8th Grade Math Flashcards: Apply Pythagorean Theorem To Problems

What this deck covers

This deck focuses on Apply Pythagorean Theorem To Problems, giving you a quick way to review the definitions, rules, and examples that matter most for 8th Grade Math.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find the length of the diagonal of a $7$ by $24$ rectangle.

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

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

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

Flashcard 3: State the Pythagorean Theorem formula for a right triangle with legs $a$, $b$ and hypotenuse $c$.

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

Flashcard 4: State the formula to find a missing leg $a$ when the hypotenuse is $c$ and the other leg is $b$.

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

Flashcard 5: State the formula to find the hypotenuse $c$ when the legs are $a$ and $b$.

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

Flashcard 6: Find the distance between points $(2,3)$ and $(5,7)$ using the Pythagorean Theorem.

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

Flashcard 7: Find the hypotenuse $c$ when the legs are $a = 3$ and $b = 4$.

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

Flashcard 8: A ladder is $10$ ft long and reaches $6$ ft high. What is the distance from the wall to the ladder base?

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

Flashcard 9: A right triangle has legs $7$ and $24$. What is the hypotenuse?

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

Flashcard 10: Find the missing leg $b$ when $c = 13$ and $a = 5$.

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

Flashcard 11: Find the missing leg $a$ when $c = 10$ and $b = 6$.

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

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

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

Flashcard 13: Find the length of the hypotenuse when the legs are $a = 8$ and $b = 15$.

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

Flashcard 14: Find the diagonal of a $6$ by $8$ rectangle using the Pythagorean Theorem.

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

Flashcard 15: A rectangle has diagonal $13$ and one side $5$. Find the other side length.

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

Flashcard 16: Identify whether the side lengths $5$, $6$, and $7$ form a right triangle using $a^2+b^2=c^2$.

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

Flashcard 17: Find the distance between points $(0,0)$ and $(6,8)$ using the Pythagorean Theorem.

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

Flashcard 18: Identify whether the side lengths $6$, $8$, and $10$ form a right triangle using $a^2+b^2=c^2$.

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

Flashcard 19: Find the diagonal of a square with side length $9$ using the Pythagorean Theorem.

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

Flashcard 20: Find the hypotenuse when the legs are $5$ and $12$.

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

Flashcard 21: Find the hypotenuse when the legs are $9$ and $12$.

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

Flashcard 22: What is the formula for the hypotenuse $c$ in terms of legs $a$ and $b$?

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

Flashcard 23: Find the missing leg when $c=13$ and one leg is $5$.

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

Flashcard 24: Find the hypotenuse when the legs are $6$ and $8$.

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

Flashcard 25: What is the distance formula between $(x_1,y_1)$ and $(x_2,y_2)$ from the Pythagorean Theorem?

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

Flashcard 26: State the Pythagorean Theorem for a right triangle with legs $a$ and $b$ and hypotenuse $c$.

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

Flashcard 27: Identify which side is the hypotenuse in a right triangle.

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

Flashcard 28: Identify whether sides $7$, $24$, and $25$ form a right triangle.

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

Flashcard 29: A square has side length $9$. What is the length of its diagonal?

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

Flashcard 30: What is the formula for a leg $a$ in terms of hypotenuse $c$ and other leg $b$?

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