8th Grade Pythagorean Theorem Practice Questions & Solutions

Solved Example

Easy Level

💡 Question 1: Finding the Hypotenuse
A right triangle has two legs measuring 6 cm and 8 cm. What is the length of its hypotenuse?

Solution & Explanation

👉 Let the legs be $a$ and $b$, and the hypotenuse be $c$.

We are given $a = 6$ cm and $b = 8$ cm.
The Pythagorean Theorem states: $a^2 + b^2 = c^2$.
Substitute the given values into the formula: \[ 6^2 + 8^2 = c^2 \]
Calculate the squares: \[ 36 + 64 = c^2 \]
Add the numbers: \[ 100 = c^2 \]
To find $c$, take the square root of both sides: \[ c = \sqrt{100} \]
Solve for $c$: \[ c = 10 \text{ cm} \]

✅ The length of the hypotenuse is 10 cm.

Solved Example

Easy Level

💡 Question 2: Finding a Leg
The hypotenuse of a right triangle is 17 inches long. One leg of the triangle is 8 inches long. Find the length of the other leg.

Solution & Explanation

👉 Let the hypotenuse be $c$ and the legs be $a$ and $b$.

We are given $c = 17$ inches and one leg, say $a = 8$ inches. We need to find the other leg, $b$.
Use the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the known values: \[ 8^2 + b^2 = 17^2 \]
Calculate the squares: \[ 64 + b^2 = 289 \]
Subtract 64 from both sides to isolate $b^2$: \[ b^2 = 289 - 64 \] \[ b^2 = 225 \]
Take the square root of both sides to find $b$: \[ b = \sqrt{225} \]
Solve for $b$: \[ b = 15 \text{ inches} \]

✅ The length of the other leg is 15 inches.

Solved Example

Medium Level

💡 Question 3: Identifying a Right Triangle
A triangle has side lengths of 7 meters, 24 meters, and 25 meters. Is this a right triangle? Explain your reasoning.

Solution & Explanation

👉 To determine if it's a right triangle, we check if the side lengths satisfy the Pythagorean Theorem ($a^2 + b^2 = c^2$).

In a right triangle, the hypotenuse $c$ is always the longest side. So, let $c = 25$ meters.
Let the other two sides be $a = 7$ meters and $b = 24$ meters.
Substitute these values into the Pythagorean Theorem: \[ 7^2 + 24^2 = 25^2 \]
Calculate the squares: \[ 49 + 576 = 625 \]
Add the numbers on the left side: \[ 625 = 625 \]
Since both sides of the equation are equal, the Pythagorean Theorem holds true for these side lengths.

✅ Yes, this is a right triangle because the sum of the squares of the two shorter sides equals the square of the longest side.

Solved Example

Medium Level

💡 Question 4: Solving with Decimals
Find the length of the hypotenuse of a right triangle whose legs measure 4.5 cm and 6 cm.

Solution & Explanation

👉 Let the legs be $a = 4.5$ cm and $b = 6$ cm. We need to find the hypotenuse $c$.

Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ (4.5)^2 + 6^2 = c^2 \]
Calculate the squares: \[ 20.25 + 36 = c^2 \]
Add the numbers: \[ 56.25 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{56.25} \]
Solve for $c$: \[ c = 7.5 \text{ cm} \]

✅ The length of the hypotenuse is 7.5 cm.

Solved Example

Medium Level

💡 Question 5: Perimeter of a Rectangle
A rectangular park has a length of 120 meters and a diagonal path that is 130 meters long. What is the perimeter of the park?

Solution & Explanation

👉 In a rectangle, the length, width, and diagonal form a right triangle.

We know the length (one leg) $a = 120$ meters.
We know the diagonal (hypotenuse) $c = 130$ meters.
We need to find the width (the other leg) $b$.
Using the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 120^2 + b^2 = 130^2 \]
Calculate the squares: \[ 14400 + b^2 = 16900 \]
Subtract 14400 from both sides: \[ b^2 = 16900 - 14400 \] \[ b^2 = 2500 \]
Take the square root of both sides: \[ b = \sqrt{2500} \] \[ b = 50 \text{ meters} \]
Now that we have the length (120 m) and width (50 m), we can find the perimeter of the rectangle.
The formula for the perimeter of a rectangle is $P = 2 \times (\text{length} + \text{width})$.
Substitute the values: \[ P = 2 \times (120 + 50) \] \[ P = 2 \times (170) \] \[ P = 340 \text{ meters} \]

✅ The perimeter of the park is 340 meters.

Solved Example

Medium Level

💡 Question 6: Distance on a Coordinate Plane
On a coordinate plane, what is the shortest distance between point P(2, 3) and point Q(8, 11)?

