1. Problem (i): Given squares with areas 7 cm², 17 cm², and 10 cm², determine if they surround a right angled triangle. 2. Step 1: Recall the Pythagorean theorem states for a right angled triangle with sides $a$, $b$, and hypotenuse $c$, the areas satisfy $a^2 + b^2 = c^2$. 3. Step 2: Check if any two areas sum to the third: - $7 + 10 = 17$ (True) - $7 + 17 = 24$ (False) - $10 + 17 = 27$ (False) 4. Step 3: Since $7 + 10 = 17$, the squares can exactly surround a right angled triangle. 5. Problem (ii): Given squares with areas 13 cm², 10 cm², and 22 cm², check if they surround a right angled triangle. 6. Step 1: Check sums: - $10 + 13 = 23$ (False) - $10 + 22 = 32$ (False) - $13 + 22 = 35$ (False) 7. Step 2: No two areas sum to the third, so they do not surround a right angled triangle. 8. Problem: Area of square on hypotenuse of right angled isosceles triangle is 24 cm². Find areas of squares on other two sides. 9. Step 1: In a right angled isosceles triangle, the legs are equal, so if each leg has length $x$, hypotenuse length is $x\sqrt{2}$. 10. Step 2: Area of square on hypotenuse is $24 = (x\sqrt{2})^2 = 2x^2$. 11. Step 3: Solve for $x^2$: $x^2 = \frac{24}{2} = 12$. 12. Step 4: Areas of squares on other two sides are each $12$ cm². 13. Problem: Area of square on hypotenuse of right angled isosceles triangle is 30 cm². Find areas of squares on other two sides. 14. Step 1: Using same logic, $30 = 2x^2$. 15. Step 2: Solve for $x^2$: $x^2 = \frac{30}{2} = 15$. 16. Step 3: Areas of squares on other two sides are each $15$ cm². 17. Problem: Three squares exactly surround a right angled triangle. Areas on two smaller sides are 9 cm² and 12 cm². Find area of third square. 18. Step 1: Sum of areas on smaller sides: $9 + 12 = 21$. 19. Step 2: Area of square on hypotenuse is $21$ cm². 20. Problem (a): Right angled triangle at L, sides QL = 12 cm, QR = 20 cm. Find length LR. 21. Step 1: Use Pythagoras: $QR^2 = QL^2 + LR^2$. 22. Step 2: $20^2 = 12^2 + LR^2$. 23. Step 3: $400 = 144 + LR^2$. 24. Step 4: $LR^2 = 400 - 144 = 256$. 25. Step 5: $LR = \sqrt{256} = 16$ cm. 26. Problem (b): Right angled triangle at Y, sides XY = 10 cm, YZ = 8 cm. Find length XZ. 27. Step 1: $XZ^2 = XY^2 + YZ^2 = 10^2 + 8^2 = 100 + 64 = 164$. 28. Step 2: $XZ = \sqrt{164} = 2\sqrt{41} \approx 12.81$ cm. 29. Problem (c): Right angled triangle at Q, sides PQ = 6 cm, PR = 10 cm. Find length QR. 30. Step 1: Since right angle at Q, $PR^2 = PQ^2 + QR^2$. 31. Step 2: $10^2 = 6^2 + QR^2$. 32. Step 3: $100 = 36 + QR^2$. 33. Step 4: $QR^2 = 64$. 34. Step 5: $QR = \sqrt{64} = 8$ cm. Final answers: (i) Yes, for areas 7, 10, 17. (ii) No, for areas 13, 10, 22. Areas on other sides for hypotenuse 24: 12 and 12. Areas on other sides for hypotenuse 30: 15 and 15. Third square area: 21. Lengths: LR = 16 cm, XZ = $2\sqrt{41}$ cm, QR = 8 cm.