Pythagoras' Theorem · Study Notes · Edexcel IGCSE 4MA1 Higher

$a^{2} + b^{2} = c^{2}$ . Find missing sides of right triangles, distances between points, diagonals.

At a glance

$a^{2} + b^{2} = c^{2}$ , where $c$ is the HYPOTENUSE.
Right angles only.
Hypotenuse: $c = \sqrt{a^{2} + b^{2}}$ .
Leg: $a = \sqrt{c^{2} - b^{2}}$ .
3D Pythagoras: $d^{2} = a^{2} + b^{2} + c^{2}$ .
Distance formula: $d = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2}}$ .
Common triples: $(3,4,5), (5,12,13), (8,15,17), (7,24,25)$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.7 — Use Pythagoras' theorem in 2D problems.
4.7 — Use Pythagoras in 3D contexts.
4.7 — Find distance between two points using Pythagoras.

2D Pythagoras

▼

$a^{2} + b^{2} = c^{2}$ . Identify hypotenuse first.

Theorem: In a right-angled triangle:

$a^{2} + b^{2} = c^{2}$

where $c$ is the HYPOTENUSE (longest side, opposite the right angle).

To find the hypotenuse: $c = \sqrt{a^{2} + b^{2}}$ .

To find a leg: $a = \sqrt{c^{2} - b^{2}}$ .

The hypotenuse is the longest side, opposite the right angle; the leg squares sum to its square.

Worked example. Right triangle, legs $3$ and $4$ .

$c^{2} = 9 + 16 = 25$ , so $c = 5$ .

Worked example. Hypotenuse $13$ , one leg $5$ .

$5^{2} + b^{2} = 169 \to b^{2} = 144 \to b = 12$ .

Common Pythagorean triples (memorise — saves time):

Triple	$a^{2} + b^{2}$
$3, 4, 5$	$9 + 16 = 25$
$5, 12, 13$	$25 + 144 = 169$
$8, 15, 17$	$64 + 225 = 289$
$7, 24, 25$	$49 + 576 = 625$

Also multiples: $6, 8, 10$ (= $2 \times 3, 4, 5$ ); $9, 12, 15$ ; etc.

Worked qualitative. Spotting a triple saves arithmetic.

$5, 12, ?$ → recognise as $5, 12, 13$ .
$9, 12, ?$ → recognise as $3, 4, 5$ scaled by $3$ → $15$ .

Edexcel tip. Identify HYPOTENUSE first. The largest side. Then decide if you're finding it or a leg.

$c$ = hypotenuse (longest, opposite right angle).
Hypotenuse: $\sqrt{a^{2} + b^{2}}$ .
Leg: $\sqrt{c^{2} - b^{2}}$ .
Memorise common triples.

3D Pythagoras

▼

$d^{2} = a^{2} + b^{2} + c^{2}$ for cuboid diagonal.

3D Pythagoras finds the long diagonal of a cuboid (or the longest distance in 3D space).

For cuboid with sides $a, b, c$ :

$d^{2} = a^{2} + b^{2} + c^{2}$

Why? Apply 2D Pythagoras twice.

Floor diagonal: $d_{\text{floor}}^{2} = a^{2} + b^{2}$ .
Long diagonal: $d^{2} = d_{\text{floor}}^{2} + c^{2} = a^{2} + b^{2} + c^{2}$ .

Worked example. Cuboid $4 \times 6 \times 12$ .

$d^{2} = 16 + 36 + 144 = 196$ .
$d = 14$ .

Find the floor diagonal first, then combine it with the height to reach the space diagonal.

Other 3D applications:

Distance from corner to opposite corner of a room.
Distance from vertex of a pyramid to a corner of the base.
Length of a string in a 3D structure.

Worked qualitative. A box is $3 \times 4 \times 12$ inside. Will a $13$ -inch ruler fit diagonally?

Diagonal: $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$ .
Ruler fits exactly along the long diagonal. Just.

Edexcel tip. State the formula explicitly. Mark schemes credit recognising 3D Pythagoras as separate from 2D.

$d^{2} = a^{2} + b^{2} + c^{2}$ .
Cuboid diagonals.
Apply 2D Pythagoras twice = same result.
Useful for 3D distance problems.

Distance between two points

▼

$d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$ .

Distance formula is just Pythagoras applied to coordinates.

For points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

$d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$

Why? Form a right triangle:

Horizontal leg: $|x_2 - x_1|$ .
Vertical leg: $|y_2 - y_1|$ .
Hypotenuse: $d$ .
Pythagoras: $d^{2} = (\Delta x)^{2} + (\Delta y)^{2}$ .

