8th Grade · 8.G.B.6 · 8.G.B.7

The Pythagorean theorem in 8th grade

The Pythagorean theorem is one of the big ideas of 8th grade math. In every right triangle, the two shorter sides and the longest side fit one rule: $a^2 + b^2 = c^2$ . Your child learns why the rule is true, how to use it to find a missing side, and how to solve Pythagorean theorem word problems about ladders, screens and boxes.

What the standards say

The official Common Core wording, quoted exactly. The plain version follows.

8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse.

8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.

What it means, in plain words

Why the rule is true, and its converse 8.G.B.6

A right triangle has one square corner. The two sides that make that corner are called the legs, $a$ and $b$ . The side across from the corner is the hypotenuse, $c$ , and it is always the longest side.

The theorem says that if you draw a square on each side, the two smaller squares together have exactly the same area as the big one. In 8th grade, children see a proof of the Pythagorean theorem made from pictures: four copies of the same triangle fit inside one large square in two different ways, and the space left over shows that $a^2 + b^2$ and $c^2$ are the same area.

The converse of the Pythagorean theorem turns the rule around. If the three sides of a triangle fit $a^2 + b^2 = c^2$ , the triangle must have a right angle. Builders use this to check that a corner is square.

Your child should be able to:

Name the legs and the hypotenuse of a right triangle
Explain with a picture why $a^2 + b^2 = c^2$ is true
Use the converse to decide whether three lengths make a right triangle

Example

Is a triangle with sides 5, 12 and 13 a right triangle?

The longest side is 13, so compare $5^2 + 12^2$ with $13^2$ .
$5^2 + 12^2 = 25 + 144 = 169$ , and $13^2 = 169$ .
They match, so the triangle has a right angle. With sides 5, 12 and 14 they would not match, because $14^2 = 196$ .

Finding the missing side of a right triangle 8.G.B.7

Most of the year’s work is finding the missing side of a right triangle. When your child knows two sides, the theorem gives the third. To find the hypotenuse, add the two squares and take the square root. To find a leg, subtract instead: $b^2 = c^2 - a^2$ .

Pythagorean theorem word problems hide the triangle inside a real situation: a ladder against a wall, the diagonal of a TV screen, a shortcut across a park. The first step is always to find the square corner and decide which side is the hypotenuse. In 8th grade the triangle can also sit inside a 3D shape, such as the slant height of a cone or the diagonal across a box.

Your child should be able to:

Find the hypotenuse when both legs are known
Find a leg when the hypotenuse and the other leg are known
Spot the right triangle in a word problem or inside a 3D shape
Round a square root sensibly when the answer is not a whole number

Example

A 13-foot ladder leans against a wall, with its foot 5 feet from the wall. How high up the wall does it reach?

The ladder is the hypotenuse, so $c = 13$ . The ground is one leg, so $a = 5$ .
$b^2 = 13^2 - 5^2 = 169 - 25 = 144$
$b = \sqrt{144} = 12$ , so the ladder reaches 12 feet up the wall.

Where children usually get stuck

Your child adds the squares even when the longest side is the one given, and gets a wall taller than the ladder.
The rule was learned as "add the squares", so the child adds whichever two numbers the problem gives without asking which side is the hypotenuse.
What helps. Before any numbers, ask where the square corner is and which side is across from it. That side is $c$ , and the answer has to make it the longest.
For legs of 3 and 4, your child writes $3 + 4 = 7$ or $6 + 8 = 14$ .
Squaring gets skipped or mixed up with doubling, and 7 or 14 looks like a reasonable length.
What helps. Draw the squares on graph paper once. A 3 by 3 square holds 9 small squares, not 6. Seeing it once makes the rule stick.
Your child gets $c^2 = 25$ and writes 25 as the length of the side.
After the arithmetic, the last step, taking the square root, gets forgotten.
What helps. Ask whether the answer makes sense. A side of 25 on a triangle with legs of 3 and 4 cannot be right, and that check catches the missing step.

