Module 23 of 45Lesson 1 of 12 · Unit G20 · Triangle Truths & the Right-Angle Rule
Geometry & Trigonometry · Unit G20 · Standard
Triangle Truths & the Right-Angle Rule
A triangle is the smallest shape you can make with straight lines, and it is the one shape that cannot be pushed out of shape without breaking — which is why roofs, gates, bridges and cell masts are all full of them. This unit is two halves of one story. First: what is true of every triangle — which lengths can be a triangle at all, how its angles behave, and what its name tells you. Then: what is true of the right-angled ones — the one rule, more than two thousand years old, that turns two sides into the third without a ruler ever touching the page.
The twentieth unit of the Geometry & Trigonometry strand, and the tenth of the Standard band. This is a merged unit: it covers both the triangle properties row and the Pythagoras’ theorem row of the official grid, because the first is the natural first half of the second. G02 gave you the angle sum; this unit adds everything else a triangle is obliged to do, and then the one rule that belongs to the right-angled ones alone.
By the end of this unit you can
- Classify a triangle by its sides and by its angles, and describe what the ticks and arcs on a diagram are telling you
- Apply the isosceles base-angle property and the angle sum, and determine an exterior angle from the two interior angles opposite it
- Determine whether three given lengths can be the sides of a triangle, and explain what the perpendicular height of a triangle is
- Calculate the hypotenuse of a right-angled triangle, giving the answer exactly as a simplified surd and to 3 significant figures
- Calculate a shorter side by subtraction, and determine whether a triangle with three given sides is right-angled
- Apply Pythagoras twice to find the space diagonal of a cuboid and the slant height of a pyramid, and discuss whether an answer is sensible in its context
The one idea behind all of it
Three lengths either close up into a triangle or they do not, and the test is quick: the two shorter ones together must be longer than the longest. Once they do close up, the triangle is fixed — its three angles are decided, and they always add to 180°. Give one of those angles a special value, 90°, and something extra becomes true: the side facing the right angle, the hypotenuse, is tied to the other two by a² + b² = c². Square the two you know, add or subtract, take the square root, and you have the one you do not — on paper, in a roof, or across the diagonal of a box.
a² + b² = c²
where c is the side facing the right angle, and the three angles always add to 180°
Two habits do most of the work in this unit. Find the hypotenuse before you write anything down — it is the side facing the right angle, it is always the longest of the three, and knowing which side it is decides whether you add or subtract. And keep the answer exact until the very end: most Pythagoras answers are surds, and a surd rounded to three figures and then used again in a second stage is an error waiting to grow.
Every length carries its unit and every angle carries its degree sign. A bare 13 is not an answer; 13 cm is. Anything that is not exact is given to 3 significant figures — rounded, never truncated — and the exact surd is given first whenever a question asks for it.
Where this unit sits
This unit is two grid rows joined at the hip. Triangle properties and Pythagoras’ theorem were listed separately in the blueprint and are built here as one unit, because the first is the natural first half of the second: you cannot ask which side is the hypotenuse until you can name the parts of a triangle, and you cannot check that an answer is possible without the triangle inequality. Unit G02 Angle Arithmetic already established that the angles of a triangle add to 180°; that fact is used here and is not retaught.
Two things it deliberately does not do. Squares and square roots — including estimating a root between two whole numbers — belong to N10 Squares, Roots & Powers, which is already built. Simplifying a surd, turning √20 into 2√5, is Algebra u19 Root and Branch, also already built; this unit uses simplified surds throughout and links back rather than teaching the simplification again. Neither is a separate page here, and neither is examined as a topic of its own.
What this unit closes and what it opens. It closes the loop left open in G14 Solid Progress, where the slant height of a cone and of a pyramid was handed to you as a stated result: here you prove it. It also names the theorem behind G16 Distance, Midpoint, Gradient — the distance formula is Pythagoras, wearing coordinates, and the right-angled triangle it uses is the one you drop underneath a slanting segment. What it opens is G21 SOH CAH TOA: Pythagoras finds a third side from two sides, and trigonometry is what you reach for the moment an angle is involved.
House rules. Every length carries its unit — a bare 13 is not an answer, 13 cm is. Every angle carries its degree sign. An answer that is not a whole number is given exactly, as a simplified surd, and then to 3 significant figures if a decimal is asked for — rounded, never truncated, which is the discipline of N04 Round It Right. And in a two-stage calculation the exact value goes into the second stage: rounding in the middle is how the third significant figure gets lost.
1
Naming a triangle, and the marks that name it
Classification by sides and by angles, and the isosceles property that links the two
A triangle answers to two names, and a question may want either. From its sides it is equilateral if all three are equal, isosceles if at least two are (so an equilateral triangle is a special case of isosceles), and scalene if no two are. From its angles it is right-angled if one is exactly 90°, obtuse-angled if one is bigger than that, and acute-angled if all three are smaller. The two namings are independent — a triangle can be isosceles and obtuse at the same time — but they are tied together by one fact worth more than any other in this concept: equal sides face equal angles. Find two equal sides and you have found two equal angles, and the other way round as well.
The tick marks are not decoration: one tick on all three edges says the three sides are the same length, two ticks on a pair says that pair is equal, and no ticks at all says no two sides are equal. Every tick in this strand is drawn from the measured length, so a tick cannot lie.
The two sides that meet at C are the same length, so the two angles that face them — the base angles at A and at B — are the same size. That is the isosceles property, and it works in both directions: equal sides give equal angles, and equal angles give equal sides.
Worked example 1 · Name it three ways
1
Take the triangle A B C with sides 6 cm, 2√3 cm, 2√3 cm. Identify which sides are equal.
two of them are 2√3 cm
2
Classify it by its sides.
isosceles
3
Calculate the two base angles — the ones facing the equal sides.
30° each
4
Calculate the third angle, and check the three add to 180°.
120°
5
Classify it by its angles.
obtuse-angled
6
Describe what a diagram would put on it.
two matching ticks and two matching arcs
A triangle has two names, not one: a name from its sides and a name from its angles, and a question may want either. The two are linked by the isosceles property — equal sides face equal angles — so finding one pair immediately gives you the other. Notice that the two names here are independent: isosceles says nothing about being obtuse, and obtuse says nothing about being isosceles.
| The name | What the sides do | What the angles do | How a diagram marks it |
|---|---|---|---|
| equilateral | all three sides equal | all three angles 60° | one tick on every side |
| isosceles | at least two sides equal (an equilateral triangle counts) | the two angles facing those sides are equal | matching ticks on the two equal sides, matching arcs on the two equal angles |
| scalene | no two sides equal | no two angles equal | no ticks at all |
| right-angled | the hypotenuse faces the right angle and is the longest | one angle exactly 90° | a small square in the corner |
| obtuse-angled | the longest side faces the obtuse angle | one angle bigger than 90°, so the other two are small | nothing special — you have to measure or calculate |
EXAMPLEThe triangle with sides 6 cm, 2√3 cm, 2√3 cm has two equal sides, so it is isosceles, and the two angles facing those sides are 30° each. The third angle is 120°, which is more than a right angle, so the same triangle is also obtuse-angled. The triangle with sides 3 cm, 6 cm, 3√3 cm has no two sides equal, so it is scalene — and because 9 + 27 = 36 it is also right-angled.
