CPhill

I'm assuming this is supposed to be :

S=12pi(3V/[16pi])^(2/3)

(A)  exponentiate both sides to the 3/2 power

S^(3/2)  = 12pi ( 3V/[ 16 pi])

S^(3/2)  = (36/16) V

S^(3/2)  = ( 9/4)V

S^(3/2)  = 2.25V

[ S^(3/2)] / 2.25   = V

(1 / 2.25)S^(3/2)   = V

[0.44]S^(3/2)  = V       [the 0.44 is the "rounded" coefficient]

(B)  12pi(3[150]/[16pi])^(2/3)  = about 162.54  sq ft

BTW - this is correct based on my interpretation of the formula......if the "pi"  in the parentheses in the formula was not supposed to be in the denominator, a clearer  form would have been  :

S = S=12pi[ (3/16)Vpi ] ^(2/3).........I did not interpret it in that way.......I interpreted "16 pi" as the  whole denominator

1/2 of the area  = 120

1/2  of the base  = 10     ......so.........

A = (1/2)BH

120 = (1/2)(10)(H)

120/5  = H   =  24

And one of the equal sides  = sqrt [ H^2 + (1/2 of the base)^2 ] =  sqrt [24^2 + 10^2]  = sqrt(676)   = 26

So....the prerimeter=   2*26 + 20  = 72 units

Let's pretend that the legs are  264 and 495 units in length

264   factors as  11 * 24  ....so  2.64 =  11 * .24

And

495  factors as  11 * 45........so   4.95  = 11 * .45

And a Pythagorean Triple triangle of  8 - 15 - 17 ...... exists....so.....   a triangle  of   .08  - .15 - .17 also exists

And notice that   3*.08  = .24     and 3*.15 = .45.......so  3* .17 =  .51   ....which implies that....

[ 3* .08]^2 + [3 * .15]^2  = [ 3 * .17]*2..........which further implies that......

[ .24]^2 + [ .45]^2   = [.51]^2    .........which further implies that....

[11* .24]^2  + [ 11 * .45] ^2   = [ 11 * .51]^2

So....the hypotenuse must be   11 * .51  = [ 10 + 1] [.51]   = 5.1  + .51   =  5.61 units in length

Let a radius drawn through the chord bisect the chord.  And, by Euclid, such a radius will meet that chord at right angles.

And the center of the circle, the intersection point of the radius and the chord and the intersection point of the radius with the circle will form a right triangle

And......by the Pythagoren Theorem,  the distance - d - that the chord lies from the center of the cirdle is :

sqrt [radius^2  - (1/2 the chord length)^2]   =

sqrt (14^2 - 6^2) = sqrt(160) units = 4*sqrt(10)  units

Look at the following pic, Mellie

Angle ABC can be found thusly :

sin CAB/ 26 = sin ABC / 24

And sin CAB = sin90 = 1

So

Sin ABC = 24/26

arcsin (24/26)  = ABC

And the altitude  AX drawn from A to to side BC will be one leg ot the right triangle AXB........so this altitude will be shorter than the altitude drawn from B  to side AC, since this altitude is the hypotenuse of triangle AXB. And AX will clearly be shorter than the remaining altitude drawn from C  to side AB.

And we can find AX thusly

AX / sinABC = AB / sin AXB

AX / sin[arcsin(24/26)]  = AB / sin AXB

But AXB = 90 so the sine of this angle  = 1   and   AB = 10.....so we have

AX = 10 * sin[arcsin(24/26)]  = 120/13 units  = about 9.23 units

And this is the shortest altitude

A = (1/2) (AB)(BC)sin120  =  (1/2)(12)(12)(sqrt(3)/2)  = 36sqrt(3)  sq units

PR = 12*sin60 = 12 * [sqrt(3)/2]  =  6 * sqrt(3)

And we can find XR thusly :

XR/ sin30  = 6/ sin60

XR   = 6sin30/ sin60

XR  = 6(1/2) / [sqrt(3) / 2]

XR  = 3*2/ sqrt(3) = 6/sqrt(3)

So PX  =  PR - XR =   6*sqrt(3) - 6/sqrt(3) =  [18 - 6] / sqrt(3)  =  12/sqrt(3)

And the area of PQX   =  (1/2) *PX* RQ  = (1/2) (12/sqrt(3)) * (6)   =  36 / sqrt(3)  square units

Here's a pic :  

A positive is always greater than a negative.

12 / 18  =    x / 28        [reduce 12 / 18  to simplest form ]

2 / 3   =  x / 28       cross-multiply

28 * 2   =  3x

56  = 3x         divide both sides by 3

56/3   =  x

2x+8y=6    divide through by 2  →  x + 4y  = 3  

-5x+-20y=-15     divide through by -5 →  x +4y = 3  

These equations are exactly the same.....so we will have infinite solution pairs for (x , y)......!!!