Try four practice questions

Real 8th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Assume a, b, and c are side lengths, and c is the longest side. Is Marcus's claim really the converse? Select the best answer.
$\displaystyle \text{Pythagorean Theorem: If the triangle is right, then } a^2 + b^2 = c^2$
$\displaystyle \text{Marcus's claim: If the triangle is right, then } a^2 + b^2 = c^2$
A
It kept the original direction instead of reversing it.
B
It reversed the order and added 'not' to both parts.
C
It changed the conclusion instead of reversing the theorem.
D
It is the correct converse written a different way.
Show the answer and how to solve it
The answer is 'It kept the original direction instead of reversing it' because Marcus simply repeated the theorem without swapping its two parts.
1. 1Start with the original theorem. It says: if the triangle is right, then $a^2 + b^2 = c^2$ .
2. 2The true converse swaps those two parts: the 'if' part becomes $a^2 + b^2 = c^2$ , and the 'then' part becomes 'the triangle is right.'
3. 3Marcus's claim has the same 'if' and 'then' as the original theorem — nothing was swapped. So it kept the original direction instead of reversing it.
Common mistake. A common mistake is thinking the claim looks slightly different so it must be the converse. Marcus's claim is identical to the original theorem — the 'if' and 'then' parts are in the same order, not reversed.
The idea underneath. A converse keeps the same condition and conclusion but swaps their order; changing them into 'not' statements creates a different statement.
2On level
Look at the right cone below. What is the missing slant height?
Slant height:units
Show the answer and how to solve it
Answer. 5
The slant height is 5 units because the radius 3 and height 4 form a 3-4-5 right triangle.
1. 1Look at the cone. The slant height, the vertical height, and the radius form a right triangle inside the cone, with the slant height as the hypotenuse.
2. 2Use s for the slant height and apply the Pythagorean Theorem with the radius and height as the two legs.
3. 3Add the squares and take the square root.
Common mistake. A common mistake is using the diameter instead of the radius. The right triangle inside the cone uses the radius (3), not the full diameter (6).
The idea underneath. In a right cone, the slant height is the hypotenuse of a right triangle whose legs are the vertical height and the radius.

3On level

Fill in each box to complete the simplification chain.

Start	$\displaystyle c^2 + 2ab$	$\displaystyle =$	$\displaystyle (a+b)^2$
Step 1	$\displaystyle c^2 + 2ab$	$\displaystyle =$	$\displaystyle a^2 +$ $\displaystyle + b^2$
Step 2	$\displaystyle c^2$	$\displaystyle =$

Show the answer and how to solve it

The missing expressions are $2ab$ and $a^{2} + b^{2}$ . Expanding $(a+b)^{2}$ gives $a^{2} + 2ab + b^{2}$ , and canceling $2ab$ from both sides leaves $c^{2} = a^{2} + b^{2}$ .

1Multiply each term in the first parentheses by each term in the second. The two $ab$ terms add to $2ab$ . Expand $(a+b)^{2}$ using the perfect-square identity.
2The first missing expression is the middle term from the expansion: $2ab$ .
3Subtract $2ab$ from both sides so the equal terms cancel.
4After canceling, the second missing expression is $a^{2} + b^{2}$ , yielding the Pythagorean theorem.

Common mistake. A common mistake is expanding $(a+b)^{2}$ as $a^{2} + b^{2}$ , forgetting the middle term $2ab$ . Another error is subtracting $2ab$ from only one side, which unbalances the equation.

The idea underneath. The proof works because expanding $(a+b)^{2}$ creates the same $2ab$ term that already appears on the other side, so those equal terms cancel and only the square terms remain.

4Stretch

Maya is finding the diagonal of the front face of this rectangular prism. What mistake did Maya make?

Maya's work is shown below.

Setup	$\displaystyle d$	$\displaystyle =$	$\displaystyle \sqrt{6^2 + 8^2}$
Simplify	$\displaystyle d$	$\displaystyle =$	$\displaystyle \sqrt{36 + 64}$
Simplify	$\displaystyle d$	$\displaystyle =$	$\displaystyle \sqrt{100}$
Finish	$\displaystyle d$	$\displaystyle =$	$\displaystyle 100$

Maya used the side lengths from a different visible face of the prism.

Maya forgot to take the square root at the end.

Maya used all 3 dimensions, so she found a space diagonal instead of a face diagonal.

Maya added the side lengths directly instead of using the Pythagorean theorem.

Maya forgot to square one of the side lengths.

Show the answer and how to solve it

The correct choice is that Maya forgot to take the square root at the end. Her work correctly reached $\sqrt{100}$ but then wrote 100 as the answer instead of 10.

1Look at the prism. The front face uses the 6 cm width and the 8 cm height.
2Now look at Maya's work. The setup and simplification are correct through the line $d = \sqrt{100}$ , but the final line writes $d = 100$ instead of evaluating the square root.
3 $\sqrt{100} = 10$ , not 100. Writing the radicand as the answer skips the square root step entirely.
4The correct final step evaluates $\sqrt{100} = 10$ , so the diagonal of the front face is 10 cm.
5The mistake is that Maya forgot to take the square root at the end.

Common mistake. A common mistake is treating the value under the radical as the final answer. After reaching $d = \sqrt{100}$ , the last step must evaluate the square root: $d = 10$ , not $d = 100$ .

The idea underneath. A face diagonal is the hypotenuse of a right triangle inside one face, so you use only the two side lengths of that face.