Practice · Naming a triangle, and the marks that name it · 10 questions
P1State the name given to a triangle whose three sides are all the same length.Fluency
Answerequilateral
P2A triangle has sides of 5 cm, 5 cm and 8 cm. Classify it by its sides.Fluency
Answerisosceles
P3A triangle has sides of 6 cm, 8 cm and 11 cm. Classify it by its sides.Fluency
Answerscalene
P4The angles of a triangle are 30°, 30° and 120°. Classify it by its angles.Fluency
Answerobtuse-angled
P5A triangle has sides of 6 cm, 2√3 cm, 2√3 cm. Classify it by its sides and calculate its two base angles.Applied
Answerisosceles; the two base angles are 30° eachTwo of the sides are 2√3 cm and the third is different, so two sides are equal and the triangle is isosceles. The two angles facing the equal sides are therefore equal, and since the apex angle is 120° the two of them share the remaining 60°, which is 30° each. Equal sides face equal angles: that one sentence is the whole isosceles property, and it works in both directions.
P6A triangle has sides of 3 cm, 6 cm, 3√3 cm. Classify it by its sides and by its angles.Applied
Answerscalene, and right-angledNo two of the three sides are equal, so it is scalene. Squaring the two shorter sides gives 9 + 27 = 36, and the longest side squared is 6² = 36 as well, so the converse of Pythagoras’ theorem says there is a right angle — and it faces the 6 cm side. A triangle carries two names, one from its sides and one from its angles, and neither name tells you the other.
P7Two of the angles of a triangle are 55° and 70°. Determine the third angle and classify the triangle by its sides and by its angles.Applied
Answer55°; isosceles and acute-angledThe three angles add to 180°, so the third is 180 − 55 − 70 = 55°. Two of the angles are now equal, so the two sides facing them are equal and the triangle is isosceles. Every angle is less than 90°, so it is acute-angled. This is the isosceles property used backwards: equal angles give equal sides just as surely as equal sides give equal angles.
P8Explain why a triangle cannot have two right angles.Applied
Answertwo right angles already use the whole 180°, leaving nothing for the third angleThe three angles of any triangle add to exactly 180°. Two right angles come to 90 + 90 = 180° between them, so the third angle would have to be 0° — and an angle of 0° means the two sides lie on top of one another and the shape has no inside. Drawing it makes the same point: two perpendiculars to the same line are parallel, and parallel lines never meet to make a third corner.
P9An isosceles triangle has an apex angle of x°. Determine an expression for each of its base angles.Challenge
Answer(180 − x) ÷ 2 degrees eachThe two base angles are equal, so call each of them b. The angle sum gives x + b + b = 180, so 2b = 180 − x and b = (180 − x) ÷ 2. Check it on a case you already know: an apex of 40° gives (180 − 40) ÷ 2 = 70°, which is the answer to Worked example 2. Notice the expression also tells you the apex can never reach 180°, because the base angles would then be 0°.
P10Show that every equilateral triangle is also isosceles and acute-angled.Challenge
Answerall three angles are 60°, so it has two equal sides and no angle of 90° or moreEquilateral means all three sides are equal, so in particular two of them are equal, which is what isosceles asks for — every equilateral triangle is isosceles, though the reverse is not true. For the angles: equal sides face equal angles, so all three angles are the same, and since they add to 180° each one is 180 ÷ 3 = 60°. Every angle is less than 90°, so the triangle is acute-angled, and this is the only shape of triangle whose angles are fixed before you draw it.
2
Angles inside, and the angle just outside
The angle sum, the exterior-angle property, and chains that use both
Every triangle spends exactly 180° on its three angles, which is why knowing two of them always gives you the third. Now carry one side on past a vertex and look at the angle that opens up outside the triangle: that is the exterior angle, and it is equal to the two interior angles at the other two vertices added together. It is not a new fact — the interior and exterior angles at a vertex sit on a straight line and so make 180° between them, exactly like the three interior angles — but it turns a two-step calculation into one step, and in a long chain that saving is where the marks are.
Carry the side AB on past B to D and the angle DBC that opens up outside the triangle is the exterior angle at B. It is 120°, which is exactly what the two angles at the far end — 90° at A and 30° at C — add up to. That is not a coincidence, and it saves a step every time.
Worked example 2 · Isosceles hunt: one chain, three steps
1
A triangle is isosceles with an apex angle of 40°. Identify which two angles must be equal.
the two base angles
2
The three angles add to 180°. Calculate what is left for the two base angles together.
140°
3
Determine each base angle.
70° each
4
The base is now carried on past one of the base vertices. Determine the exterior angle there.
110°
5
Verify it with the exterior-angle property: add the two interior angles at the other end.
40° + 70° = 110°
6
Explain which of the two routes you would use in an exam.
the exterior-angle property — it is one step
The exterior angle at a vertex equals the two interior angles at the other two vertices added together. It is not a new fact — it follows at once from the angle sum, because the interior and exterior angles at one vertex make a straight line — but it turns a two-step calculation into a one-step one, and in a long angle chain that saving is where the marks are.
| The triangle | Its three angles | They add to | Its name by angles |
|---|---|---|---|
| the equilateral of the first figure | 60° + 60° + 60° | 180° | acute-angled |
| the isosceles A B C above | 30° + 30° + 120° | 180° | obtuse-angled |
| the 30-60-90 triangle | 90° + 60° + 30° | 180° | right-angled |
| the 3-4-5 triangle | 90° + 36.9° + 53.1° | 180° | right-angled |
| the 5-12-13 triangle | 90° + 67.4° + 22.6° | 180° | right-angled |
EXAMPLETwo angles of a triangle are 40° and 75°, so the third is 180 − 40 − 75 = 65°. An isosceles triangle has an apex of 40°, so its base angles share 140° and each is 70°. And if two interior angles are 43° and 75°, the exterior angle at the third vertex is 118° — found in one step by adding, or in two by finding the third interior angle first and taking it from 180°.
Practice · Angles inside, and the angle just outside · 10 questions
P11Two angles of a triangle are 40° and 75°. Calculate the third angle.Fluency
Answer65°
P12A right-angled triangle has an angle of 28°. Calculate its third angle.Fluency
Answer62°
P13In triangle A B C the angle at A is 47° and the angle at B is 68°. Determine the exterior angle at C.Fluency
Answer115°
P14The exterior angle at one vertex of a triangle is 130°. Determine the interior angle at that vertex.Fluency
Answer50°
P15An isosceles triangle has an apex angle of 40°. Calculate its two base angles.Applied
Answer70° eachThe two base angles are equal because the two sides facing them are equal. The three angles add to 180°, so the two base angles share 180 − 40 = 140° between them, which is 70° each. Always take the apex away first and then halve; halving 180 first is the slip that turns this into 50°.
P16An isosceles triangle has base angles of 64°. Calculate its apex angle.Applied
Answer52°Both base angles are 64°, so together they use 128° of the 180° available and the apex gets what is left: 180 − 128 = 52°. This is the reverse of P15, and the check is the same in both directions — add your three angles up and see 180°.
P17In an isosceles triangle the exterior angle at one of the base vertices is 150°. Determine all three interior angles.Applied
Answer30°, 30° and 120°The interior angle at that base vertex is 180 − 150 = 30°, because an interior angle and its exterior angle make a straight line. The other base angle is equal to it, so it is 30° too, and the apex is 180 − 30 − 30 = 120°. Check with the exterior-angle property instead: the exterior angle 150° should equal the two opposite interior angles, 30 + 120 = 150°. It does.
