Geometry Questions!!

1.

Let O be the center of the semi-circle

Draw CE  perp  to AB

Connect OC  = r

OE = 8.5

EB = r - 8.5

CB = 3

Using the Pythagorean Theorem

Height of triangle OCE   = Height of triangle ECB

CE^2  =  CE^2

r^2  - (OE)^2  = (CB)^2  - ( EB)^2

r^2 - (8.5)^2  =  (3)^2 - (r - 8.5)^2

r^2  - 72.25  =  9 - r^2 + 17r - 72.25

r^2 - 72.25  =  -r^2 + 17r - 63.25

2r^2 - 17r -9  = 0

(2r + 1) ( r -  9)  =  0

r =  9 

2. 

Area of Sector  = pi *r^2 *  (100  / 360)

250 pi  =  pi * r^2 * (5/18)

r^2  = 250 (18 / 5)

r^2  =  50 * 18

r^2  = 100 * 9

r^2 = 900

r  = 30

3.

2* area ABC =  AD * BC                  2* area ABC =  BE * AC

2* area ABC =  12 * BC                   2* area ABC  = 16 * AC

area ABC = 6 BC                             area ABC  =  8 AC

6 BC =  8 AC

AC = (8/6) BC = (4/3) BC

Triangle inequality

AB + BC >  AC

AB + BC >  (4/3)BC

AB  >  (4/3)BC - BC

AB > (1/3)BC

Then

2 [ ABC ]  =   BC  *  AD

2[ ABC ]   = BC * 12

2 [ ABC ] = (1/3)BC *  36

Since AB > (1/3)BC

2[ ABC ]  < AB *  36

2[ ABC ] < AB * CF

Then CF  < 36

So

Max integer value for CF =  35