GCSE Pythagoras’ theorem Practice Questions

State Pythagoras' theorem.

In a right-angled triangle the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides: $a^2 + b^2 = c^2$.

Write the theorem out for this triangle.

The hypotenuse $c$ is the side opposite the right angle; the two shorter sides are $3$ cm and $4$ cm.

State the length of the hypotenuse.

The hypotenuse is $5$ cm.

State Pythagoras' theorem.

In a right-angled triangle the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides: $a^2 + b^2 = c^2$.

Write the theorem out for this triangle.

The hypotenuse is $25$ cm, so $25^2$ sits on its own; the two shorter sides are $7$ cm and $x$ cm.

Select the correct equation.

To find a shorter side, subtract the square of the known shorter side from the square of the hypotenuse.

Work out the horizontal and vertical steps from A to B.

The steps are $6$ across and $8$ up, which are the two shorter sides of a right-angled triangle with $AB$ as its hypotenuse.

Apply the theorem to that triangle.

The distance is the hypotenuse, so its SQUARE is the sum of the squares of the steps. The distance itself is the square root of that.

Take the square root of both sides.

The square root has to be taken at the end, so the whole sum must sit under the root sign.

Evaluate the correct expression.

The distance $AB$ is $10$ units.

Note why leaving out the root is wrong.

That is the SQUARE of the distance, not the distance: it is far too big.

Select the correct calculation.

Square the steps, add them, then take the square root: $\sqrt{6^2 + 8^2} = 10$.

State Pythagoras' theorem.

In a right-angled triangle the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides: $a^2 + b^2 = c^2$.

Work out the square of the hypotenuse.

The two shorter sides square to $40$, so $c = \sqrt{40}$ — and $40$ is not a square number, so the answer is a surd.

Find the largest square factor of the number under the root.

$4$ is a square number and $10$ has no square factors left, so $4 \times 10$ is the split to use.

Split the surd and take the root of the square factor.

$\sqrt{4} = 2$, so the surd simplifies to $2\sqrt{10}$.

Check the simplified surd by squaring it.

Squaring $2\sqrt{10}$ gives back $40$, so the simplification is right.

Square each of the other options to rule them out.

Squaring an option must give back $40$. None of these does, so none of them is the hypotenuse.

Check the surd cannot be simplified any further.

No prime is repeated in $2 \times 5$, so $10$ has no square factor and $\sqrt{10}$ is in its simplest form.

Check the size of the answer as a decimal.

The hypotenuse is about $6.32$ cm, which is bigger than both $2$ cm and $6$ cm, as it must be.

Note the standard mistake with surds.

The square root of a sum is not the sum of the square roots. The whole total must go under the root sign first.

Select the exact length of the hypotenuse.

The exact hypotenuse is $2\sqrt{10}$ cm.

State the converse of Pythagoras' theorem.

A triangle is right-angled exactly when the square of its longest side equals the sum of the squares of the other two sides. So square the three lengths and test $a^2 + b^2 = c^2$, with $c$ the longest.

Square the two shorter sides and add them.

The two shorter sides are $5$ cm and $9$ cm, and their squares add to $106$.

Square the longest side.

The longest side is $10$ cm, and $10^2 = 100$.

Compare the two results.

$106$ is greater than $100$, so the two are not equal.

Say what that means for the triangle.

Because the squares do not balance, the triangle is not right-angled. In fact the angle opposite the longest side is acute.

Rule out the equality option.

The equality would mean a right angle, and the numbers say otherwise.

Rule out the inequality the other way round.

The comparison only works one way, and it is $106 > 100$.

Rule out the options that use the wrong pair of sides.

Any test has to put the LONGEST side on its own. Pairing $5$ with $10$ and comparing with $9$ gets the roles the wrong way round, and the numbers do not work either.

Rule out the last option.

The squares of the two largest sides certainly do not come to less than the square of the smallest.

Check the three lengths do make a triangle at all.

The two shorter sides together beat the longest side, so the triangle exists — it just is not right-angled.

Work out what the longest side would have to be for a right angle.

A right angle would need the longest side to be about $10.3$ cm, and it is $10$ cm instead.

Say what the comparison means for the largest angle.

When the sum of the squares of the two shorter sides is greater than the square of the longest side, the angle opposite the longest side is acute, smaller than $90^\circ$.

Note that scaling the triangle would not change the verdict.

Doubling every side multiplies every square by $4$, so the comparison — and so the shape of the triangle — stays exactly the same.

Note the standard mistake of picking the wrong hypotenuse.

The longest side is $10$ cm, so that is the only one that could be a hypotenuse. Testing with $9$ cm or $5$ cm on its own would be meaningless.

Select the correct statement.

$106 > 100$, so the correct statement is $5^2 + 9^2 > 10^2$.