Apply Pythagorean Theorem to Problems - 8th Grade Math

1. Find the missing leg $b$ if $c=10$ and $a=6$ in a right triangle.

   $b=8$. $10^2-6^2=100-36=64$, so $b=\sqrt{64}=8$.

2. Find the distance between points $(0,0)$ and $(6,8)$ using the Pythagorean Theorem.

   $10$. Distance formula uses Pythagorean theorem: $\sqrt{6^2 + 8^2} = 10$.

3. Identify whether $c$ is a leg or the hypotenuse in the equation $a^2+b^2=c^2$.

   $c$ is the hypotenuse. In the Pythagorean theorem, $c$ represents the longest side.

4. A ladder is $10$ ft long and reaches $6$ ft high. What is the distance from the wall to the ladder base?

   $8$ ft. Base distance: $\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = 8$ ft.

5. What is the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ from the Pythagorean Theorem?

   $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Treats horizontal and vertical distances as legs of a right triangle.

6. Identify whether the side lengths $6$, $8$, and $10$ form a right triangle using $a^2+b^2=c^2$.

   Yes, because $6^2 + 8^2 = 10^2$. Check: $36 + 64 = 100$, which equals $10^2$.

7. State the formula for a missing leg $a$ in terms of hypotenuse $c$ and leg $b$.

   $a=\sqrt{c^2-b^2}$. Rearrange $a^2+b^2=c^2$ to isolate $a^2$, then take square root.

8. Find the missing leg when $c=17$ and one leg is $15$.

   $8$. $\sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8$.

9. A square has side length $9$. What is the length of its diagonal?

   $9\sqrt{2}$. Diagonal forms hypotenuse with two sides: $\sqrt{9^2 + 9^2} = 9\sqrt{2}$.

10. Identify whether the side lengths $5$, $6$, and $7$ form a right triangle using $a^2+b^2=c^2$.

    No, because $5^2 + 6^2 \ne 7^2$. Check: $25 + 36 = 61 \ne 49$, so not a right triangle.

11. Find the missing leg when $c=13$ and one leg is $5$.

    $12$. $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.

12. Find the diagonal of a $6$ by $8$ rectangle using the Pythagorean Theorem.

    $10$. Diagonal forms hypotenuse: $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10$.

13. State the Pythagorean Theorem formula for a right triangle with legs $a,b$ and hypotenuse $c$.

    $a^2+b^2=c^2$. In right triangles, the sum of squares of legs equals the square of the hypotenuse.

14. Find $c$ for a right triangle with legs $a=5$ and $b=12$.

    $c=13$. $5^2+12^2=25+144=169$, so $c=\sqrt{169}=13$.

15. State the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ using the Pythagorean Theorem.

    $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The Pythagorean theorem applied to coordinate differences.

16. Find the distance between $(2,1)$ and $(6,4)$.

    $5$. $\sqrt{(6-2)^2 + (4-1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

17. Find the missing leg $b$ when $c = 13$ and $a = 5$.

    $b = 12$. $b^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $b = 12$.

18. Find the diagonal of a square with side length $9$ using the Pythagorean Theorem.

    $9\sqrt{2}$. Square diagonal: $\sqrt{9^2 + 9^2} = \sqrt{162} = 9\sqrt{2}$.

19. Find the distance between points $(1,1)$ and $(4,5)$ on a coordinate plane.

    $5$. $\sqrt{(4-1)^2+(5-1)^2}=\sqrt{9+16}=\sqrt{25}=5$.

20. A ladder is $10$ ft long and reaches $8$ ft high. How far is the base from the wall?

    $6\text{ ft}$. $\sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ ft.

21. Find the distance between points $(2,3)$ and $(8,3)$ on a coordinate plane.

    $6$. Points have same $y$-coordinate, so distance is $|8-2|=6$.

22. Identify which side is the hypotenuse in a right triangle.

    The side opposite the $90^\circ$ angle. The hypotenuse is always the longest side in a right triangle.

23. Find the length of the hypotenuse when the legs are $a = 8$ and $b = 15$.

    $17$. $\sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$.

24. Identify the error: A student wrote $c=\sqrt{a^2-b^2}$ for the hypotenuse. What is correct?

    $c=\sqrt{a^2+b^2}$. The hypotenuse formula adds, not subtracts, the squared legs.

25. A right triangle has legs $7$ and $24$. What is the hypotenuse?

    $25$. $\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.

