geometry

Let C  = (0,0)      let B =  (8,6)

Then  BC =   sqrt ( 8^2 + 6^2)  = sqrt (100)  = 10

Draw  OM perpendicular to  BC

This will form two right triangles OBM  and OCM

OB  = 8    OM  = r    BM =  sqrt ( OB^2  -r^2)  =  sqrt  ( 8^2  - r^2)

OC =  6    OM  = r    MC = sqrt (OC^2 - OM^2) = sqrt ( 6^2 - r^2)

So

BM  +  MC  = BC

sqrt ( 8^2  - r^2)  + sqrt ( 6^2  -r^2)  =  10

sqrt (64 - r^2)  +  sqrt (36 - r^2)   =10

sqrt (64 -r^2)  = 10   - sqrt (36 -r^2)         square both sides

64  -r^2  = 100  - 20sqrt (36 - r^2)  +  36 -r^2

64  =  136  - 20 sqrt (36 - r^2)

-72  = -20  sqrt (36 - r^2)

72/20 = sqrt (36 -r^2)

18/5   =sqrt (36 - r^2)      square both sides

324/25  =  36 - r^2

-576/25  = -r^2

r^2  =576/25      take the  positive  root

r  = 24/5  cm

PQRS is a rectangle with PQ=16 cm and QR=12 cm. If ABCD is a rhombus formed PQRS, what is the radius of the circle inscribed in rhombus ABCD?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AO = 6 cm        BO = 8 cm 

∠ABO = tan^-1(6 / 8)

Radius EO = sin(∠ABO) * BO = 4.8 cm

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ or

AB = sqrt(AO² + BO²) = 10 cm

Radius      EO = (AO * BO) / AB = 4.8 cm