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

Given: AB = AC = 13 cm, BC = 10 cm.

Let D be the foot of the perpendicular from A to BC.

To Prove: The length of AD (perpendicular from A to BC).

Construction: Join AD.

Proof: Since Δ ABC is an isosceles triangle (AB = AC), the perpendicular AD from A to BC will bisect BC. Thus,

BD = DC = $\dfrac{\text{BC}}{2} = \dfrac{10}{2}$ = 5 cm

In right-angled triangle ADB,

AB² = AD² + BD²

⇒ 13² = AD² + 5²

⇒ 169 = AD² + 25

⇒ AD² = 169 - 25

⇒ AD² = 144

⇒ AD = $\sqrt{144}$ = 12

Hence, the length of the perpendicular from A to BC is 12 cm.