CPhill

https://web2.0calc.com/questions/help-with-triangle_26

DE = EC

EA  =  sqrt [ AC^2 -  EC^2] =  sqrt [5^2 - 4^2] = sqrt [ 9 ]=   3

2 * area ADC  = area of parallelogram

2 * (1/2) DC * EA   =

DC * EA  =

8 * 3  =  

24

Let adjacent  sides be x, y

So.....

2x + 2y  =  40

x + y  = 20

y = 20 - x

By the Pythagorean Theorem

x^2  + y^2  = (10 sqrt 2) ^2

x^2 + ( 20 - x)^2  =  200

x^2 + x^2  - 40x + 400 =  200

2x^2  - 40x + 200  = 0

x^2 - 20x + 100  = 0

(x -10)^2  = 0           take positive root

x  - 10  =  0

x = 10

y = 20 - 10  =  10

Area = xy =  10 * 10  =   100

Area =  2 * (1/2) * 12^2 * sin (20°)  =  144 * sin (20°)  ≈  49.25

The ray will strike X  and then , because angle A = angle B  it will travel to  vertex D

The  total path length =   2 sqrt [ .5^2  + 1^2 ] =  2 sqrt [ 5/4]  = sqrt 5

Semi-perimeter  =  [ 15 + 9 + 10 ] /  2  =  17

Area  =   sqrt [ 17 * 2 * 7 * 8 ]  = 4 sqrt 119

Shortest altitude is  drawn  to the longest side

4 sart 119 =  (1/2) AB * Altityude

4 sqrt 119 = (1/2) 15 * Altitude

[ 4 sqrt 119  ]  /  [ 1/2 * 15 ]  =   [8 sqrt 119] / 15 =  Altitude  ≈  5.82

Without explanation , side of  square =

[  product of leg lengths  ]    [ sum of leg lengths ] = 

[ 6 * 6 ] / [ 6 + 6 ] = 

36  /  12     =

3

https://web2.0calc.com/questions/hexagon_12

AM =  sqrt [ 20^2 - 15^2]  =  sqrt [175 ]  =  5sqrt (7)

DM = AM / 2  =   2.5 sqrt (7)

Draw ED

Triangle AED ≈ Triangle ABM

AD / AM  = ED / BM

(1/2)AM / AM  = ED / BM

1/2  = ED / BM

ED = (1/2) BM

Triangle EDT ≈ BMT

Since ED = 1/2 BM

Then the height of triangle EDT  is 1/2 the height of triangle BMT

So the height of triangle BMT =  [ 2 /( 1 + 2)  ] * DM  = (2/3)DM

And the  height of triangle BME = DM

So  the area of triangle BME = (1/2)(BM)((DM)

And  the area  of triangle BMT = (1/2)(BM)(2/3)(DM)  = (1/3) (BM)(DM)

So  the area of triangle BET = [ BME ]  - [ BMT ]   = 

(1/2)(BM)(DM)   -  (1/3) (BM)(DM) =  (1/6) (BM) (DM)  =

(1/6) (15) ( 2.5 sqrt (7) )  = 

6.25 sqrt 7 =

25sqrt (7)  / 4

Ooops!!!....I found the area of CBD !!!

See my corrected answer....