Question

If $0 = 9,{\rm{\;b}} = 4,{\rm{\;c}} = 8$ then the distance between the middle point of BC & the foot of the perpendicular from A is
(A) 2
(B)1
(C) $\dfrac{8}{3}$
(D) $\dfrac{7}{3}$

Answer 1

Hint:
We have to use the relation between the angle of the triangle and its sides. We know such a formula which is the cos rule in trigonometry.
$\cos {\rm{c}} = \dfrac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$ to find d. then find DE which is the distance between the middle point of BC & the foot of the perpendicular from A.

Complete step by step solution:
We have a triangle ABC where D is the mid-point of line BC & E is the point for the foot of a perpendicular from A.
Now, DC CD CE
 $\dfrac{1}{2}{\rm{a}} - {\rm{b}}\cos {\rm{c}}$$ = \dfrac{{\rm{a}}}{2} - {\rm{b}}\left( {\dfrac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}} \right)$by using formula
 $\cos {\rm{c}} = \dfrac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$
 $ = \dfrac{{\rm{a}}}{2} - \dfrac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{a}}}}$
 $ = \dfrac{{{{\rm{a}}^2} - {{\rm{a}}^2} - {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{a}}}}$
 $\therefore {\rm{DE}} = \dfrac{{{{\rm{c}}^2} - {{\rm{b}}^2}}}{{2{\rm{a}}}}$
Now, we know that $0 = 9,{\rm{\;b}} = 4,{\rm{\;c}} = 8$

$\therefore {\rm{DE}} = \dfrac{{{8^2} - {4^2}}}{{2 \times 9}} = \dfrac{{64 - 16}}{{18}}$
$ = \dfrac{{48}}{{13}} = \dfrac{8}{3}$

Note:
Length of AD is found using Stewart theorem.
i.e. ${\rm{A}}{{\rm{D}}^2} = \dfrac{{2{{\rm{b}}^2} + 2{{\rm{c}}^2} - {{\rm{a}}^2}}}{4}$
Where AD is the median. Once AD is found, DE can be easily determined.