Hard GCSE Pythagoras’ theorem Questions

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 $5^2 + 9^2 > 10^2$.

Note that the distance from the origin comes from Pythagoras.

The horizontal and vertical steps from the origin $(0, 0)$ to a point $(x, y)$ are $x$ and $y$, and they meet at a right angle, so $d^2 = x^2 + y^2$.

Note that the squared distances can be compared directly.

Distances are never negative, so whichever point has the biggest $x^2 + y^2$ is the furthest away. There is no need to take any square roots.

Work out the squared distance to (3, 8).

The point $(3, 8)$ is $\sqrt{73}$ from the origin.

Work out the squared distance to (8, 2).

The point $(8, 2)$ is $\sqrt{68}$ from the origin.

Work out the squared distance to (7, 4).

The point $(7, 4)$ is $\sqrt{65}$ from the origin.

Work out the squared distance to (1, 8).

The point $(1, 8)$ is $\sqrt{65}$ from the origin.

Work out the squared distance to (6, 5).

The point $(6, 5)$ is $\sqrt{61}$ from the origin.

Compare the five squared distances.

The largest is $73$, and no other point matches it, so there is a clear winner.

Work out the actual distance to the furthest point.

The furthest point is about $8.54$ units from the origin.

Work out the distance to the runner-up.

The next furthest point is about $8.25$ units away, so the gap is real and not a rounding effect.

Note the standard mistake.

The point with the biggest single coordinate need not be the furthest away: both coordinates count, through $x^2 + y^2$.

Note what the answer means geometrically.

All the points at the winning distance lie on the circle $x^2 + y^2 = 73$, and every other point given lies inside it.

Check the winner against the smallest of the five.

The nearest point has $x^2 + y^2 = 61$, a long way below $73$, so the ordering is not close anywhere near the ends.

Note that the signs of the coordinates would not matter.

Squaring removes any minus signs, so a point and its reflections in the axes are all the same distance from the origin.

Select the point furthest from the origin.

$(3, 8)$ has the largest value of $x^2 + y^2$, namely $73$, so it is the furthest from the origin.

Draw the right-angled triangle with AB as its hypotenuse.

Going across and then up from $A$ to $B$ makes a right-angled triangle whose hypotenuse is $AB$, so the distance is found with Pythagoras.

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

Subtracting the coordinates gives the two legs of the triangle: $9$ across and $6$ up. Their signs do not matter, because both get squared.

Square the length of the horizontal step.

$9 \times 9 = 81$.

Square the length of the vertical step.

$6 \times 6 = 36$.

Add the two squares together.

Pythagoras' theorem says the square of the hypotenuse is the sum of the squares of the other two sides, so $d^2 = 117$. This is the SQUARE of the length, not the length.

Take the positive square root.

A length cannot be negative, so take the positive square root of $117$.

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

$9$ is a square number and $13$ has no square factors left, so $9 \times 13$ is the split to use.

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

$\sqrt{9} = 3$, so the surd simplifies to $3\sqrt{13}$.

Check the simplified surd by squaring it.

Squaring $3\sqrt{13}$ gives back $117$, so the simplification is right.

Check the surd cannot be simplified any further.

No prime is repeated in $13$, so $13$ has no square factor and $\sqrt{13}$ is in its simplest form.

Check what the answer is measured in.

The points are given on a coordinate grid, so the distance is measured in grid units, not in centimetres.

Check that the hypotenuse is the longest side.

The hypotenuse must be longer than either of the other two sides, and $117$ is bigger than both $81$ and $36$, so it is.

Check the triangle inequality.

The two shorter sides together must be longer than the hypotenuse. Squaring both, $225 > 117$, so they are.

Check the answer by substituting it back.

The squares of the two shorter sides add to $117$, which is exactly the square of the answer, so the answer fits the theorem.

State the distance between the two points.

The distance $AB$ is $3\sqrt{13}$ units.

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 and rearrange it.

The hypotenuse is $21$ cm, so it goes on its own on one side of the theorem. The missing side is one of the SHORTER sides, so subtract.

Square the length of the hypotenuse.

$21 \times 21 = 441$.

Square the length of the known shorter side.

$13 \times 13 = 169$.

Subtract to find the square of the missing side.

Rearranging $a^2 + b^2 = c^2$ gives $x^2 = c^2 - a^2$: to find a SHORTER side you subtract, you do not add.

Take the positive square root.

A length cannot be negative, so take the positive square root of $272$.

Work out the square root and round it.

$\sqrt{272} = 16.49242\ldots$, and rounding to $2$ decimal places gives $16.49$.

Check the rounded answer by squaring it.

Squaring the rounded answer gives $271.9201$, which is very close to $272$, so the rounding is right.

Estimate the size of the answer before working it out.

$272$ lies between the square numbers $256$ and $289$, so the answer lies between $16$ and $17$. Any answer outside that range would be wrong.

Check the units.

Every length in the question is in cm, so the missing side is in cm too.

Check the missing side is shorter than the hypotenuse.

The hypotenuse is the longest side, so the answer must be less than $21$, and $272 < 441$ confirms it.

Check the answer by putting it back into the theorem.

The squares of the two shorter sides add to $441$, the square of the hypotenuse, so the answer fits.

Note the standard mistake in this type of question.

Adding the squares here would make the missing side longer than the hypotenuse, which is impossible. Subtract.

Note the size of the exact answer as a decimal.

The exact answer $16.49$ is about $16.5$, which is a sensible length for this triangle.

State the length of the missing side.

The other shorter side is $16.49$ 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: $a^2 + b^2 = c^2$.

Note that AC is shared by the two triangles.

$AC$ is the hypotenuse of triangle $ABC$ and one of the shorter sides of triangle $ACD$, so finding $AC$ links the two triangles together.

Work out the square of AC in the first triangle.

Angle $ABC$ is the right angle, so $AB$ and $BC$ are the two shorter sides and $AC^2 = 113$.

Keep AC as its square for the moment.

$AC^2 = 113$ is all the next triangle needs, so there is no point rounding $\sqrt{113}$ here — rounding early would spoil the final answer.

Apply the theorem in the second triangle.

Angle $ACD$ is the right angle, so $AD$ is the hypotenuse of triangle $ACD$ and $CD$ is a shorter side. Subtract.

Substitute the two squares.

$15^2 = 225$, and taking away $AC^2 = 113$ leaves $CD^2 = 112$.

Take the positive square root.

A length cannot be negative, so take the positive square root of $112$.

Work out the square root and round it.

$\sqrt{112} = 10.583\ldots$, and rounding to $1$ decimal place gives $10.6$.

Check the rounded answer by squaring it.

Squaring the rounded answer gives $112.36$, which is very close to $112$, so the rounding is right.

Estimate the size of the answer before working it out.

$112$ lies between the square numbers $100$ and $121$, so the answer lies between $10$ and $11$. Any answer outside that range would be wrong.

Check the units.

Every length in the question is in cm, so CD is in cm too.

Check that CD is shorter than AD.

$AD$ is the hypotenuse of triangle $ACD$, so it must be the longest side of it, and $112 < 225$ confirms that $CD$ is shorter.

Check the answer by rebuilding the second triangle.

The two shorter sides of triangle $ACD$ square to $225$ together, which is $AD^2$, so the answer fits the theorem.

Note the length of AC itself.

$AC = \sqrt{113}$ cm, though the working never needed it as a decimal — only its square.

State the length of CD.

$CD = 10.6$ cm.