Solution & Explanation

👉 The shortest distance between two points on a coordinate plane can be found using the Pythagorean Theorem by creating a right triangle.

First, find the horizontal distance (difference in x-coordinates), which will be one leg of the right triangle: \[ \Delta x = |8 - 2| = 6 \text{ units} \] Let this be $a$.
Next, find the vertical distance (difference in y-coordinates), which will be the other leg: \[ \Delta y = |11 - 3| = 8 \text{ units} \] Let this be $b$.
The shortest distance between P and Q is the hypotenuse $c$ of this right triangle.
Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 6^2 + 8^2 = c^2 \]
Calculate the squares: \[ 36 + 64 = c^2 \]
Add the numbers: \[ 100 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{100} \] \[ c = 10 \text{ units} \]

✅ The shortest distance between point P(2, 3) and point Q(8, 11) is 10 units.

Solved Example

Real World Example

🌍 Question 7: Ladder Against a Wall
A 13-foot ladder is leaning against a wall. The base of the ladder is 5 feet away from the base of the wall. How high up the wall does the ladder reach?

Solution & Explanation

👉 The ladder, the wall, and the ground form a right triangle.

The length of the ladder is the hypotenuse $c = 13$ feet.
The distance from the wall to the base of the ladder is one leg $b = 5$ feet.
The height the ladder reaches up the wall is the other leg $a$, which we need to find.
Using the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the known values: \[ a^2 + 5^2 = 13^2 \]
Calculate the squares: \[ a^2 + 25 = 169 \]
Subtract 25 from both sides: \[ a^2 = 169 - 25 \] \[ a^2 = 144 \]
Take the square root of both sides: \[ a = \sqrt{144} \] \[ a = 12 \text{ feet} \]

✅ The ladder reaches 12 feet high up the wall.

Solved Example

Real World Example

🌍 Question 8: TV Screen Size
A television screen is advertised by the length of its diagonal. If a TV screen is 24 inches wide and 18 inches tall, what is its diagonal size (to the nearest inch)?

Solution & Explanation

👉 The width, height, and diagonal of a TV screen form a right triangle.

The width is one leg $a = 24$ inches.
The height is the other leg $b = 18$ inches.
The diagonal is the hypotenuse $c$, which we need to find.
Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 24^2 + 18^2 = c^2 \]
Calculate the squares: \[ 576 + 324 = c^2 \]
Add the numbers: \[ 900 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{900} \] \[ c = 30 \text{ inches} \]

✅ The diagonal size of the TV screen is 30 inches.

8th Grade Math (Algebra I): Pythagorean Theorem Practice Questions

Example 1:

💡 Question 1: Finding the Hypotenuse
A right triangle has two legs measuring 6 cm and 8 cm. What is the length of its hypotenuse?

Solution:

👉 Let the legs be $a$ and $b$, and the hypotenuse be $c$.

We are given $a = 6$ cm and $b = 8$ cm.
The Pythagorean Theorem states: $a^2 + b^2 = c^2$.
Substitute the given values into the formula: \[ 6^2 + 8^2 = c^2 \]
Calculate the squares: \[ 36 + 64 = c^2 \]
Add the numbers: \[ 100 = c^2 \]
To find $c$, take the square root of both sides: \[ c = \sqrt{100} \]
Solve for $c$: \[ c = 10 \text{ cm} \]

✅ The length of the hypotenuse is 10 cm.

Example 2:

💡 Question 2: Finding a Leg
The hypotenuse of a right triangle is 17 inches long. One leg of the triangle is 8 inches long. Find the length of the other leg.

Solution:

👉 Let the hypotenuse be $c$ and the legs be $a$ and $b$.

We are given $c = 17$ inches and one leg, say $a = 8$ inches. We need to find the other leg, $b$.
Use the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the known values: \[ 8^2 + b^2 = 17^2 \]
Calculate the squares: \[ 64 + b^2 = 289 \]
Subtract 64 from both sides to isolate $b^2$: \[ b^2 = 289 - 64 \] \[ b^2 = 225 \]
Take the square root of both sides to find $b$: \[ b = \sqrt{225} \]
Solve for $b$: \[ b = 15 \text{ inches} \]

✅ The length of the other leg is 15 inches.

Example 3:

💡 Question 3: Identifying a Right Triangle
A triangle has side lengths of 7 meters, 24 meters, and 25 meters. Is this a right triangle? Explain your reasoning.

Solution:

👉 To determine if it's a right triangle, we check if the side lengths satisfy the Pythagorean Theorem ($a^2 + b^2 = c^2$).

In a right triangle, the hypotenuse $c$ is always the longest side. So, let $c = 25$ meters.
Let the other two sides be $a = 7$ meters and $b = 24$ meters.
Substitute these values into the Pythagorean Theorem: \[ 7^2 + 24^2 = 25^2 \]
Calculate the squares: \[ 49 + 576 = 625 \]
Add the numbers on the left side: \[ 625 = 625 \]
Since both sides of the equation are equal, the Pythagorean Theorem holds true for these side lengths.

