Question

ABC is an isosceles triangleAB=AC=13 cm and BC=10 cm. Calculate the length of the perpendicular from A to BC.

Answer 1

AB =AC= 13CM
BC = 10CM
LET D IS A POINT WHERE THE PERPENDICULAR FROM A MEETS BC.
PERPENDICULAR AD = 12CM
$= \sqrt{ {13}^{2} - ({ \frac{10}{2} )}^{2} } \\ = \sqrt{169 - 25} \\ = \sqrt{144} \\ = 12$
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