1. **Problem (d):** Show that a triangle with sides 10 cm, 5 cm, and 11 cm is not right-angled. 2. **Step 1:** Recall the Pythagorean theorem for right-angled triangles: $$a^2 + b^2 = c^2$$ where $c$ is the longest side. 3. **Step 2:** Identify the longest side: 11 cm. 4. **Step 3:** Calculate squares of the sides: $$10^2 = 100$$ $$5^2 = 25$$ $$11^2 = 121$$ 5. **Step 4:** Check if $$10^2 + 5^2 = 11^2$$ $$100 + 25 = 125 \neq 121$$ 6. **Step 5:** Since $$125 \neq 121$$, the triangle is not right-angled. 7. **Problem (e):** Find the area of a triangle with sides 8 cm, 8 cm, and 10 cm, expressing the answer as $$5\sqrt{n}$$ cm². 8. **Step 1:** Use Heron's formula for the area of a triangle with sides $a$, $b$, and $c$: $$s = \frac{a+b+c}{2}$$ $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ 9. **Step 2:** Calculate semi-perimeter: $$s = \frac{8 + 8 + 10}{2} = \frac{26}{2} = 13$$ 10. **Step 3:** Calculate the area: $$\text{Area} = \sqrt{13(13-8)(13-8)(13-10)} = \sqrt{13 \times 5 \times 5 \times 3}$$ 11. **Step 4:** Simplify inside the square root: $$\sqrt{13 \times 5 \times 5 \times 3} = \sqrt{13 \times 25 \times 3} = \sqrt{975}$$ 12. **Step 5:** Factor 975 to simplify the square root: $$975 = 25 \times 39$$ 13. **Step 6:** Simplify: $$\sqrt{975} = \sqrt{25 \times 39} = 5\sqrt{39}$$ 14. **Final answer:** The area is $$5\sqrt{39}$$ cm².