If $a = 9, b = 4, c = 8$ then the distance between the middle point of $B C$ and the foot of the perpendicular from $A$ is

2

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\frac{8}{3}

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\frac{7}{3}

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Solution

The correct option is D $\frac{8}{3}$

We have a triangle $A B C$ where $D$ is the midpoint of line $B C$ and $E$ is the point for foot of a perpendicular from $A$

Now $D E = C D - C E$
$= \frac{1}{2} a - b cos C$

$= \frac{a}{2} - b (\frac{a^{2} + b^{2} - c^{2}}{2 a b})$ by using formula $cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

$= \frac{a}{2} - \frac{a^{2} + b^{2} - c^{2}}{2 a}$

$= \frac{a^{2} - a^{2} - b^{2} + c^{2}}{2 a}$

$∴ D E = \frac{c^{2} - b^{2}}{2 a}$
Now we know a=9, b=8, c=4
$=> D E = \frac{8^{2} - 4^{2}}{2 \times 9}$ $= \frac{64 - 16}{18}$ $= \frac{48}{18}$ $= \frac{8}{3}$

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If a=9,b=4,c=8 then the distance between the middle point of BC and the foot of the perpendicular from A is

If $a = 9, b = 4, c = 8$ then the distance between the middle point of $B C$ and the foot of the perpendicular from $A$ is