Problem: Each side of triangle A B C A B CA B C is 12 121 2 units. D DD is the foot of the perpendicular dropped from A AA on B C B CB C , and E EE is the midpoint of A D A DA D . The length of B E B EB E , in the same unit, is:
Answer Choices:
A. 18 \sqrt{18}1 8 ​
B. 28 \sqrt{28}2 8 ​
C. 6 66
D. 63 \sqrt{63}6 3 ​
E. 98 \sqrt{98}9 8 ​
Solution:
B E ‾ 2 = B D ‾ 2 + D E ‾ 2 , B D ‾ = 6 , D E ‾ = 1 2 D A ‾ = 1 2 ⋅ 6 3 = 3 3 \overline{B E}^{2}=\overline{B D}^{2}+\overline{D E}^{2}, \overline{B D}=6, \overline{\mathrm{DE}}=\dfrac{1}{2} \overline{\mathrm{DA}}=\dfrac{1}{2} \cdot 6 \sqrt{3} =3 \sqrt{3}B E 2 = B D 2 + D E 2 , B D = 6 , D E = 2 1 ​ D A = 2 1 ​ ⋅ 6 3 ​ = 3 3 ​ \quad therefore B E ‾ 2 = 36 + 27 \overline{\mathrm{BE}}^{2}=36+27B E 2 = 3 6 + 2 7 \quad therefore \overline{\mathrm{BE}}=\sqrt