Problem:
In β³ABC,AB=6,AC=8,BC=10, and D is the midpoint of BC. What is the sum of the radii of the circles inscribed in β³ADB and β³ADC?
Answer Choices:
A. 5β
B. 411β
C. 22β
D. 617β
E. 3
Solution:
By the converse of the Pythagorean Theorem, β BAC is a right angle, so BD=CD=AD=5, and the area of each of the small triangles is 12 (half the area of β³ABC). The area of β³ABD is equal to its semiperimeter, 21ββ (5+5+6)=8, multiplied by the radius of the inscribed circle, so the radius is 812β=23β. Similarly, the radius of the inscribed circle of β³ACD is 34β. The requested sum is 23β+34β=(D)617ββ.