+0  
 
0
13
3
avatar+1679 

A semicircle inscribed inside a quarter circle. Find the length of a chord of the quarter circle that is also a tangent of the semicircle.

 

 Jan 9, 2024
 #1
avatar+36919 
+1

The answer I get is 36 cm ...... It took a while for me to figure a way to get the answer.....there is likely an easier way but here is my (long -hopefully-correct) solution:

See image :

 

Use arc tan to find the 63.435 angles then find the 53.13 angle     ( 63.435+63.435 + 53.13 = 180 ° ) 

           then use sin and cos functions for angle 53.13 °  to find the coordinates of T

              then use the coordinates of B and T to find the equation of the line BP to be   y = - 3/4 x + 30 

                   the equation of the Quarter circle is    x^2 + y^2 = 900    or    y = sqrt(900-x^2) 

                         equate this to the line equation and solve for the 'x' coordinate of  point P     then use the line equation and this value of the 'x' coordinate  to find the value of the 'y' coordinate for point P

  

           Now you have the coordiantes of the two endpoints of the chord line 

                        B    and   P  ====>    then finish by using the distance formula to find  BP = 36 cm 

 

 

There really must be an easier way !!!    Whew !   smileycheeky

 Jan 9, 2024
 #2
avatar+128732 
+1

I don't find an easier way, either, EP.

 

I did it with just some  Geometry

 

Here's my solution  (but it  might not  be  any  easier)

 

Like EP, I found BT to be 30

 

We can find the coordinates of T  by finding the intersection of two circles  (this is the key !!!)

 

One with a center of (0,30) and a  radius of 30

One with a center of (15,0) and a  radius of 15

 

The equations are

 

x^2  + ( y - 30)^2  = 900    →  x^2 + y^2 - 60y   =  0   (1)

(x - 15)^2 + y^2  = 225     →  x^2 - 30x + y^2  = 0     (2)

 

Subtract these

 

-60y + 30x  =  0

x = 2y

Sub  this into (2)

 

4y^2 - 60y + y^2  = 0

5y^2 -60y  =  0

y - 12  = 0

y =12

 

x = 2(12)   = 24

 

T = (24,12)

 

Slope   of   line through BP = ( 30 - 12) / ( 0 - 24) = -3/4

 

Equation of this  line  is  y =(-3/4)x + 30

 

Now, as EP  said, put this into  the  equation of the circle  x^2 + y^2  =900  and find the x coordinateof P  as  28.8

And  -(3/4)(28.8) + 30  =   8.4 = the y coordinate of P

 

TP = sqrt [ (28.8 -24)^2  + (12 - 8.4)^2 ]  = sqrt [ 4.8^2  + 3.6^2 ]  =  sqrt (36)   = 6

 

Nice job, EP!!!

 

cool cool cool

 Jan 9, 2024
 #3
avatar+36919 
+1

Thanx, Chris !   

     I was wondering if I had this correct.....now I know it likely is correct.....or maybe we BOTH got it wrong ! HaHa 

~EP

ElectricPavlov  Jan 10, 2024

1 Online Users