ΔABC is an equilateral triangle. Point P is on base BC such that PC = (1/3) BC, if AB = 6 cm find AP.

\(\triangle ABC\) is an equilateral triangle. Point P is on base BC such that \(PC = \frac{1}{3} BC\), if AB = 6 cm find AP.

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

1 answer

Answer: AP = \(2\sqrt{7}\) cm

Draw AD ⊥ BC. In an equilateral triangle, D is the midpoint of BC, so DC = 3 cm.

\[AD = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}\ \text{cm}\]

PC = \(\frac{1}{3} \times 6\) = 2 cm, so PD = DC − PC = 3 − 2 = 1 cm.

In right-angled \(\triangle ADP\),

\[AP^2 = AD^2 + PD^2 = 27 + 1 = 28\] \[AP = \sqrt{28} = 2\sqrt{7}\ \text{cm}\]

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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