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Please help

 Mar 7, 2024
 #1
avatar+128794 
+1

First one

Altitude YW =  sqrt [ XZ^2  - (YZ /2)^2 ]  = sqrt  [12^2  - (6sqrt 3)^2  ] = sqrt [ 144  - 108] = sqrt [ 36] = 6

 

sin (YZW  )  =  YW / YZ  = 6 / 12   = 1/2

 

Draw  WN  where N is the tangent point of  the semi-circle to YZ

 

radius of  semi-circle / sin (YZW) = WZ / sin (WNZ)

 

r  / (1/2)  = (6sqrt 3) / sin 90°

 

r = (1/2) (6sqrt 3)  =  3sqrt 3  = sqrt (27)

 

Area of semi-circle   = (1/2)pi r^2   = (1/2)pi (27)  =  13.5 pi

 

cool cool cool

 Mar 7, 2024
edited by CPhill  Mar 7, 2024
 #2
avatar+85 
0

I dont think thats correct

Akhaim1  Mar 7, 2024
 #3
avatar+128794 
+1

It's correct

 

Here's the graph :

 

 

 

cool cool cool

CPhill  Mar 7, 2024
 #4
avatar+128794 
+1

Second one

Let  O be the center of the circle

Let  M be the mid-point  of UV

Let d be the distance from O to M

We can form a right triangle such that

d^2 + (UV/2)^2  = r^2

d^2 + 19^2  =  r^2      (1)

 

Let the midpoint of YZ = N

Let the distance from O to N =  d + 4

We can  form another right triangle such that

(d + 4)^2 + (YZ/2)^2  = r^2 

(d + 4)^2 + 11^2  = r^2      (2)

 

Equate (1) , (2)

 

d^2 + 19^2   =  (d + 4)^2  + 11^2

 

d^2 + 361  = d^2 + 8d + 16 + 121

 

8d =  361  - 16 - 121

 

8d =  224

 

d =  224/8  = 28

 

28^2 +19^2   =r^2

 

1145 =  r^2

 

And we can form a third right triangle such that

 

(d + 2)^2 + (WX / 2)^2  = r^2

 

(28 + 2)^2  + WX^2 / 4  = 1145

 

30^2  + WX^ 2 /4 = 1145

 

900 + WX^2/4 = 1145

 

WX^2 / 4  = 1145 - 900

 

WX^2 / 4  = 245

 

WX^2   =  245 * 4 = 980

 

WX  = sqrt 980   ≈ 31.3

 

 

cool cool cool

 Mar 7, 2024
 #5
avatar+85 
+1

My bad, you are right. The second one is also correct

 Mar 8, 2024

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