P18In triangle A B C the exterior angle at C is 118° and the angle at A is 43°. Determine the angle at B.Applied
Answer75°The exterior angle at C equals the two interior angles at the other two vertices added together, so 118° = 43° + B and B = 118 − 43 = 75°. The long route works too and is worth doing once as a check: the interior angle at C is 180 − 118 = 62°, and then B = 180 − 43 − 62 = 75°. Same answer, one more line.
P19Show that the three exterior angles of any triangle add to 360°.Challenge
Answereach exterior angle is 180° minus its interior angle, and 3 × 180 − 180 = 360°Call the interior angles a, b and c, so that a + b + c = 180. The exterior angle at each vertex is 180 minus the interior angle there, so the three exterior angles are 180 − a, 180 − b and 180 − c. Adding them gives 540 − (a + b + c) = 540 − 180 = 360°. The argument never used a particular triangle, so it holds for every one — and it is the same 360° you get by walking right round the outside of the shape and finishing facing the way you started.
P20A triangle has angles of 90°, 60° and 30°. Show that the exterior angle at the 60° vertex is equal to the two interior angles opposite it added together.Challenge
Answerthe exterior angle is 120°, and 90 + 30 = 120°The interior angle at that vertex is 60°, so the exterior angle beside it is 180 − 60 = 120°, because the two of them lie along a straight line. The two interior angles at the other vertices are 90° and 30°, and 90 + 30 = 120° as well. The reason is the angle sum: the three interior angles use up 180°, and the straight line at the vertex also measures 180°, so what is left over on the line must be exactly what is left over in the triangle.
3
Can three lengths even be a triangle?
The triangle inequality, the perpendicular height, and what fixes a triangle completely
Not every set of three lengths is a triangle. Lay the longest one down and swing the other two from its ends: if they cross, you have a triangle; if they stop short of each other, you have nothing. The test is the two shorter lengths added must be greater than the longest — strictly greater, because if they are exactly equal the three lengths lie flat in a straight line and enclose no area at all. This concept also settles two words that are often confused. The perpendicular height of a triangle runs from a vertex to the opposite side at a right angle; the median runs from a vertex to the midpoint of the opposite side. They are different lines, and only the first one is the height an area calculation wants.
Swing an arc of 3 cm from A and one of 4 cm from B and they stop short of each other, with a gap between them. Three lengths only make a triangle when the two shorter ones together are longer than the longest — and 3 + 4 is not more than 9.
The green dashed line from R meets PQ at a right angle: that is the perpendicular height, and F is its foot. The purple dashed line from R meets PQ at its midpoint M, marked with matching ticks: that is the median. They are two different lines to two different points, and only the first one is the height an area calculation wants.
Worked example 3 · Can these be a triangle?
1
Four sets of lengths. Take 3 cm, 4 cm, 9 cm first and identify the longest.
9 cm
2
Calculate what the other two add to, and compare.
3 + 4 = 7, which is less than 9
3
Determine whether they make a triangle.
no — the arcs never meet
4
Now 5 cm, 12 cm, 13 cm. Determine the same thing.
yes — 5 + 12 = 17, more than 13
5
Now 2 cm, 2 cm, 4 cm, which is the awkward one.
no — 2 + 2 = 4 exactly, so it lies flat
6
And 7 cm, 10 cm, 15 cm.
yes — 7 + 10 = 17, more than 15
Only one comparison matters: the two shorter lengths added together against the longest. If the sum is bigger, the triangle exists; if it is smaller, the two short sides cannot reach; and if it is exactly equal, the three lengths lie flat in a straight line and enclose nothing at all. That last case is the one students miss, so read the sign carefully: it must be greater than, not greater than or equal to.
| The three lengths | The longest | The other two added | A triangle? |
|---|---|---|---|
| 3, 4, 9 cm | 9 cm | 7 cm | no |
| 5, 12, 13 cm | 13 cm | 17 cm | yes |
| 2, 2, 4 cm | 4 cm | 4 cm | no |
| 7, 10, 15 cm | 15 cm | 17 cm | yes |
| 6, 6, 6 cm | 6 cm | 12 cm | yes |
EXAMPLE3 cm, 4 cm and 9 cm: the two shorter add to 7 cm, which is less than 9 cm, so no triangle. 5 cm, 12 cm and 13 cm: 5 + 12 = 17 cm, which beats 13 cm, so yes. 2 cm, 2 cm and 4 cm: 2 + 2 = 4 cm exactly, so the three lengths lie flat and there is still no triangle. And if two sides are 8 cm and 5 cm, the third must be more than 3 cm and less than 13 cm.
Practice · Can three lengths even be a triangle? · 10 questions
P21Determine whether 3 cm, 4 cm and 9 cm can be the sides of a triangle.Fluency
Answerno
P22Determine whether 5 cm, 12 cm and 13 cm can be the sides of a triangle.Fluency
Answeryes
P23Determine whether 2 cm, 2 cm and 4 cm can be the sides of a triangle.Fluency
Answerno
P24Determine whether 7 cm, 10 cm and 15 cm can be the sides of a triangle.Fluency
Answeryes
P25Determine whether 9 cm, 12 cm and 20 cm can be the sides of a triangle, and explain your reasoning.Applied
Answeryes — 9 + 12 = 21 cm, which is greater than 20 cmThe longest length is 20 cm, so that is the one the other two have to beat. Together they come to 9 + 12 = 21 cm, which is greater than 20 cm, so the two shorter sides can reach each other and the triangle closes up. It closes only just: 21 is barely more than 20, so the triangle is a long thin sliver. The other two comparisons, 9 + 20 against 12 and 12 + 20 against 9, both pass automatically and prove nothing.
P26Two sides of a triangle are 8 cm and 5 cm. Determine the range of possible lengths for the third side.Applied
Answergreater than 3 cm and less than 13 cmThe third side must be short enough for the other two to reach round it, so it must be less than 8 + 5 = 13 cm. It must also be long enough that the 5 cm side is not left stranded: 5 plus the third side must beat 8, so the third side must be more than 8 − 5 = 3 cm. The answer is the open range between the difference and the sum, and neither end is included — at exactly 3 cm or exactly 13 cm the three lengths lie flat in a line.
P27Explain why testing the two shorter sides against the longest is enough, and the other two comparisons need not be made.Applied
Answerany comparison that includes the longest side passes automaticallyTake three lengths and call the longest one c. The comparison a + c > b is certainly true, because c on its own is already at least as big as b and a is positive on top of that; the same argument makes b + c > a certainly true. So those two comparisons can never fail and can never tell you anything. Only a + b > c — the two shorter ones against the longest — is capable of failing, so it is the only one worth making.
P28A triangle P Q R has PQ = 8 cm and a perpendicular height of 5 cm drawn from R. Describe where that height meets PQ, and explain how it differs from the median from R.Applied
Answerthe height meets PQ at a right angle, at the foot of the perpendicular; the median meets PQ at its midpointThe perpendicular height from R is the line from R that crosses PQ at 90°, and where it lands is called the foot — it can be anywhere along PQ, and in an obtuse triangle it can even land outside PQ altogether, on the extension. The median from R is a different line: it runs to the midpoint of PQ, cutting it into two equal pieces, and it usually does not meet PQ at a right angle. The two lines coincide only when the triangle is isosceles with PQ as its base. The area of a triangle uses the height, never the median.