The gaps in x and y are the legs; the straight-line distance is the hypotenuse.

Worked example. Distance from $A(2, 3)$ to $B(7, 15)$ .

$\Delta x = 5$ , $\Delta y = 12$ .
$d = \sqrt{25 + 144} = \sqrt{169} = 13$ .

3D extension. For points in 3D:

$d = \sqrt{(\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}}$ .

Worked qualitative. What's the distance from origin to $(3, 4)$ ?

$d = \sqrt{9 + 16} = 5$ .
Directly $5$ , since $0$ -to-point distance.

Edexcel tip. Always show the differences explicitly. $\Delta x = 7 - 2 = 5$ . Mark scheme credits this step.

Pythagoras with $\Delta x, \Delta y$ .
$d = \sqrt{(\Delta x)^{2} + (\Delta y)^{2}}$ .
Extends to 3D.
Always compute differences.

Quick recap

$a^{2} + b^{2} = c^{2}$ .
Hypotenuse: opposite right angle.
Find leg: $c^{2} -$ other².
3D: $d^{2} = a^{2} + b^{2} + c^{2}$ .
Distance formula: Pythagoras with coordinates.
Memorise common triples.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$a^{2} + b^{2} = c^{2}$ .
Hypotenuse: $\sqrt{a^{2} + b^{2}}$ .
Leg: $\sqrt{c^{2} - b^{2}}$ .
3D: $d^{2} = a^{2} + b^{2} + c^{2}$ .
Triples: $(3,4,5), (5,12,13), (7,24,25), (8,15,17)$ .

How it’s examined

Pythagoras every Higher Tier paper (3-7 marks). Often combined with circles, cones, or coordinate geometry. Examiner reports flag (1) using on non-right triangles, (2) wrong direction (adding when should subtract), (3) forgetting square root.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on pythagoras theorem, written exactly the way a tutor would explain them at the board.

1Find the hypotenuse
Foundation• Pythagoras, hypotenuse
▼
Question
Right triangle: legs $3$ and $4$ . Find the hypotenuse.
Step-by-step solution
1. Step 1
  $a^{2} + b^{2} = c^{2}$ .
2. Step 2
  $3^{2} + 4^{2} = 9 + 16 = 25$ . So $c = 5$ .
Answer
$c = 5$
2Find a missing leg
Foundation• Adapted from 4MA1/1H Jan 2024 Q9• Pythagoras, leg
▼
Question
Hypotenuse $13$ , one leg $5$ . Find the other leg.
Step-by-step solution
1. Step 1
  $5^{2} + b^{2} = 13^{2}$ .
2. Step 2
  $25 + b^{2} = 169 \implies b^{2} = 144 \implies b = 12$ .
Answer
$b = 12$
Examiner tip
Subtract from hypotenuse² when finding a leg. $b^{2} = c^{2} - a^{2}$ .
3Distance between two points
Higher• distance
▼
Question
Find the distance between $A(2, 3)$ and $B(7, 15)$ .
Step-by-step solution
1. Step 1
  Form a right triangle: horizontal leg $\Delta x = 5$ , vertical leg $\Delta y = 12$ .
2. Step 2
  Pythagoras: $d^{2} = 5^{2} + 12^{2} = 25 + 144 = 169$ .
3. Step 3
  $d = 13$ .
Answer
$d = 13$
43D Pythagoras
Higher• Adapted from 4MA1/2H May/Jun 2024 Q18• 3D, Pythagoras
▼
Question
Cuboid $4 \times 6 \times 12$ . Find the long diagonal.
Step-by-step solution
1. Step 1
  Floor diagonal: $4^{2} + 6^{2} = 52$ , so floor diag $= \sqrt{52}$ .
2. Step 2
  Long diag: $52 + 12^{2} = 52 + 144 = 196$ . So $d = 14$ .
3. Step 3
  Or directly: $d^{2} = 4^{2} + 6^{2} + 12^{2} = 16 + 36 + 144 = 196$ . $d = 14$ .
Answer
$d = 14$
Examiner tip
3D Pythagoras: $d^{2} = a^{2} + b^{2} + c^{2}$ . Same as 2D, just three terms.
5Pythagoras in an isosceles triangle
Higher• Pythagoras, isosceles
▼
Question
Isosceles triangle with two equal sides $13$ cm and base $10$ cm. Find the perpendicular height.
Step-by-step solution
1. Step 1
  Drop perpendicular from apex to base. Bisects base: half = $5$ cm.
2. Step 2
  Right triangle with hypotenuse $13$ , leg $5$ :
  $h^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$
3. Step 3
  $h = 12$ cm.
Answer
$h = 12$ cm
6Explaining a wrong hypotenuse [1 mark]
Getting started• Paper 1H/2H style — 1 mark• pythagoras, hypotenuse, explain
▼
Question
The two shorter sides of a right-angled triangle are $5$ cm and $7$ cm.

