1. Problem: Find length AP in the first triangle where AP is a median, BC = 18, and $AB^2 + AC^2 = 260$. 2. P is midpoint of BC, so $BP = \frac{1}{2} BC = \frac{1}{2} \times 18 = 9$. 3. Using Apollonius theorem: $$AB^2 + AC^2 = 2AP^2 + 2BP^2$$ Substitute values: $$260 = 2AP^2 + 2(9)^2 = 2AP^2 + 162$$ Simplify: $$2AP^2 = 260 - 162 = 98$$ $$AP^2 = 49$$ $$AP = 7$$ 4. The 7th sum explanation requested is about calculating $AP = 2\sqrt{7}$ cm in the second problem. Second problem setup: $\Delta ABC$ is equilateral with side 6 cm. Point P is on BC such that $PC = \frac{1}{3} BC = 2$ cm. Drop perpendicular AM from A to BC, making AM perpendicular to BC at M. Since $BC=6$, and M is midpoint of BC in an equilateral triangle, $$MC = \frac{1}{2} BC = 3\text{ cm}$$ The triangle $AMC$ is a 30°-60°-90° triangle because $\angle C=60^\circ$ and $\angle AMC=90^\circ$. By 30°-60°-90° triangle theorem: $$AM = \frac{\sqrt{3}}{2} AC = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3} \text{ cm}$$ $$MC = 3 \text{ cm}$$ Point P divides BC such that $PC=2$ cm, and $MC=3$ cm. So segment $MP = MC - PC = 3 - 2 = 1$ cm. In $\Delta AMP$, $\angle AMP = 90^\circ$ by construction. By Pythagoras theorem: $$AP^2 = AM^2 + MP^2 = (3\sqrt{3})^2 + 1^2 = 27 + 1 = 28$$ $$AP = \sqrt{28} = 2\sqrt{7} \text{ cm}$$ Answer: $AP = 2\sqrt{7}$ cm.