P29The sides of a triangle are x cm, (x + 3) cm and (x + 7) cm. Determine the smallest whole-number value of x for which the triangle exists.Challenge
Answerx = 5The longest side is x + 7, so the comparison that can fail is x + (x + 3) > x + 7. That simplifies to 2x + 3 > x + 7, then to x > 4, so the smallest whole number that works is x = 5. Check it: the sides are 5 cm, 8 cm and 12 cm, and 5 + 8 = 13, which is greater than 12 — just. Check x = 4 as well and see it fail: 4 + 7 = 11, exactly equal to 11, so the three lengths lie flat and enclose nothing.
P30Explain which sets of three measurements fix a triangle completely, and discuss why knowing all three angles does not.Challenge
Answerthree sides, two sides and the angle between them, two angles and a side, or a right angle with the hypotenuse and one side; three angles fix only the shape, not the sizeThree sides fix a triangle: build it with compasses and there is only one place the arcs can cross. Two sides with the angle between them fix it, because the third corner is then decided. Two angles and any one side fix it, because the third angle follows from the angle sum and the side sets the scale. And in a right-angled triangle the hypotenuse together with one other side fixes it, because Pythagoras delivers the third side. Three angles fix none of it: they decide the shape but not the size, so all the triangles with angles of 90°, 60° and 30° are enlargements of one another — which is exactly why AAA is a similarity condition in G18 and not a congruence one.
4
a² + b² = c²: finding the hypotenuse
Name the side opposite the right angle first, then square, add and root
Give a triangle a right angle and something extra becomes true of it. Name the side facing the right angle — the hypotenuse, always the longest of the three — and call it c, and call the two shorter sides a and b. Then a² + b² = c², always, in every right-angled triangle there has ever been. To find the hypotenuse, square the other two, add, and take the square root. The number under the root is usually not a square number, so the answer is usually a surd: leave it as a simplified surd and it is exact, write it as a decimal and it is rounded, however many figures you give.
This is the theorem as a picture rather than as a formula. The two small squares, built on the two sides that meet at the right angle, hold 16 cm² and 9 cm² between them. The big square, built on the side opposite the right angle, holds 25 cm² — exactly the other two added together.
Name the hypotenuse before you write anything down: it is the side facing the right angle, BC, and it is always the longest of the three. Squaring the other two and adding gives 25 + 144 = 169, and 169 is a square number, so this answer needs no rounding at all.
Here the two squares add to 20, and 20 is not a square number. The exact length of QR is the simplified surd printed on it; 4.47 cm is that same number rounded to 3 significant figures, and the two are not interchangeable.
Worked example 4 · Find the hypotenuse, exactly and then roundly
1
A right-angled triangle has shorter sides of 2 cm and 4 cm. Identify the hypotenuse.
the side facing the right angle
2
Calculate the two squares and add them.
20 under the root
3
20 is not a square number, so the root is not a whole number. Simplify it — that move is Algebra u19’s.
√20 = 2√5
4
State the exact length.
2√5 cm
5
Calculate the same length to 3 significant figures.
4.47 cm
6
Verify the answer is sensible: is it longer than 4 cm and shorter than 6 cm?
yes — 4.47 sits between them
The exact answer is the surd; the decimal is that number rounded. If a question says exact, or in surd form, or leave your answer in the form a√b, a decimal earns nothing. And the sense check at the end is free: a hypotenuse is always longer than either other side and always shorter than the two of them added, so an answer outside that window is wrong before you check anything else.
| The two shorter sides | Under the root | Exact | To 3 s.f. |
|---|---|---|---|
| 3 cm and 4 cm | 25 | 5 cm | 5.00 cm |
| 5 cm and 12 cm | 169 | 13 cm | 13.0 cm |
| 9 cm and 12 cm | 225 | 15 cm | 15.0 cm |
| 2 cm and 4 cm | 20 | 2√5 cm | 4.47 cm |
| 1 cm and 1 cm | 2 | √2 cm | 1.41 cm |
What the hypotenuse is
- the side facing the right angle, never one of the two that make it
- always the longest of the three sides
- always shorter than the other two added together
- the one you get by adding the two squares
What it never is
- the side that happens to look longest on a diagram that is not to scale
- the two shorter sides added — that is the way round the corner, not across it
- found by subtracting, unless the hypotenuse is the side you already know
- a number without a unit after it
EXAMPLEShorter sides of 3 cm and 4 cm give 9 + 16 = 25, so the hypotenuse is 5 cm and no rounding is needed. Shorter sides of 2 cm and 4 cm give 4 + 16 = 20, and 20 is not a square number, so the exact hypotenuse is 2√5 cm — which is 4.47 cm to 3 significant figures. A square of side 6 cm has a diagonal of 6√2 cm, because 36 + 36 = 72.
Practice · a² + b² = c²: finding the hypotenuse · 10 questions
P31A right-angled triangle has shorter sides of 3 cm and 4 cm. Calculate its hypotenuse.Fluency
Answer5 cm
P32A right-angled triangle has shorter sides of 5 cm and 12 cm. Calculate its hypotenuse.Fluency
Answer13 cm
P33A right-angled triangle has shorter sides of 9 cm and 12 cm. Calculate its hypotenuse.Fluency
Answer15 cm
P34A right-angled triangle has shorter sides of 8 cm and 15 cm. Calculate its hypotenuse.Fluency
Answer17 cm
P35A right-angled triangle has shorter sides of 2 cm and 4 cm. Determine its hypotenuse exactly, and then to 3 significant figures.Applied
Answer2√5 cm (4.47 cm to 3 s.f.)Squaring and adding gives 4 + 16 = 20, so the hypotenuse is √20. Twenty is 4 × 5 and 4 is a square number, so the root simplifies to 2√5 cm — that is the exact answer. To 3 significant figures it is 4.47 cm. Both are correct answers to different questions, and only the first one is exact.
P36A square has sides of 6 cm. Determine the exact length of its diagonal.Applied
Answer6√2 cm (8.49 cm to 3 s.f.)The diagonal cuts the square into two right-angled triangles whose shorter sides are both 6 cm, so the diagonal is the hypotenuse of one of them. Squaring and adding gives 36 + 36 = 72, and √72 simplifies to 6√2 cm, or 8.49 cm to 3 significant figures. Every square behaves this way: the diagonal is always the side length multiplied by √2.
P37A gate is 1.6 m wide and 1.2 m high. Calculate the length of the diagonal brace that runs corner to corner.Applied
Answer2 mThe gate is a rectangle, so the brace is the hypotenuse of a right-angled triangle with shorter sides 1.6 m and 1.2 m. Squaring and adding gives 2.56 + 1.44 = 4, and √4 = 2, so the brace is exactly 2 m. The numbers were chosen to be a 3-4-5 triangle scaled by 0.4, which is why the answer is so clean — and why the brace must be ordered at 2 m, not 2.8 m.
P38A right-angled triangle has shorter sides of 7 cm and 24 cm. Calculate its hypotenuse.Applied
Answer25 cmSquaring and adding gives 49 + 576 = 625, and 625 is a square number: √625 = 25. The hypotenuse is 25 cm, and no rounding is needed at all. 7, 24, 25 is worth recognising on sight, alongside 3-4-5, 5-12-13 and 8-15-17.
P39A right-angled triangle has two shorter sides of 3√2 cm each. Show that its hypotenuse is a whole number of centimetres.Challenge
Answer6 cmSquaring a surd removes the root sign: (3√2)² = 9 × 2 = 18. Both shorter sides give 18, so the sum is 18 + 18 = 36, and √36 = 6. The hypotenuse is exactly 6 cm. Two irrational sides have produced a rational hypotenuse, which is a useful reminder that a surd in the working does not mean a surd in the answer.
