CPhill

Not as bad as you might  think.....

CE = 2sqrt 61  = sqrt 244

AD = sqrt 601 

Note  that we have this system

AD^2  = (BC/2)^2  + (AB)^2

CE^2  = (BC)^2 + (AB/2)^2         simplify

601  = BC^2/4 + AB^2

244 = BC^2 + (AB/2)^2

601 = BC^2 /4  + AB^2

244 = BC^2 + AB^2/4                mutiply the top equation through by -4

-2404 = -BC^2- 4AB^2

244  = BC^2  + AB^2/4     add these

-2160  =  -(15/4)AB^2

(-2160) * (-4/15)  =AB^2  = 576

AB = sqrt (576)  = 24

To find BC

244 = BC^2 + (24/2)^2

244 = BC^2 + 144

100 = BC^2

10= BC

Hypotenuse  =  sqrt [ 24^2 + 10^2 ] =  sqrt [676 ]  =  26

https://web2.0calc.com/questions/help-geometry_154

https://web2.0calc.com/questions/triangle_41241

https://web2.0calc.com/questions/triangle_19840

  A

30

                X

B 90                 C 60

            12

AB =  12sqrt 3

Law of Sines

AB  /  sin BXA  = BX / sin BXA

12sqrt 3 / sin 75  = BX / sin 30

BX  =  6sqrt 3 / sin 75  

[ ABX ]  = (1/2) BX * AB * sin (45) =  

(1/2) (6sqrt 3 / sin 75) ( 12sqrt 3) sin (45)

108 sin (45) / sin(75)  =

216 / (1 + sqrt 3)  ≈   79.06

Think about this...let m  =  1

So   a  =  (1,0)    and b =(0, 1)

These are the same  length  and  are perpendicular

A                        B

                          4

                          P

                          1

D     1    Q  4     C

AB  =  5

AP^2  =  5^2 + 4^2  =  41

AD  = 5

AQ^2  =  5^2 + 1^2  =  26

PQ^2  = 4^2 + 1^2  = 17

Law  of Cosines

PQ^2  = AP^2 + AQ^2  -  2 (AP* AQ) cos (PAQ)

[PQ^2 - AP^2 - AQ^2 ] / [ -2 sqrt (AP * AQ) ]  = cos PAQ

[ 17  - 41 - 26 ] / [ -2 sqrt (26 * 41) ]  = cos PAQ  = 25 / sqrt 1066 =  cos PAQ

sin PAQ  =  sqrt [ 1066 - 25^2 ] / sqrt 1066  = sqrt (441) / sqrt 1066  =   21  / sqrt 1066

The triangle is equilateral...AD is a bisector and  the altitude of ABC

                      A

                     24

B                   D                 C

Angle  BAD  = 30

BD  =  24/sqrt 3

BC  =  2 * 24 /sqrt 3 =   48/ sqrt 3 =   16sqrt 3

[ABC ]  = (1/2) BC * AD  =  (1/2) (16sqrt 3) (24)  =  192 sqrt 3

(a)

Let  a  = ( 0, m)

Let b = (m , 0)

Note that these are equal in length

(a + b)  =  (m , m)

(a - b)  =  (-m , m)

If  two vectors are perpendicular, their dot product  =  0

dot product =   (m , m)  ( dot )  (-m , m)  = -m^2  + m^2  =   0

Area =  L * B =     2B * B  = 2B^2

Yellow and Purple area =  (2/5 + 1/3) * 2B^2  =  (22/15)B^2

Pink area  = (1/2) ( 2B^2  - 22/15 *B^2 ) =  (4/15)B^2

So

Area - Painted Area =   240

2B^2 - (22/15 + (4/15))B^2  =  240

[4/15) B^2 =  240

B^2  = (240* 4 /15)  

B^2  = 960 / 15

B^2  =  64

B = 8

2B =  16

Perimeter  =  2 [ B + 2B ]  =   2 [ 8 + 16 ] =    48 cm

https://web2.0calc.com/questions/area_72

We know the three sides of PQR.....the other info doesn't matter

Semi-perimeter  =  [ 36 + 26+ 22 ]  /  2 =  84 / 2 =  42

Heron's Formula :

[ PQR ]  =  sqrt [ 42  * 6 * 16 * 20 ] =  48sqrt (35)   ≈ 283.97

B = (6/5)R

G = (3/2)B  →  G = (3/2)(6/5)R   =  (9/5)R

Total  number of shirts at te beginning  =  R + B + G  = R + (6/5)R  + (9/5)R =  4R

So  after he gives away  12 red shirts, we have

[R -12 ]  =  (1/10) 4R

R - 12 =  (2/5)R

R - (2/5)R  = 12

(3/5)R  =12

R =12 *(5/3)  =  20

The total shirts  he  had when he started =  4 (20)  = 80

After giving away 12 red shirts  he has  80  - 12  =   68   left

Side of the square =  [product of leg lengths ] /  [ sum of leg lengths ] =  [ 1 * 3 ] /  [ 1 + 3] =   3 / 4

Area of square  =  [3/4]^2 =  9 / 16