For this explanation, refer to the drawing at:

    http://aycu16.webshots.com/image/27615/2000693903621692109_rs.jpg

In order to solve for “x”, first solve for “z”, where:

    x = z + 4

Using the pythagorean theorem on the triangle:

    40^2 + z^2 = w^2

    z^2 = w^2 – 40^2

    z = sqrt(w^2 – 40^2)

Substituting this into the first equation:

    x = sqrt(w^2 – 40^2) + 4

Questions #5:

The main difficulty I had with this problem was

interpretting the description to create the drawing.

I assumed that the shoreline was perpendicular to

the line from the “boat” to point “A”. See:

    http://aycu13.webshots.com/image/27772/2000936854809150959_rs.jpg

If this is correct, then this reduces to solving a

right triangle, again using the Pythagorean theorem.

In this case:

    14^2 = 6^2 + Total^2

    160 = Total^2

    12.65 = Total

Refering to the sketch:

    d + x = Total

    d + x = 12.65

    x = 12.65 – d

Thus, the distance from point “P” to “L” is “12.65 -d”.

Question #6:

The “trick” to this problem is to discover the length of

the cylinder. We know that it is bounded by the sphere,

but it can’t be quite as long as the diameter of the sphere

(because in cross-section, we’re fitting a rectangle into

a circle). See the sketch at:

    http://aycu03.webshots.com/image/30362/2001583973896536371_rs.jpg

In this cross-section, the length of the rectangle (cylinder)

is 2*x, whereas the diameter of the circle (sphere) is 2*R.

Once again using the Pythagorean theorem:

    x^2 + r^2 = R^2

    x^2 = R^2 – r^2

    x = sqrt(R^2 – r^2)

Since the length of the cylinder is:

    length = 2 * x

    length = 2 * (sqrt(R^2 – r^2))

    length = 2 * sqrt(R^2 – r^2)

And the volume of a cylinder is:

    volume = (cross-section-area) * (length)

    volume = (pi * r^2) * (length)

    volume = (pi * r^2) * (2 * sqrt(R^2 – r^2))

    volume = pi * r^2 * 2 * sqrt(R^2 – r^2)

    volume = 2 * pi * r^2 * sqrt(R^2 – r^2)

Question #7:

Refer to the drawing at:

    http://aycu12.webshots.com/image/29411/2003520012729067389_rs.jpg

Since the original piece of wire was 32cm, we know

that the perimeter of the square and the rectangle

total to 32.

    perimeter of square = x + x + x + x

    perimeter of square = 4x

    perimeter of rectangle = y + (y-2) + y + (y-2)

    perimeter of rectangle = 4y + -4

    perimeter of square + perimeter of rectangle = 32

    4x + (4y – 4) = 32

    4x + 4y – 4 = 32

    4x + 4y = 36

    4y = 36 – 4x

    y = 9 – x

Use this value for “y” to derive an equation for the

sum of the area of the two shapes:

    Area = (area of the square) + (area of the rectangle)

    Area = x^2 + (y * (y-2))

    Area = x^2 + ((9-x) * ((9-x)-2))

    Area = x^2 + ((9-x) * (7-x))

    Area = x^2 + (63 – 7x – 9x + x^2)

    Area = 2x^2 -16x + 63

.