Golden rectangle in triangle

A Golden Rectangle  is  one where  the  sides are in proportion  of  1  : Phi

Where Phi =  (1 + sqrt 5 ) / 2  ≈  1.618

Find   the area  of  the ltriangle

Semiperimeter, S =   (7 + 11 + 14)  /2  =  16

Area of  triangle=  sqrt  [16 8 (16 - 70 (16 - 11) (16 - 14) ]    =   sqrt [ 16 * 9 * 5 * 2 ]  =  12 sqrt (10)

We  can find  the   height of the  triangle  thusly

Area  = (1/2) AC *  height

12sqrt (10)  = 7 * height

(12/7) sqrt 10  =  height     ....call  this BD

Let the rectangle  be MNOP   where  AP  +  PO  +  OC  =   14

Let  the  height of  the  rectangle = a     and  let PO  =   a * Phi

We can  find   DC   as      sqrt  [ 11^2  - (12/7 * sqrt 10)^2 ]  = 67/7

And we  can  find  DA  as  sqrt [ 7^2  - (12/7 * sqrt 10)^2 ] =  31 / 7

By similar    triangles

OC  / a    =   DC / BD

OC  = a  *  ( 67/7)/ [ (12sqrt 10 / 7  ]    =   67 a /  [ 12 sqrt 10] 

AP   /  a  =  DA / BD

AP  =  31a  /[ 12sqrt 10 ]

So

AP  +  PO  +  OC    =14

31 a / [ 12sqrt10 ]  +  (1 + sqrt 5] / 2 * a  +  67a / [12sqrt 10 ]  =  14

Solving for  a  we get  that  a  ≈  3.3329

So  the  area of  the rectangle       a *  a * Phi  =    (3.3329)^2 * 1.618    ≈  17.97  units ^2

Yikes !!!!!!!     

The golden rectangle is inscribed in triangle ABC so that the longer side of a rectangle lies on the side AC. If AB = 7, BC = 11, and AC = 14, then what's the area of a rectangle?

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Using the law of cosines, we find that ∠A ≈ 50.754º and ∠C ≈ 29.526º.

To calculate an x, we can use this equation:

(x / tanA) + (x * Φ) + (x / tanC) = 14                 Φ = 1/2(1 + √5)

x ≈ 3.333

Rectangle area = x²Φ ≈ 17.974