CPhill

We need to write the equation of a plane, here, Julius.....

Since the prism is bisected  by the xy plane,

Let D  = (0, 0 ,-3 )

Let  A  = (0, 0, 3)

Let E  = (4, 0,3)

Let G   =(4,0,-3)

We can form the following  vectors  DG  = (4,0, 0)    and  DE  = (4, 0, 6)

[These aren't the only two possibilities......we just need any two vectors formed by the points]

Note that all the points lie in the xz  plane....so...we need a vector that is orthagonal to this plane

The cross-product of the two vectors will be orthagonal to both of these vectors...so we have

                         i          j           k       i         j

DG X DE   =    4         0          0       4        0

                        4        0           6       4        0       

 ( [ 0 * 6]i   + [ 0 * 4 ] j  + [4 * 0] k   -  (  [ 4 * 0 ] k  + [ 0 *0] i  +  [ 6 * 4]j )   =

0i   + 0 j   +  0 k   - 0 k - 0i  - 24 j

So....we have

0i  -24j - 0k       =  { 0, -24, 0 }    this is the vector  that is orthagonal to both vectors ....[and hence,  the plane containing those vectors]

And  the equation of the plane becomes

0 (x - 4) -24 (y - 0)  + 0 (z - 3)   =  0

-24y    =  0 

4.  We can use the secant-tangent theorem to solve this, ACG...

We have something like this  (not to scale) :

PA * PB  = PT^2

PA * PB  = (AB - PA)^2

PA * ( AB + PA)  = AB^2 - 2PA*AB + PA^2

PA*AB + PA^2 = AB^2  - 2PA*AB + PA^2

3*AB + 3^2  = AB^2  - 2*3*AB  + 3^2

3AB + 9  = AB^2 - 6AB + 9

3AB  = AB^2 - 6AB     rearrange as

AB^2  - 9AB  = 0       factor

AB( AB - 9)  =  0

Seting each factor to 0 and solving for AB  we have that either

AB  = 0  (reject)    or  AB  = 9   (accept)

So.....PB  = AB + PA

PB  = 9 + 3

PB  = 12

Looking at the triangle on the left, we can write its base as 50 - x

And, using the Pythagorean Theorem , we can solve this :

(50 - x)^2  + 30^2  =  x^2       simplify

x^2 - 100x + 2500 + 900 = x^2       subtract x^2  from both sides

-100x  +  3400  =  0

3400 = 100x       divide both sides by 100

34  =  x  =  AB

The midpoint  of this segment is  ( 3, 5)

And the slope between the two segment endpoints  is  [7 - 3] / [ 5 - 1]   = 4/4  = 1

So.......the perpedicular bisector  will have a negative reciprocal slope  = -1

And the equation of this bisector is given  by :

y = -1(x - 3) + 5

y = -x + 3 + 5

y = -x + 8

So

x + y  =  8     

And  "b"  =  8

2.

We can find angle BAC as follows

6^2  = 2(5)^2  - 2(5^2) cos(BAC)

[36 - 50]  / (-50)  = cos(BAC)

-14/-50  = cos(BAC)

7/25 = cos (BAC)

arccos ( 7/25) = BAC

And since BAC is an inscribed angle, the BOC has twice its measure =  2arccos(7/25)

And   cos [ 2 arccos (7/25) ]  =    - 527/ 625

And OC  = OB

So......using the Law of Cosines

6^2  = 2(OC)^2 - 2(OC)^2 cos [2arccos(7/25) ]

36 = 2(OC)^2 - 2(OC)^2 [ -527/ 625]

36 = 2(OC)^2 + 2(OC)^2 [ 527/625]

18 = OC^2 [ 1 + 527/625]

18 = OC^2 [ 1152/ 625]

18 * 625 / 1152  = OC^2  =  11250/ 1152  =  625/ 64

So.....the area of triangle OBC  =

(1/2) OC^2 * sin (BOC)

(1/2) (625/64) * sin [ 2 arccos(7/25) ]  = 

(625/ 128) * sin [ 2 arccos (7/25) ]  =

(625 / 128) * (336/625) =

336/128    =

( 21 /  8  )   units^2

"C"  is correct

We have a cone topped by a hemisphere.....the total volume is

pi * r^2 * h / 3   +  (2/3) pi*r^3   =

pi  [ 4^2 * 10 / 3   + (2/3)* 4^3 ]  =

pi  [ 160/ 3  + 128/3 ] =

pi [ 288/3 ]  =

96 pi   cm^3

Volume of a sphere  =

(4/3) pi * radius^3  =

(4/3) * 3.14 * 5^3  ≈

523 cm^3

We need to find the radius of the sphere

Surface area  =  4 * pi * r^2

113.04  = 4 * pi * r^2       divide both sides by 4 pi

113.04 / [ 4 pi]  =  r^2

√ [ 113.04/ [ 4pi] ]  = r

So...the volume is given by

(4/3) pi *  r^3

(4/3)pi * (√ [ 113.04/ [ 4pi] ])^3  =  

35.97 pi   in^3

The circumference  of Cylinder A   = 3m

So...the radius  is given as

(3) / (2 * 3.14)   = r

The circumference of Cylinder B  = 5m

So..... the radius is given as

(5) / (2 * 3.14)  = r

So.....the volume of Cylinder A   =

(3.14) * [ 3 / (2*3.14)]^2  * 5  =  3.58 m^3

The volume of Cylinder B  =

(3.14) [ (5) / (2 * 3.14) ]^2 * 3  =  5.97 m^3

Cylinder B  has the greater volume