive from the unit circle, where cosine is represented as an x value and sine is represented as a y value.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(16°) = x/4
4cos(16°) = x ≈ 3.84504678375
sine = oppsite / hypotenuse
sinθ = y/r
sin(16°) = y/4
4sin(16°) = y ≈ 1.10254942327<span>
So (4, 16°) </span>⇒ (3.84504678375, 1.10254942327).
Let's convert (-2, 177°) to Cartesian coordinates.
Whenever you have a negative radius, that means to put the point opposite where it would have been if it had a positive radius. (see attached)
In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).
Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.
cosine = adjacent / hypotenuse
cosθ = x/r
cos(357°) = x/2
2cos(357°) = x ≈ 1.99725906951
sine = opposite / hypotenuse
sinθ = y/r
sin(357°) = y/2
2sin(357°) = y ≈ -0.10467191248
So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).
Now we must use the distance formula with our two points.