✅ Yes, this is a right triangle because the sum of the squares of the two shorter sides equals the square of the longest side.

Example 4:

💡 Question 4: Solving with Decimals
Find the length of the hypotenuse of a right triangle whose legs measure 4.5 cm and 6 cm.

Solution:

👉 Let the legs be $a = 4.5$ cm and $b = 6$ cm. We need to find the hypotenuse $c$.

Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ (4.5)^2 + 6^2 = c^2 \]
Calculate the squares: \[ 20.25 + 36 = c^2 \]
Add the numbers: \[ 56.25 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{56.25} \]
Solve for $c$: \[ c = 7.5 \text{ cm} \]

✅ The length of the hypotenuse is 7.5 cm.

Example 5:

💡 Question 5: Perimeter of a Rectangle
A rectangular park has a length of 120 meters and a diagonal path that is 130 meters long. What is the perimeter of the park?

Solution:

👉 In a rectangle, the length, width, and diagonal form a right triangle.

We know the length (one leg) $a = 120$ meters.
We know the diagonal (hypotenuse) $c = 130$ meters.
We need to find the width (the other leg) $b$.
Using the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 120^2 + b^2 = 130^2 \]
Calculate the squares: \[ 14400 + b^2 = 16900 \]
Subtract 14400 from both sides: \[ b^2 = 16900 - 14400 \] \[ b^2 = 2500 \]
Take the square root of both sides: \[ b = \sqrt{2500} \] \[ b = 50 \text{ meters} \]
Now that we have the length (120 m) and width (50 m), we can find the perimeter of the rectangle.
The formula for the perimeter of a rectangle is $P = 2 \times (\text{length} + \text{width})$.
Substitute the values: \[ P = 2 \times (120 + 50) \] \[ P = 2 \times (170) \] \[ P = 340 \text{ meters} \]

✅ The perimeter of the park is 340 meters.

Example 6:

💡 Question 6: Distance on a Coordinate Plane
On a coordinate plane, what is the shortest distance between point P(2, 3) and point Q(8, 11)?

Solution:

👉 The shortest distance between two points on a coordinate plane can be found using the Pythagorean Theorem by creating a right triangle.

First, find the horizontal distance (difference in x-coordinates), which will be one leg of the right triangle: \[ \Delta x = |8 - 2| = 6 \text{ units} \] Let this be $a$.
Next, find the vertical distance (difference in y-coordinates), which will be the other leg: \[ \Delta y = |11 - 3| = 8 \text{ units} \] Let this be $b$.
The shortest distance between P and Q is the hypotenuse $c$ of this right triangle.
Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 6^2 + 8^2 = c^2 \]
Calculate the squares: \[ 36 + 64 = c^2 \]
Add the numbers: \[ 100 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{100} \] \[ c = 10 \text{ units} \]

✅ The shortest distance between point P(2, 3) and point Q(8, 11) is 10 units.

Example 7:

🌍 Question 7: Ladder Against a Wall
A 13-foot ladder is leaning against a wall. The base of the ladder is 5 feet away from the base of the wall. How high up the wall does the ladder reach?

Solution:

👉 The ladder, the wall, and the ground form a right triangle.

The length of the ladder is the hypotenuse $c = 13$ feet.
The distance from the wall to the base of the ladder is one leg $b = 5$ feet.
The height the ladder reaches up the wall is the other leg $a$, which we need to find.
Using the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the known values: \[ a^2 + 5^2 = 13^2 \]
Calculate the squares: \[ a^2 + 25 = 169 \]
Subtract 25 from both sides: \[ a^2 = 169 - 25 \] \[ a^2 = 144 \]
Take the square root of both sides: \[ a = \sqrt{144} \] \[ a = 12 \text{ feet} \]

✅ The ladder reaches 12 feet high up the wall.

Example 8:

Solution:

👉 The width, height, and diagonal of a TV screen form a right triangle.

The width is one leg $a = 24$ inches.
The height is the other leg $b = 18$ inches.
The diagonal is the hypotenuse $c$, which we need to find.
Apply the Pythagorean Theorem: $a^2 + b^2 = c^2$.
Substitute the values: \[ 24^2 + 18^2 = c^2 \]
Calculate the squares: \[ 576 + 324 = c^2 \]
Add the numbers: \[ 900 = c^2 \]
Take the square root of both sides: \[ c = \sqrt{900} \] \[ c = 30 \text{ inches} \]

✅ The diagonal size of the TV screen is 30 inches.

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💡 8th Grade Math (Algebra I): Pythagorean Theorem Practice Questions

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