CPhill

Triangles AZY and  ACY have the same height

But base AC is 5/2  of base AZ..so area of ACY  = (5/2) AZY

Similary....triangles ZYC and  AZY  have the same height

But base ZC is  (3/2) base AZ...so  area of ZYC  = (3/2) AZY

And triangles ACY  and  YBC  have the same  height

But base AY is 3 times base  BY....so area of ACY = (3)YBC

So.....

ACY  = (3)YBC    (1)

ACY = (5/2)AZY  (2)

ZYC = (3/2)AZY    (3)

From (1)  and (2)

(3)YBC = (5/2)AZY

(6/5)YBC  = AZY    (4)

And from (3)

(2/3)ZYC  = AZY  (5)

So...from  (4)  and (5)

(6/5)YBC =  (2 / 3)  ZYC

So

[ YBC ] / [ ZYC ]  =    [  2/3] / [ 6/5]

[ YBC] / [ZYC ] =  (2/3) * (5/6)   =  10/ 18  =  5 / 9

Excellent,  tertre  !!!

Draw QT  and QS

Angles TQP, SQT  and  SQR  will trisect the interior angle PQR

And PQR  = 108

So  ...angle  SQT  = 108 / 3   = 36

But...by symmetry....angle XQS  =  (1/2)angle SQT  = (1/2)36  = 18°

AWX  and AYZ  are under the same height

So....they are to each other as teir bases

So  WX / YZ  =  30/20   = 3/2

WX =  (3/2)YZ

So

WX + XY   = 9    ⇒  (3/2)YZ + XY  = 9       (1)

XY + YZ  =  7     ⇒  YZ + XY  =  7         (2)

Subtract  (2)  from (1)  and we have that

(1/2)YZ  = 2

YZ  = 4

So XY = 3

And WX = (3/2)YZ  = (3/2)4  = 6

So......the height of   AWX  can be found as

30  = (1/2) WX * height

30  = 3 * height

10 =  height  of  AWX  = height of  AXY

So...the area of AXY  =  (1/2)XY * 10  =  (1/2)(3)(10) =  15 units^2

We can solve this with a proportion

32/ 60  =  L / 15         divide top bottom of the fraction on the left side by 4

8 / 15  = L  / 15

L = 8 ft = shadow cast by radio tower

See my corrected answer, shreyas1

The rectangle at the top left has the perimeter of 24

The rectangle at the top  right has the perimeter of 42

The rectangle at the  bottom right has the perimeter of 32

So...we are trying to find the perimeter of the rectangle at the bottom left

Since the top rectangles are under the same height...call this height x

Call the width of the top left rectangle, y   and the width of the top/bottom right rectangles, z

Call the  height of the bottom left/right rectangles, w

So  

2x + 2y  = 24           2x + 2z =  42

x + y  = 12     (1)              x + z  = 21    (2)

Subtract (1) from (2)  and we have

 So    z - y = 9  ⇒   z  = y + 9    (3)

So...for the bottom right rectangle

2 ( w + z )  = 32

w + z  = 16

z = 16 - w      (4)

So...equate  (3)  and (4)

y + 9 =  16 - w

y + w  = 7

So  the perimeter of the bottom left rectangle  =

2 ( y + w)  =

2 (7)  =  14

HAHAHA!!!...on these geometry problems....I'm never too sure, either...!!!

Since triangles CBD  and  CAD  are on the same base (CD)...then their areas are to each other as their heights

So  

[ CAD ] / [ CBD ]  =  12 / 7

So

[ CAD ]  = (12/7) [ CBD ]

So

[ CBDA ] + [CBD ]  =  [ CAD ]

15   +   [CBD ]  =  (12/7) [ CBD ]

15  =  (12/7) [ CBD ] - [CBD ]

15  =  (12/7) [CBD ] - (7/7)[CBD]

15 = (5/7) [CBD ]

(7/5)(15)  = CBD

21  =  CBD

But  CAD  =  (12/7) [ CBD ]  = (12/7) (21) =  12 * 3  = 36  = [CDA]

You are correct  !!!

 YA + AZ + BZ + BX + + CX + CY  =78

(CY + YA)  + (AZ + BZ)  +  ( BX  + CX)  = 78

But AC  = (1/3)(CY + YA + AC)    ⇒  AC - (1/3)AC  = (1/3) (CY + YA) ⇒

(2/3)AC  = (1/3)(CY + YA)  ⇒  2AC  = (CY + CA)

Similarly

2AB = (AZ + BZ)

And

2BC  = (BX + CX)

So

2 ( AC + AB + BC)  = (CY + CA) + ( AZ + BZ) + (BX + CX)

2 ( AC + AB + BC)  =  78

(AC + AB + BC)  = 39  =  perimeter of ABC

CORRECTED ANSWER......