Tariq says, "The hypotenuse is $12$ cm."

Explain why Tariq is wrong. [1]
Step-by-step solution
1. Step 1
  Tariq has added the lengths, $5 + 7 = 12$ . Pythagoras adds the squares of the lengths.
2. Step 2
  B1 Give a correct reason, supported by the correct value if you work it out.
  $h = \sqrt{5^{2} + 7^{2}} = \sqrt{74} = 8.60\ldots \text{ cm}, \text{ not } 12 \text{ cm}$
Answer
He added the sides instead of their squares; the hypotenuse is $\sqrt{74} = 8.60$ cm
Examiner tip
Examiner tip: A second reason also works: any side of a triangle must be shorter than the other two added together, so $12$ cm is impossible.

Common mistake: Writing only "he used the wrong method". An explanation must say what is wrong (the sides were added, not their squares) or give the correct length.
7Side of a rhombus from its diagonals [2 marks]
Getting started• Paper 1H/2H style — 2 marks• pythagoras, rhombus
▼
Question
The diagonals of a rhombus have lengths $16$ cm and $30$ cm.

Work out the length of one side of the rhombus. [2]
Step-by-step solution
1. Step 1
  M1 The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with shorter sides $8$ cm and $15$ cm.
  $s^{2} = 8^{2} + 15^{2} = 64 + 225 = 289$
2. Step 2
  A1 Square root.
  $s = 17 \text{ cm}$
Answer
$17$ cm
Examiner tip
Examiner tip: State the property you use: the diagonals of a rhombus bisect each other at right angles. That is why the triangle is right-angled and why its sides are half the diagonals.

Common mistake: Using the full diagonals, $\sqrt{16^{2} + 30^{2}} = 34$ . That is twice the side, because the full diagonals are not the sides of one small triangle.
8Perimeter of a right-angled trapezium [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• pythagoras, trapezium, perimeter
▼
Question
$ABCD$ is a trapezium with $AB$ parallel to $DC$ and angle $DAB = 90°$ .

$AB = 19$ cm, $DC = 12$ cm and $AD = 8.4$ cm.

Calculate the perimeter of the trapezium. Give your answer correct to 3 significant figures. [3]
Step-by-step solution
1. Step 1
  M1 Drop a perpendicular from $C$ to $AB$ , meeting it at $E$ . Then $CE = AD = 8.4$ cm and $EB = 19 - 12 = 7$ cm.
  $EB = 19 - 12 = 7$
2. Step 2
  M1 $BC$ is the hypotenuse of triangle $CEB$ .
  $BC = \sqrt{7^{2} + 8.4^{2}} = \sqrt{49 + 70.56} = \sqrt{119.56} = 10.934\ldots$
3. Step 3
  A1 Add the four sides, using the unrounded $BC$ .
  $19 + 12 + 8.4 + 10.934\ldots = 50.334\ldots = 50.3 \text{ cm}$
Answer
$50.3$ cm (awrt $50.3$ )
Examiner tip
Examiner tip: Sketch the trapezium and mark the rectangle $AECD$ . The right-angled triangle is the piece left over, and its base is the difference of the parallel sides.

Common mistake: Using $19$ or $12$ as the base of the triangle instead of their difference, $7$ cm.
9A missing coordinate from a given distance [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• pythagoras, coordinates, distance
▼
Question
$A$ is the point $(-3, k)$ and $B$ is the point $(5, 2)$ .

The length of $AB$ is $17$ units.

Find the two possible values of $k$ . [4]
Step-by-step solution
1. Step 1
  M1 The horizontal distance is $5 - (-3) = 8$ . The vertical distance is $k - 2$ .
  $5 - (-3) = 8$
2. Step 2
  M1 Use Pythagoras with $AB = 17$ as the hypotenuse.
  $8^{2} + (k - 2)^{2} = 17^{2}$
3. Step 3
  A1 Rearrange.
  $(k - 2)^{2} = 289 - 64 = 225$
4. Step 4
  A1 Take both square roots.
  $k - 2 = \pm 15 \;\Rightarrow\; k = 17 \text{ or } k = -13$
Answer
$k = 17$ or $k = -13$
Examiner tip
Examiner tip: The question says "two possible values", which is a clue that the square root has a positive and a negative value. Point $A$ can be above or below $B$ .