P40A rectangle measures 24 cm by 10 cm. Calculate the length of its diagonal, and hence determine the perimeter of the triangle formed by the diagonal and two sides of the rectangle.Challenge
Answer26 cm; the perimeter is 60 cmThe diagonal is the hypotenuse of a right-angled triangle whose shorter sides are 24 cm and 10 cm. Squaring and adding gives 576 + 100 = 676, and √676 = 26, so the diagonal is 26 cm. The triangle it makes with the two sides has a perimeter of 24 + 10 + 26 = 60 cm. Follow through on your own diagonal if it was wrong: the second part is worth its own marks.
5
Finding a shorter side, and asking whether it is right at all
The subtraction case, and the converse that turns three lengths into a right angle
The same rule run the other way finds a shorter side. If you know the hypotenuse and one short side, you subtract: c² − a² = b². Deciding between adding and subtracting is the whole game, and it is decided by one question — is the side I am looking for the one facing the right angle, or not? Then there is the rule read backwards, its converse: give me three lengths and no diagram at all, and if the two smaller squares add to the largest square, the triangle must be right-angled, with the right angle facing the longest side. If they do not add up, it is not — and which way the comparison falls even tells you whether the biggest angle is obtuse or acute.
The ladder itself is the hypotenuse, because it faces the right angle where the wall meets the ground. The unknown here is a shorter side, so this one is a subtraction: square the ladder, take away the square of the foot distance, and root what is left.
Nobody said this triangle was right-angled. Squaring the two shorter sides gives 64 + 225 = 289, and the longest side squared is 17² = 289 as well — so the angle at A really is a right angle. Running the theorem backwards like this is called its converse.
Worked example 5 · The ladder against the wall
1
A 6.5 m ladder leans against an upright wall with its foot 2.5 m from the wall. Identify the hypotenuse.
the ladder itself
2
Determine whether this is an addition or a subtraction.
a subtraction — the unknown is a shorter side
3
Calculate the two squares.
6.5² = 42.25 and 2.5² = 6.25
4
Calculate the difference, which is the square of the height.
36
5
Determine how far up the wall the ladder reaches.
6 m
6
Discuss whether the answer is sensible.
6 m is less than the 6.5 m ladder, as it must be
Everything in this question turns on one decision made in the first five seconds: the ladder is the hypotenuse, because it faces the right angle where the wall meets the ground. Adding the two given numbers’ squares here would give a height of about 7 m — a ladder reaching higher than its own length, which is impossible. The sense check catches that instantly, so always make it.
| The hypotenuse | The known shorter side | The subtraction | The other shorter side |
|---|---|---|---|
| 13 cm | 5 cm | 144 | 12 cm |
| 10 cm | 6 cm | 64 | 8 cm |
| 25 cm | 7 cm | 576 | 24 cm |
| 6.5 cm | 2.5 cm | 36 | 6 cm |
| 7 cm | 3 cm | 40 | 2√10 cm |
EXAMPLEA hypotenuse of 13 cm with a short side of 5 cm gives 169 − 25 = 144, so the other side is 12 cm. A hypotenuse of 7 cm with a short side of 3 cm gives 49 − 9 = 40, so the other side is 2√10 cm, or 6.32 cm to 3 significant figures. And for the converse: 8, 15, 17 gives 64 + 225 = 289 = 17², so that triangle is right-angled, while 7, 8, 12 gives 49 + 64 = 113 against 144, so that one is not.
Practice · Finding a shorter side, and asking whether it is right at all · 10 questions
P41A right-angled triangle has a hypotenuse of 13 cm and one shorter side of 5 cm. Calculate the other shorter side.Fluency
Answer12 cm
P42A right-angled triangle has a hypotenuse of 10 cm and one shorter side of 6 cm. Calculate the other shorter side.Fluency
Answer8 cm
P43A right-angled triangle has a hypotenuse of 25 cm and one shorter side of 7 cm. Calculate the other shorter side.Fluency
Answer24 cm
P44A right-angled triangle has a hypotenuse of 6.5 cm and one shorter side of 2.5 cm. Calculate the other shorter side.Fluency
Answer6 cm
P45A 6.5 m ladder leans against an upright wall with its foot 2.5 m from the wall. Calculate how far up the wall it reaches.Applied
Answer6 mThe ladder faces the right angle between the wall and the ground, so the ladder is the hypotenuse and the unknown height is a shorter side — which makes this a subtraction. Squaring gives 42.25 for the ladder and 6.25 for the foot distance, and 42.25 − 6.25 = 36. The square root of 36 is 6, so the ladder reaches 6 m up the wall. Sense check: 6 m is less than the 6.5 m ladder, as it has to be.
P46A right-angled triangle has a hypotenuse of 7 cm and one shorter side of 3 cm. Determine the other shorter side exactly, and then to 3 significant figures.Applied
Answer2√10 cm (6.32 cm to 3 s.f.)Subtracting the squares gives 49 − 9 = 40, so the side is √40. Forty is 4 × 10, so the root simplifies to 2√10 cm, which is 6.32 cm to 3 significant figures. Check the answer is sensible: it is shorter than the 7 cm hypotenuse and longer than the 3 cm side, which is where a second shorter side has to sit.
P47Determine whether a triangle with sides 8 cm, 15 cm and 17 cm is right-angled.Applied
Answeryes — 64 + 225 = 289, and 17² = 289Use the converse of the theorem: square the two shorter sides and add, then compare with the square of the longest. Here 8² + 15² = 64 + 225 = 289, and 17² = 289 as well. The two are equal, so the triangle is right-angled, and the right angle is the one facing the 17 cm side. Always square the longest side on its own; putting it in with the other two is the slip that turns this test into nonsense.
P48Determine whether a triangle with sides 7 cm, 8 cm and 12 cm is right-angled.Applied
Answerno — 49 + 64 = 113, but 12² = 144Squaring the two shorter sides and adding gives 7² + 8² = 49 + 64 = 113, while the longest side squared is 12² = 144. They are not equal, so the triangle is not right-angled. It is a real triangle — 7 + 8 = 15, which beats 12 — but the angle facing the 12 cm side is bigger than a right angle, because the square of that side is more than the other two squares put together. That is a useful extra: when c² is greater, the triangle is obtuse; when it is smaller, the triangle is acute.
P49A rectangle has a diagonal of 26 cm and one side of 24 cm. Calculate its other side, and hence determine its area.Challenge
Answer10 cm; the area is 240 cm²The diagonal is the hypotenuse of a right-angled triangle whose other two sides are the sides of the rectangle. Subtracting the squares gives 676 − 576 = 100, and √100 = 10, so the other side is 10 cm. The area of the rectangle is then 24 × 10 = 240 cm². Notice the units change between the two parts: a length in centimetres, an area in square centimetres.
P50Show that a triangle with sides 9 cm, 40 cm and 41 cm is right-angled, and state which angle is the right angle.Challenge
Answer81 + 1600 = 1681 = 41², so it is right-angled; the right angle faces the 41 cm sideSquare the two shorter sides and add: 9² + 40² = 81 + 1600 = 1681. Square the longest side on its own: 41² = 1681. The two agree exactly, so by the converse of Pythagoras’ theorem the triangle is right-angled. The right angle is always the one opposite the longest side, so it is the angle between the 9 cm and 40 cm sides. 9, 40, 41 is a genuine Pythagorean triple, and the Criterion B task on this page shows where it comes from.
