Find the area of the shaded region in the above diagram.
The area of the shaded region is equal to the area of the semicircle with diameter 3 plus the area of the semicircle with diameter 4 plus the area of the right triangle minus the area of the semicircle with diameter 5. We have
DC is the diameter of the circle with center at A.BE is perpendicular to DC. If DE=15 and EC=9, find BE.
Since △DEB∼△DBC, we have
DEBE=BDBC⟹15BE=BDBC⟹BE=15BDBC(1)
and
BDBE=DCBC⟹BDBE=24BC⟹BE=24(BD)(BC).(2)
Substituting (1) into (2) gives
15BDBC15(24)BD=24(BD)(BC)=(BD)2=360=610.
Using the Pythagorean theorem, we have
BC=DC2−BD2=242−(610)2=216=66.
Finally,
BE=15BDBC=15(61066)=10156.
Rationalizing the denominator gives
BE=10156⋅1010=103015=315.□
Challenge Problems
69771π+69775697+726794917425+4927888
BC is tangent to both a circle with center at A and a circle with center at D. The area of the circle with center at A is 225π and the area of the circle with center at D is 36π.
If BC=16, find the distance between the centers of the two circles.