Question

ABC is an isosceles triangle, where sides AB and AC are equal. If AB = AC = 8 cm and BC =12 cm, find the minimum distance of vertex A from the side BC.

• A. $2\sqrt{7}$ cm
• B. $3\sqrt{7}$ cm
• C. $4\sqrt{7}$ cm
• D. $5\sqrt{7}$ cm

Answer 1

The correct option is A $2\sqrt{7}$ cm

Length of the perpendicular drawn from vertex A is the minimum distance of vertex A from its opposite side.




The perpendular drawn divides the opposite side in two equal parts. Therefore, the length of side, BD = 6 cm

Now, using pythagorus theorem, in triangle ABD:
$A{D}^2+B{D}^2=A{B}^2$
$A{D}^2+6^2=8^2$
$A{D}^2+36=64$
$A{D}^2=64-36$
$A{D}^2=28$
$AD=\sqrt{28}$
$AD=2\sqrt{7}$

$\therefore$ the minimum distance of vertex A from the side BC is $AD=2\sqrt{7}$.