26. What is the hypotenuse in a right triangle: the side opposite which angle?

    Opposite the $90^\circ$ angle. The hypotenuse is the longest side in a right triangle.

27. Find the distance between points $(2,3)$ and $(5,7)$ using the Pythagorean Theorem.

    $5$. $\sqrt{(5-2)^2 + (7-3)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.

28. Identify which side is the hypotenuse in a right triangle when using $a^2 + b^2 = c^2$.

    The side opposite the $90^\circ$ angle, labeled $c$. The hypotenuse is always the longest side, opposite the right angle.

29. A rectangle has diagonal $13$ and one side $5$. Find the other side length.

    $12$. $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.

30. State the formula to find a missing leg $a$ when the hypotenuse is $c$ and the other leg is $b$.

    $a = \sqrt{c^2 - b^2}$. Rearrange $a^2 + b^2 = c^2$ to isolate $a^2$, then take the square root.

31. State the Pythagorean Theorem for a right triangle with legs $a$ and $b$ and hypotenuse $c$.

    $a^2 + b^2 = c^2$. Relates the squares of all three sides in a right triangle.

32. Find the hypotenuse when the legs are $6$ and $8$.

    $10$. $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

33. Find the hypotenuse $c$ when the legs are $a = 3$ and $b = 4$.

    $c = 5$. $3^2 + 4^2 = 9 + 16 = 25$, so $c = \sqrt{25} = 5$.

34. Find the distance between $(0,0)$ and $(3,4)$.

    $5$. $\sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

35. Find the hypotenuse when the legs are $5$ and $12$.

    $13$. $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

36. What is the formula for the hypotenuse $c$ in terms of legs $a$ and $b$?

    $c = \sqrt{a^2 + b^2}$. Rearrange the Pythagorean theorem to solve for the hypotenuse.

37. A rectangle is $9$ by $12$. What is the diagonal length?

    $15$. Diagonal $=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$.

38. State the formula to find the hypotenuse $c$ when the legs are $a$ and $b$.

    $c = \sqrt{a^2 + b^2}$. Take the square root of the sum of the squared legs.

39. Find the distance between $(-1,-2)$ and $(2,2)$.

    $5$. $\sqrt{(2-(-1))^2 + (2-(-2))^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

40. Find $c$ for a right triangle with legs $a=6$ and $b=8$.

    $c=10$. $6^2+8^2=36+64=100$, so $c=\sqrt{100}=10$.

41. What is the formula for a leg $a$ in terms of hypotenuse $c$ and other leg $b$?

    $a = \sqrt{c^2 - b^2}$. Rearrange $a^2 + b^2 = c^2$ to isolate $a$.

42. Find the distance between points $(0,0)$ and $(6,8)$ on a coordinate plane.

    $10$. Distance $=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$.

43. State the Pythagorean Theorem formula for a right triangle with legs $a$, $b$ and hypotenuse $c$.

    $a^2 + b^2 = c^2$. The sum of the squares of the legs equals the square of the hypotenuse.

44. A square has side length $6$. What is the diagonal length?

    $6\sqrt{2}$. Square diagonal $=\sqrt{6^2+6^2}=\sqrt{72}=6\sqrt{2}$.

45. Find the missing leg $a$ when $c = 10$ and $b = 6$.

    $a = 8$. $a^2 = 10^2 - 6^2 = 100 - 36 = 64$, so $a = 8$.

46. Find the hypotenuse $c$ when the legs are $a = 5$ and $b = 12$.

    $c = 13$. $5^2 + 12^2 = 25 + 144 = 169$, so $c = \sqrt{169} = 13$.

47. Identify whether sides $7$, $24$, and $25$ form a right triangle.

    Yes, because $7^2+24^2=25^2$. Check if $a^2 + b^2 = c^2$: $49 + 576 = 625$ ✓

48. State the formula for the hypotenuse length $c$ in terms of legs $a$ and $b$.

    $c=\sqrt{a^2+b^2}$. Take the square root of the sum of squared legs.

49. Find the length of the diagonal of a $7$ by $24$ rectangle.

    $25$. $\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$.

50. Find the missing leg $a$ if $c=13$ and $b=5$ in a right triangle.

    $a=12$. $13^2-5^2=169-25=144$, so $a=\sqrt{144}=12$.

51. Find the hypotenuse when the legs are $9$ and $12$.

    $15$. $\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$.

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