6
Pythagoras twice over: inside a solid
Across the floor and then up — the space diagonal and the slant height
Inside a box, the theorem gets used twice. To go from one corner of a cuboid to the opposite one, first cross the floor: that diagonal is the hypotenuse of a right-angled triangle whose sides are the length and the width. Then stand up: the space diagonal is the hypotenuse of a second right-angled triangle whose sides are the floor diagonal and the height. The same two-stage move gives the slant height of a pyramid, which is exactly the number G14 Solid Progress handed you without proof. One warning matters more than all the others here: feed the exact first answer into the second stage, surd and all. Round in the middle and the final answer drifts.
Two right-angled triangles, one after the other. The green line runs across the floor of the box and is the hypotenuse of a 6 by 3 triangle. The red line runs from the same corner to the opposite one, and it is the hypotenuse of a triangle whose two sides are the green line and the height. Feed the exact green answer into the second stage, never a rounded one.
The slant height is the red line: from the apex straight down the middle of a triangular face to the middle of a base edge. It is the hypotenuse of a right-angled triangle whose other two sides are the perpendicular height and half the base edge — 12 cm and 5 cm here, which is why the slant height comes out at exactly 13 cm.
Worked example 6 · The space diagonal of a box
1
A box measures 6 cm by 3 cm by 2 cm. Identify the two right-angled triangles you will need.
one on the floor, one standing up
2
Calculate the diagonal across the floor of the box.
3√5 cm
3
Explain why you should not round that answer yet.
it is the input to the next stage
4
The second triangle has that diagonal as one side and the height as the other. Calculate what goes under the root.
49
5
Determine the space diagonal.
7 cm
6
Verify it in one line: add all three squares.
36 + 9 + 4 = 49, and √49 = 7
Pythagoras used twice, and the second use is fed the exact output of the first. Here the floor diagonal is 3√5 cm, an irrational number; rounding it to 6.71 cm and squaring that gives 45.0241, not 45, and the final answer drifts off 7. Carrying the surd is not fussiness, it is what keeps the second stage honest — and it is why the one-line check, the sum of all three squares, comes out exact.
Worked example 7 · Show that the slant height really is 13 cm
1
A pyramid has a square base of side 10 cm and a perpendicular height of 12 cm. Identify where the slant height runs.
apex to the middle of a base edge
2
Determine the horizontal distance from the foot of the height to that middle point.
half the base edge, 5 cm
3
Identify the right-angled triangle those two lines make.
legs of 12 cm and 5 cm
4
Calculate what goes under the root.
169
5
Show that the slant height is 13 cm.
13 cm
6
Explain why the slant height is not the same as the edge from the apex to a corner.
a corner is further out than a mid-edge
This closes a loop that G14 Solid Progress left open: the slant height of a pyramid or a cone was handed to you there as a stated result so that surface area could be done, and here it is proved. Half the base edge, not the whole base edge, is the number that goes into the triangle — and the slant height is always shorter than the edge to a corner, because a corner is further from the centre than the middle of a side is.
| The box | Across the floor | Under the second root | The space diagonal |
|---|---|---|---|
| 6 by 3 by 2 cm | 3√5 cm | 49 | 7 cm |
| 3 by 4 by 12 cm | 5 cm | 169 | 13 cm |
| 2 by 3 by 6 cm | √13 cm | 49 | 7 cm |
| 1 by 2 by 2 cm | √5 cm | 9 | 3 cm |
| 4 by 4 by 7 cm | 4√2 cm | 81 | 9 cm |
EXAMPLEA box 6 cm by 3 cm by 2 cm has a floor diagonal of 3√5 cm, and feeding that exact value into the second stage gives 45 + 4 = 49 and a space diagonal of exactly 7 cm. A cube of side 4 cm has a space diagonal of 4√3 cm, or 6.93 cm to 3 significant figures. A pyramid on a square base of 10 cm with a height of 12 cm has a slant height of 13 cm, because half the base edge is 5 cm and 144 + 25 = 169.
Practice · Pythagoras twice over: inside a solid · 10 questions
P51A box measures 3 cm by 4 cm by 12 cm. Calculate the diagonal across its 3 cm by 4 cm floor.Fluency
Answer5 cm
P52A box measures 3 cm by 4 cm by 12 cm. Calculate its space diagonal.Fluency
Answer13 cm
P53A box measures 6 cm by 3 cm by 2 cm. Determine the exact diagonal across its 6 cm by 3 cm floor.Fluency
Answer3√5 cm (6.71 cm to 3 s.f.)
P54A box measures 6 cm by 3 cm by 2 cm. Calculate its space diagonal.Fluency
Answer7 cm
P55A box measures 2 cm by 3 cm by 6 cm. Calculate its space diagonal.Applied
Answer7 cmStage one, across the floor: 2² + 3² = 4 + 9 = 13, so the floor diagonal is √13 cm — keep it as a surd. Stage two, standing up: that diagonal and the 6 cm height are the two shorter sides of a second right-angled triangle, and 13 + 36 = 49. The square root of 49 is 7, so the space diagonal is exactly 7 cm. The one-line check adds all three squares: 4 + 9 + 36 = 49.
P56A box measures 1 cm by 2 cm by 2 cm. Calculate its space diagonal.Applied
Answer3 cmAcross the 1 cm by 2 cm floor first: 1 + 4 = 5, so the floor diagonal is √5 cm. Then upwards with the 2 cm height: 5 + 4 = 9, and √9 = 3, so the space diagonal is 3 cm. Rounding √5 cm to 2.24 first and squaring gives 5.0176, and the final answer drifts off 3 — which is exactly why the surd is carried.
P57A pyramid has a square base of side 10 cm and a perpendicular height of 12 cm. Calculate its slant height.Applied
Answer13 cmThe slant height runs from the apex to the middle of a base edge, and the foot of the perpendicular height is at the centre of the base, which is half the base edge away from that middle point: 10 ÷ 2 = 5 cm. So the right-angled triangle has shorter sides of 12 cm and 5 cm, giving 144 + 25 = 169 and a slant height of 13 cm. Half the base edge, not the whole one, is the number that goes in.
P58A pyramid has a square base of side 16 cm and a perpendicular height of 15 cm. Calculate its slant height.Applied
Answer17 cmHalf the base edge is 16 ÷ 2 = 8 cm, so the right-angled triangle has shorter sides of 15 cm and 8 cm. Squaring and adding gives 225 + 64 = 289, and √289 = 17, so the slant height is 17 cm. This is the calculation G14 Solid Progress needed when it worked out the surface area of a pyramid and was handed the slant height ready-made.
P59A cube has sides of 4 cm. Determine the exact length of its space diagonal.Challenge
Answer4√3 cm (6.93 cm to 3 s.f.)Across the floor first: 16 + 16 = 32, so the floor diagonal is 4√2 cm. Then upwards: 32 + 16 = 48, and √48 simplifies to 4√3 cm, or 6.93 cm to 3 significant figures. Every cube behaves this way: the space diagonal is always the side length multiplied by √3, just as the face diagonal is the side multiplied by √2.
P60A pyramid has a square base of side 16 cm and a perpendicular height of 14 cm. Calculate the length of the edge running from the apex to one corner of the base.Challenge
Answer18 cmThis time the triangle reaches a corner, not the middle of an edge, so the horizontal distance is half the base diagonal. The base diagonal is 16√2 cm because 256 + 256 = 512, so half of it is 8√2 cm — and its square is 128, which is the number that matters. Then 128 + 196 = 324, and √324 = 18, so the edge is 18 cm. Compare it with P58: the corner is further from the centre than the middle of a side is, so this edge is the longer of the two.