Common mistake: Giving only $k = 17$ . The negative root, $k - 2 = -15$ , gives the second answer and is needed for the final A mark.
10Height and volume of a cuboid from its space diagonal [5 marks]
Stretch• Paper 1H/2H style — 5 marks• 3d pythagoras, cuboid, volume
▼
Question
A cuboid has length $14$ cm and width $5$ cm.

The space diagonal of the cuboid (from one corner to the opposite corner through the inside) is $17$ cm.

Work out the volume of the cuboid. Give your answer correct to 3 significant figures. [5]
Step-by-step solution
1. Step 1
  M1 The square of the space diagonal is the sum of the squares of the length, width and height $h$ .
  $14^{2} + 5^{2} + h^{2} = 17^{2}$
2. Step 2
  M1 Rearrange for $h^{2}$ .
  $h^{2} = 289 - 196 - 25 = 68$
3. Step 3
  A1 Square root, and keep the full value.
  $h = \sqrt{68} = 8.2462\ldots$
4. Step 4
  M1 Volume $=$ length $\times$ width $\times$ height, using the unrounded height.
  $V = 14 \times 5 \times 8.2462\ldots = 577.23\ldots$
5. Step 5
  A1 Round to 3 significant figures.
  $V = 577 \text{ cm}^{3}$
Answer
$577$ cm³ (awrt $577$ )
Examiner tip
Examiner tip: Store $\sqrt{68}$ in your calculator memory. Rounding the height to $8.25$ cm first gives $577.5$ , which rounds to $578$ and loses the final A mark.

Common mistake: Treating $17$ cm as a side and adding it into the squares, or using only two dimensions: $17^{2} - 14^{2} = 93$ gives the diagonal of a side face, not the height.

Question 1
Paper 1H/2H style — 1 mark1 mark
A right-angled triangle has shorter sides of length $3x$ cm and $4x$ cm.

Write down an expression, in terms of $x$ , for the length of the hypotenuse. [1]
Model answer
$(3x)^{2} + (4x)^{2} = 9x^{2} + 16x^{2} = 25x^{2}$

Answer: $5x$ cm
Why this scores
Mark scheme:
- B1 for $5x$ oe (e.g. $\sqrt{25x^{2}}$ )
Why this scores full marks: The squares are taken with brackets, so the $3$ and the $4$ are squared as well as the $x$ . Writing $3x^{2} + 4x^{2} = 7x^{2}$ gives the wrong answer $\sqrt{7}x$ .
Question 2
Paper 1H/2H style — 2 marks2 marks
In a right-angled triangle, the hypotenuse is $20.5$ cm and one of the other sides is $4.5$ cm.

Work out the length of the third side. [2]
Model answer
$x^{2} = 20.5^{2} - 4.5^{2} = 420.25 - 20.25 = 400$

$x = \sqrt{400}$

Answer: $20$ cm
Why this scores
Mark scheme:
- M1 for $20.5^{2} - 4.5^{2}$ oe
- A1 for $20$ cao
Why this scores full marks: The hypotenuse is given, so the squares are subtracted. The answer $20$ cm is shorter than the hypotenuse, as it must be. Adding would give $20.99$ cm, a side longer than the hypotenuse.
Question 3
Paper 1H/2H style — 2 marks2 marks
A wheelchair ramp rises $0.45$ m vertically over a horizontal distance of $3.6$ m.

Calculate the length of the sloping surface of the ramp. Give your answer correct to 3 significant figures. [2]
Model answer
$L^{2} = 3.6^{2} + 0.45^{2} = 12.96 + 0.2025 = 13.1625$

$L = \sqrt{13.1625} = 3.6280\ldots$

Answer: $3.63$ m
Why this scores
Mark scheme:
- M1 for $3.6^{2} + 0.45^{2}$ oe
- A1 for $3.63$ (awrt $3.63$ )
Why this scores full marks: The sloping surface is the hypotenuse. The script squares $0.45$ correctly as $0.2025$ (not $0.9$ , which is $0.45 \times 2$ ) and gives the answer to the accuracy asked.
Question 4
Paper 1H/2H style — 3 marks3 marks
An isosceles triangle has a base of $16$ cm and a perpendicular height of $11$ cm.