Now construct some for yourself
Construct a triangle with sides of 5 cm, 8 cm and 12 cm on the blank squared paper below: rule the 12 cm side first, then swing a 5 cm arc from one end and an 8 cm arc from the other. Then try 3 cm, 4 cm and 9 cm the same way and watch the arcs fail to meet. Finally construct a right-angled triangle with shorter sides of 6 cm and 8 cm along the grid lines, measure its longest side, and check your measurement against 10 cm.
Number the axes one square to one centimetre each way. Rule the longest side first, along a grid line, then swing the two other lengths from its ends with a pair of compasses. If the arcs cross, the triangle exists; if they do not, the three lengths have failed the inequality and no amount of redrawing will help.
The right-angle routine
Four moves, whichever side the question turns out to want.
1
Sketch it and mark the right angle before anything elseEven when a diagram is given, redraw it big. Put the square in the right-angle corner, then find the side facing that corner and label it c. Everything else in the calculation follows from getting that one label right.
2
Decide whether you are adding or subtractingIf the unknown is the hypotenuse, you add the two squares. If the unknown is one of the shorter sides, you subtract the square of the known short side from the square of the hypotenuse. There is no third case, and choosing wrongly is the single commonest error in the topic.
3
Work with the squares, and take the root lastWrite the squares down as numbers — 25 and 144, not 5² and 12² left standing — combine them, and only then take the square root. If the number under the root is not a square number, the exact answer is the simplified surd, and the decimal comes after it, never instead of it.
✓
Check the answer against the picture and the unitsThe hypotenuse must come out longer than either other side and shorter than the two of them added; if it does not, you have added where you should have subtracted. Then check the unit is on the answer and, in a real context, that the number is a size a real thing could be.
The mistakes that cost the most marks
1Adding the two given sides when one of them is already the hypotenuse
Wrong“The ladder is 6.5 m and its foot is 2.5 m out, so the height is √(6.5² + 2.5²) = 6.96 m.” The two numbers have been squared and added, out of habit.
RightAdding is only right when the unknown is the hypotenuse. Here the ladder is the hypotenuse — it faces the right angle between the wall and the ground — so the unknown is a shorter side and the working is a subtraction: 42.25 − 6.25 = 36, and the height is 6 m. The wrong answer, 6.96 m, is a ladder reaching higher up a wall than the ladder is long.
Fix it with one habit, before any arithmetic: find the right angle, then find the side facing it, then ask whether that side is the one you are looking for. If it is, add. If it is not, subtract. Writing the letter c on the hypotenuse of your sketch costs two seconds and removes the decision entirely — and the sense check afterwards, that the hypotenuse must be the longest of the three, catches whatever is left.
And never decide which side is the hypotenuse by looking at the picture. The figure below is drawn with the vertical direction stretched, so the side that appears longest on the page is not the longest at all. The hypotenuse is fixed by the right angle and by nothing else: it is the side facing it.
The vertical direction here is stretched, so the drawing is no guide at all to which side is longest. The hypotenuse is decided by the right angle and by nothing else: it is the side facing it, BC, whatever the picture looks like.
2Testing the triangle inequality on only one pair of sides
Wrong“3 cm, 4 cm and 9 cm: well, 4 + 9 = 13, which is bigger than 3, so it is a triangle.” One comparison has been made, and it happens to be the one that was always going to pass.
RightThe comparison that matters is the two shorter sides against the longest. Any pair that includes the longest side will pass automatically, so testing that pair proves nothing at all. Here 3 + 4 = 7, which is less than 9, so the two short sides cannot reach each other and there is no triangle. The picture is two compass arcs that stop short of one another with a 2 cm gap between them.
So sort the three lengths first, or at least spot the longest by eye, and make the one comparison that can fail. Watch the boundary case as well: 2, 2 and 4 give 2 + 2 = 4, exactly equal rather than greater, and the three lengths lie flat along one straight line enclosing no area. The rule is strictly greater than — equal is not good enough, and an exam question will test exactly that.
3Writing a rounded decimal where the exact surd was asked for
Wrong“The shorter sides are 2 cm and 4 cm, so the hypotenuse is 4.47 cm.” The number is right, but the question asked for the exact length and 4.47 is a rounded one.
RightExact means the surd. The two squares add to 20, so the hypotenuse is √20, which simplifies to 2√5 cm. That is an exact number; 4.47 cm is the same number rounded to 3 significant figures, and the two are not interchangeable. Give the surd whenever the question says exact, in surd form, or in the form a√b.
There is a practical reason as well as a marks reason, and it bites hardest in the three-dimensional questions. A rounded value carried into a second stage drags its error with it: the floor diagonal of a 6 by 3 box is 3√5 cm, and rounding it to 6.71 before squaring gives 45.0241 instead of 45, so the space diagonal comes out at 7.002 instead of exactly 7. Keep the surd all the way through, round once at the very end, and round rather than chop.
Tasks to submit
Criterion B · Investigating patterns
The triple factory
Some right-angled triangles come out in whole numbers: 3, 4, 5 is the famous one, and 5, 12, 13 is the next. Sets of three whole numbers that fit a² + b² = c² are called Pythagorean triples, and hunting for them one at a time is slow work. There is a machine that makes them. Choose any two whole numbers m and n with m bigger than n, and build three lengths from them: m² − n², 2mn and m² + n². That is the factory. Your job is to find out what it produces, whether everything it produces really is a right-angled triangle, and which of its products are genuinely new rather than an old one scaled up. It is not a rule you have met, it is in none of the units before this one, and the last part of it has an answer that will surprise you.
The pair (2, 1) produces 3, 4, 5 and the pair (3, 2) produces 5, 12, 13. Both are right-angled, and neither needed a square root to find: the factory hands you three whole numbers that fit the theorem exactly.
Part 1 · structured
The first two rows are given so that the column headings are unambiguous: a and b are the two shorter sides in either order, and c is the longest.
| (m, n) | a | b | c | a² + b² | c² | Primitive? |
|---|---|---|---|---|---|---|
| (2, 1) | 3 | 4 | 5 | 25 | 25 | yes |
| (3, 2) | 5 | 12 | 13 | 169 | 169 | yes |
| (4, 1) | ||||||
| (4, 3) | ||||||
| (5, 2) | ||||||
| (3, 1) | ||||||
| (4, 2) |
- Predict, before you calculate anything, whether you think every pair (m, n) will produce a right-angled triangle, and whether you think every Pythagorean triple can be produced this way. Write both predictions down; you will be asked to compare them with what you find.
- Complete the table for the seven pairs given. Work out the three lengths from the pair, then square the two shorter ones and add, then square the longest one, and finally decide whether the triple is primitive — that is, whether its three numbers have no common factor bigger than 1. The first two rows are filled in for you.
- Identify the pattern. Describe what is always true of the two columns headed a² + b² and c², and explain what that tells you about every triangle the factory makes.
- Determine a general rule for which pairs (m, n) give a primitive triple and which give one that is just a smaller triple scaled up. Look hard at (3, 1) and at (4, 2) before you commit yourself. State the rule as a condition on m and n, not as a list of cases, and use it to predict what the pair (5, 4) will produce and whether it will be primitive.
- Test and verify your rules. Generate three more triples from pairs of your own choosing, at least one of which you expect to be primitive and at least one of which you expect not to be. Check every one against a² + b² = c², and check your primitivity rule against each. Then discuss whether any pair could ever fail to make a triangle at all.