Calculate the perimeter of the triangle. Give your answer correct to 3 significant figures. [3]
Model answer
The height bisects the base, so each half is $8$ cm.

$s^{2} = 8^{2} + 11^{2} = 64 + 121 = 185$

$s = \sqrt{185} = 13.601\ldots$

Perimeter $= 16 + 2 \times 13.601\ldots = 43.202\ldots$

Answer: $43.2$ cm
Why this scores
Mark scheme:
- M1 for $8^{2} + 11^{2}$ oe
- M1 for $16 + 2 \times$ their $\sqrt{185}$
- A1 for $43.2$ (awrt $43.2$ )
Why this scores full marks: The script halves the base before using Pythagoras and remembers there are two equal sloping sides. Scripts that add only one sloping side get $29.6$ cm.
Question 5
Paper 1H/2H style — 3 marks3 marks
Hana walks $8.2$ km due north, then $12.5$ km due east, then $3.1$ km due south.

Calculate the direct distance from her starting point to her finishing point. Give your answer correct to 3 significant figures. [3]
Model answer
Distance north of the start: $8.2 - 3.1 = 5.1$ km

Distance east of the start: $12.5$ km

$d^{2} = 5.1^{2} + 12.5^{2} = 26.01 + 156.25 = 182.26$

$d = \sqrt{182.26} = 13.500\ldots$

Answer: $13.5$ km
Why this scores
Mark scheme:
- M1 for $8.2 - 3.1$ (= $5.1$ ) seen or used
- M1 for their $5.1^{2} + 12.5^{2}$ oe
- A1 for $13.5$ (awrt $13.5$ )
Why this scores full marks: The north and south legs point in opposite directions, so the script subtracts them before using Pythagoras. Adding them, $8.2 + 3.1 = 11.3$ , gives $16.9$ km; adding all three walking distances gives $23.8$ km, which is the distance walked, not the direct distance.
Question 6
Paper 1H/2H style — 3 marks3 marks
A radio mast at $M$ can reach any point within $7.8$ km of it.

A village $V$ is $4.7$ km due east and $6.2$ km due north of $M$ .

Is the village within range of the mast? You must show your working. [3]
Model answer
$MV^{2} = 4.7^{2} + 6.2^{2} = 22.09 + 38.44 = 60.53$

$MV = \sqrt{60.53} = 7.780\ldots$ km

$7.780\ldots < 7.8$

Answer: Yes, the village is within range.
Why this scores
Mark scheme:
- M1 for $4.7^{2} + 6.2^{2}$ oe
- A1 for $7.78$ or better (or $60.53$ compared with $7.8^{2} = 60.84$ )
- A1 dep on the previous A1 for "Yes" with a correct comparison
Why this scores full marks: The two lengths are very close, so the script keeps enough figures to compare them. Rounding $7.780\ldots$ to $7.8$ would suggest the village is exactly on the edge and makes the conclusion unclear.
Question 7
Paper 1H/2H style — 4 marks4 marks
In triangle $ABC$ , $D$ is the point on $BC$ such that $AD$ is perpendicular to $BC$ .

$AB = 13.6$ cm, $BD = 6.4$ cm and $DC = 11.2$ cm.

Calculate the length of $AC$ . Give your answer correct to 3 significant figures. [4]
Model answer
Triangle $ABD$ : $AD^{2} = 13.6^{2} - 6.4^{2} = 184.96 - 40.96 = 144$ , so $AD = 12$ cm

Triangle $ADC$ : $AC^{2} = 12^{2} + 11.2^{2} = 144 + 125.44 = 269.44$

$AC = \sqrt{269.44} = 16.414\ldots$

Answer: $16.4$ cm
Why this scores
Mark scheme:
- M1 for $13.6^{2} - 6.4^{2}$ oe
- A1 for $AD = 12$
- M1 for their $12^{2} + 11.2^{2}$ oe
- A1 for $16.4$ (awrt $16.4$ )
Why this scores full marks: The script uses the shared side $AD$ to link the two triangles. $AB$ is a hypotenuse, so squares are subtracted; $AC$ is a hypotenuse, so squares are added.
Question 8
Paper 1H/2H style — 4 marks4 marks
A right-angled triangle has an area of $54$ cm². One of the two shorter sides is $7.2$ cm.

Calculate the perimeter of the triangle. Give your answer correct to 3 significant figures. [4]
Model answer
$\dfrac{1}{2} \times 7.2 \times b = 54$ , so $3.6b = 54$ and $b = 15$ cm

$h^{2} = 7.2^{2} + 15^{2} = 51.84 + 225 = 276.84$

$h = \sqrt{276.84} = 16.638\ldots$

Perimeter $= 7.2 + 15 + 16.638\ldots = 38.838\ldots$

Answer: $38.8$ cm
Why this scores
Mark scheme:
- M1 for $\dfrac{1}{2} \times 7.2 \times b = 54$ oe
- A1 for $b = 15$
- M1 for $7.2^{2} +$ their $15^{2}$
- A1 for $38.8$ (awrt $38.8$ )
Why this scores full marks: The two shorter sides of a right-angled triangle are a base and its perpendicular height, so the area gives the missing side directly. The script then remembers to add all three sides for the perimeter.
Question 9
Paper 1H/2H style — 5 marks5 marks
A closed cylindrical tin has a height of $20$ cm and a volume of $500\pi$ cm³.

Work out the length of the longest straight rod that will fit inside the tin. Give your answer correct to 3 significant figures. [5]
Model answer
$\pi r^{2} \times 20 = 500\pi$

$r^{2} = 25$ , so $r = 5$ cm and the diameter is $10$ cm

The longest rod goes from the edge of the base, across a diameter, to the opposite edge of the top.

$L^{2} = 10^{2} + 20^{2} = 100 + 400 = 500$

$L = \sqrt{500} = 22.36\ldots$

Answer: $22.4$ cm
Why this scores
Mark scheme:
- M1 for $\pi r^{2} \times 20 = 500\pi$ oe
- A1 for $r = 5$
- M1 for using the diameter, $2 \times$ their $r$ , with the height $20$
- M1 for their $10^{2} + 20^{2}$
- A1 for $22.4$ (awrt $22.4$ )
Why this scores full marks: The script uses the cylinder volume formula from the formula sheet to find the radius, then sees that the longest rod lies in a vertical cross-section through the axis: a rectangle $10$ cm wide and $20$ cm tall. Using the radius instead of the diameter gives $20.6$ cm, which loses the last three marks.
Question 10
Paper 1H/2H style — 5 marks5 marks
An isosceles triangle has two equal sides of length $(x + 5)$ cm and a base of length $2x$ cm.

The perpendicular height of the triangle is $(x + 1)$ cm.

(a) Show that $x^{2} - 8x - 24 = 0$ [2]

(b) Work out the area of the triangle. Give your answer correct to 3 significant figures. [3]
Model answer
(a) The height bisects the base, so the right-angled triangle has sides $x$ , $(x + 1)$ and hypotenuse $(x + 5)$ .

$x^{2} + (x + 1)^{2} = (x + 5)^{2}$

$x^{2} + x^{2} + 2x + 1 = x^{2} + 10x + 25$

$x^{2} - 8x - 24 = 0$

(b) $x = \dfrac{8 \pm \sqrt{(-8)^{2} - 4(1)(-24)}}{2} = \dfrac{8 \pm \sqrt{160}}{2}$

$x = 10.324\ldots$ (the other root, $-2.32\ldots$ , is negative so is rejected)

Area $= \dfrac{1}{2} \times 2x \times (x + 1) = x(x + 1) = 10.324\ldots \times 11.324\ldots = 116.92\ldots$

Answer: $117$ cm²
Why this scores
Mark scheme:
- M1 for $x^{2} + (x + 1)^{2} = (x + 5)^{2}$ oe
- A1 for a complete, correct expansion and simplification to $x^{2} - 8x - 24 = 0$
- M1 for correct substitution into the quadratic formula (or completing the square)
- A1 for $x = 10.32\ldots$
- A1 for $117$ (awrt $117$ )
Why this scores full marks: In (a) the script uses half the base, $x$ , not $2x$ , and shows every line of the expansion. In (b) it keeps the full value of $x$ : rounding to $x = 10.3$ gives $116.39$ , which rounds to $116$ and loses the final mark.

4MA1 Pythagoras' Theorem

At a glance

What you’ll learn

2D Pythagoras

3D Pythagoras

Distance between two points

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Find the hypotenuse

2Find a missing leg

3Distance between two points

43D Pythagoras

5Pythagoras in an isosceles triangle

6Explaining a wrong hypotenuse [1 mark]

7Side of a rhombus from its diagonals [2 marks]

8Perimeter of a right-angled trapezium [3 marks]

9A missing coordinate from a given distance [4 marks]

10Height and volume of a cuboid from its space diagonal [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10