- Justify the factory algebraically. Expand (m² − n²)² + (2mn)² and expand (m² + n²)², and explain why the two expansions must be equal for every m and every n. Then explain why that single piece of algebra proves something no amount of table-filling could: that the factory works for numbers nobody has ever tested.
The blank squared paper · for drawing your triples
Number the axes yourself, one square to one centimetre each way. Draw the two shorter sides of each triple along the grid lines so that the right angle is exact, then join them and check the third side against the number the factory predicted.
Print the PDF if you would rather work on paper — the same blank grid is on the Criterion B page there. Draw the two shorter sides of each triple along the grid lines so that the right angle is exact, then measure the third side and check it against the number the factory predicted.
Part 2 · open
Now you choose what to investigate. Here are three openings — take one, or propose your own. You decide which cases to try, how many you need, how to record what you find and how you will test it.
- Which triples are missing? Investigate whether every primitive triple can be produced by the factory. Try to find the pair (m, n) that produces 20, 21, 29 and the pair that produces 33, 56, 65, and discuss what your search suggests about the completeness of the machine.
- Odd and even. Investigate the parity of the three numbers in a primitive triple — how many are odd, how many are even, and whether that ever changes. Determine a rule and explain it from the forms m² − n², 2mn and m² + n².
- Nearly square. Investigate the triples for which the two shorter sides differ by exactly 1 — 3, 4, 5 and 20, 21, 29 are the first two. Find the next one or two, determine how the pairs (m, n) that produce them are related to each other, and discuss whether the list ever stops.
Whichever you choose, take it through the same ladder: predict, identify the pattern, determine a general rule, test and verify it on a case you have not already worked out, and justify why it must be true.
Submit: your completed table, the two predictions you wrote before you calculated anything, every triple you generated with the pair that produced it, your primitivity rule, your test on pairs of your own choosing, and the algebra of step 6 — as a scan or a typed document. Marked out of 8 against Criterion B. Completing the table earns the lower bands; the primitivity rule, the test on pairs you were not given, and the algebraic justification are what reach levels 5–8. Part 2 must show your own choice of method.
Remember, you should communicate in an organized and coherent manner.
Criterion D · Applying maths in real-life contexts
How long is a rafter?
A family in Mogoditshane is putting a roof on a house they are building themselves. It is a simple gable roof: the walls go up, a ridge beam runs along the middle, and pairs of straight rafters lean from the wall plates up to the ridge. The builder has written out a sheet of measurements, and it records a great deal more than you need. Two questions have to be answered from it. How long must each rafter be, from wall plate to ridge? And the rafters have already been ordered at 4.2 m — does that length leave a sensible overhang past the wall, or has the family bought the wrong timber? Everything you need is a length somebody could measure with a tape, and the shape that answers both questions is a right-angled triangle you will have to find for yourself in the drawing.
The builder’s sheet (illustrative figures, written for this task)
- The house: a family house being built in Mogoditshane, roofed with a simple gable roof
- The roof shape: two flat slopes meeting at a ridge that runs along the middle of the house
- The span: the distance from one wall plate straight across to the other is 7.2 m
- The rise: the ridge sits 1.5 m above the level of the wall plates
- The rafters: 24 of them have been ordered, each 4.2 m long
- The spacing: the rafters are to be set 0.6 m apart along the ridge
- The length of the house: 12 m from end to end along the ridge
- The walls: the wall plates sit 2.7 m above the floor
- The ceiling: a flat ceiling is to be fixed at 2.4 m inside
- The plot: the plot is 900 m²
- The overhang: the family want the roof to project past the wall on each side so that rain runs clear of the wall face
- How it was measured: with a 5 m steel tape, every reading written down to the nearest centimetre
| What was measured | The measurement |
|---|---|
| Span, wall plate to wall plate | 7.2 m |
| Ridge rise above the wall plates | 1.5 m |
| Rafters ordered | 24 lengths of 4.2 m |
| Rafter spacing along the ridge | 0.6 m |
| Length of the house along the ridge | 12 m |
| Wall height, floor to wall plate | 2.7 m |
| Ceiling height inside | 2.4 m |
| Plot area | 900 m² |
These figures are illustrative. The house, the builder’s sheet and the rafter order are an invented illustrative example written for the task. It is not taken from any business, price list or publication, and no organisation, person or source is named or implied. What is authentic is the shape of the problem: a roof really is two right-angled triangles back to back, rafters really are ordered longer than the calculated length to give an overhang, and a tape really is read to the nearest centimetre.
L and R are the two wall plates and T is the ridge. The two rafters are equal, so they carry matching ticks. The green dashed line is the rise from wall-plate level to the ridge, and it meets the span at a right angle exactly halfway along — which is what splits the roof into two right-angled triangles.
Metres along the bottom and metres up the side. Use the same scale on both, because this is a cross-section of a real roof and two different scales would make every rafter length on it a lie.
- Identify which entries on the builder’s sheet are needed to answer the two questions and which are not, and explain, item by item, why each of the others cannot change either answer — deal with the rafter spacing, the length of the house, the wall height, the ceiling height and the plot area by name. Then say which single surplus entry you would still want to know about before the first rafter is cut, and why. (Strand i — identifying the elements of the authentic situation)
- Select the method you will use. Explain where the right-angled triangle is in this roof, which of its three sides is the hypotenuse and how you know, and why the number that goes into the calculation is half the span rather than the whole of it. Say in advance how you will check that your answer is sensible. (Strand ii — selecting adequate mathematical strategies)
- Apply them. Sketch the roof in cross-section on the blank axes, marking the span, the rise and the right angle. Calculate the length of one rafter from wall plate to ridge, the total length of rafter timber the family will use, and the overhang the ordered length leaves at each end. State the unit on every answer. (Strand iii — applying the strategies to reach a solution)
- Explain the degree of accuracy of your answers. Every measurement was written down to the nearest centimetre, so each one could be out by up to half a centimetre. Explain what that does to a rafter length worked out from two of them, whether the answer deserves to be written to the centimetre, and why an answer given to four decimal places would be dishonest here. (Strand iv — explaining the degree of accuracy of the solution)
- Discuss whether the ordered 4.2 m rafters are the right choice. Say what overhang they leave once the rafter reaches the ridge, discuss whether that is enough to keep rain off the wall, and discuss what else you would want to allow for before cutting — the notch where the rafter sits on the wall plate, the thickness of the ridge beam, and the fact that timber is sold in fixed lengths. Your recommendation may be to keep the order or to change it, but it must follow from your own numbers. (Strand v — explaining whether the solution makes sense in context)
Submit: your cross-section sketch with both axes labelled and scaled, your list of which sheet entries matter and which do not, the rafter length, the total rafter timber and the overhang with their units, and written answers for strands i, iv and v. Marked out of 8 against Criterion D at Standard level, where the command terms are explain and discuss: a solution that gives the numbers and says nothing about them cannot reach the upper bands. Give every length in metres.
When you are ready
The mastery quiz is 10 questions drawn at random from a bank of 25. You need 70% to pass and you have three attempts. Type a length as a number with no unit, like 13; type an angle as a number with no degree sign, like 70; and type an exact surd as 2sqrt5 or 2√5. Work through the practice questions first — especially Concept 5, where the difference between adding and subtracting is the whole question, and Concept 6, where the exact answer from the first stage has to be carried into the second.
Printable version
Download Unit G20 as a PDF — the same lesson, laid out for printing and for working offline.
Download Unit G20 as a PDF — the same lesson, laid out for printing and for working offline.