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Pythagoras theorem and right angled triangle problems - Mathematics Year 8

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Introduction

An acute triangle divided by a perpendicular height into two right-angled subtriangles. — The right-angle mark is at the foot of h. It licenses a right-triangle calculation in either subtriangle, not in the unmarked whole triangle. No numerical side lengths are given by this image.

Which side must be squared on its own, and when should you subtract rather than add? Identify the right angle before using Pythagoras, then connect a diagram, an equation and a checked length.

You will solve six right-triangle problems, including diagonal distances, and explain two plausible errors. Exact roots, units and rounding matter; an on-screen drawing does not replace stated geometric givens.

Identify the hypotenuse and connect square areas

A right-angled triangle has one angle of 90°. Its hypotenuse is the side opposite that angle and is the longest side. The other two sides meet at the right angle and are called legs here. Rotate the diagram and the hypotenuse remains opposite the right angle; it is not defined as the side drawn at the top, bottom or a particular slope.

For perpendicular legs a and b, and hypotenuse c, Pythagoras gives $a^{2} + b^{2} = c^{2}$ . The equation connects areas of squares built on the three sides. The sum of the two smaller square areas equals the area of the hypotenuse square. The side lengths themselves are not being added: a+b is generally longer than the straight-line hypotenuse.

Identify before substituting. Mark the right angle, name the opposite side c, record the two known lengths in the same unit and decide whether the unknown is c or a leg. In the introductory illustration, the full triangle is acute. The drawn height creates a right angle at its foot, splitting the full base into two parts. A subtriangle uses its own base segment and height, not automatically the entire base b. Do not infer those segment lengths from the screenshot.

Source example, hypotenuse missing. Let the perpendicular legs be 6 cm and 8 cm. Then c²=6²+8²=36+64=100 cm². Taking the positive square root gives c=10 cm. A length is positive here, so −10 is not a second physical side. Check that 10 is longer than both 6 and 8, then substitute: 36+64=100. A rough bound also helps: the diagonal is between the longer leg 8 and the two-leg route 6+8=14.

The squared equation can be explained by a square-area rearrangement. A large square of side a+b contains four congruent right triangles, each of area ab/2. The remaining square has area (a+b)²−4(ab/2)=a²+b² and side c in that construction. This connects the area model with the symbolic rule rather than treating the theorem as a calculator instruction.

The theorem in this course requires a stated right angle. For an arbitrary triangle with only two side lengths known, neither adding the squares nor measuring a rough picture justifies the missing side. If the right-angle condition is absent, state the missing information instead of inventing it. The course evaluates reasoning about given models, not practical surveying accuracy.

Rearrange for a leg and interpret diagonal distances

Source example, a shorter leg missing. A right triangle has hypotenuse 13 cm and one leg 5 cm. With unknown leg x, x²+5²=13², so x²=169−25=144 and x=12 cm. Check 5²+12²=25+144=169=13² and confirm 13 is longest. The subtraction is obtained by rearranging the same equation, not by memorising an unrelated rule.

An alternative is to recognise the 5–12–13 triple and verify it exactly. Recognition is efficient for a familiar triple, while the squared equation works for unfamiliar positive lengths too. Guessing a triple without checking its scale and the right-angle condition is not reliable. Retain the equation even when mental arithmetic provides the answer.

First-error analysis. A learner writes x²=13²+5²=194 for that missing leg. The first error assigns the sum to a leg rather than to the stated hypotenuse. The result √194 is larger than 13, contradicting the given hypotenuse's role. A second learner writes c=6+8=14 in the original example. That adds lengths rather than square areas; 14²=196 does not equal 36+64. Correct the first invalid relationship before correcting a final numerical answer.

For a rectangle 7 cm by 24 cm, its diagonal is the hypotenuse of a right triangle made by two adjacent sides. The diagonal satisfies d²=49+576=625 and is 25 cm. Going around those two edges would cover 31 cm; that is a different path. State which distance the context requests.

On coordinates, points (1, 2) and (7, 10) have horizontal displacement 6 and vertical displacement 8. These perpendicular displacements give straight-line distance 10 coordinate units. Check through differences before substituting, rather than squaring the raw coordinates 1, 2, 7, 10. A coordinate unit is not automatically a centimetre or kilometre without a stated map scale.

Observe how the original Corbettmaths tutorial identifies the hypotenuse. Its visible 16-and -9 example is separate from our 13-and -5 and 6-and -8 cases. Predict whether the unknown is longer or shorter than the hypotenuse before watching the calculation. Original: corbettmaths, Pythagoras - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Use a calculator only after selecting the equation. Keep the unrounded squared value or exact root until the final step; early rounding can make an independent substitution appear inconsistent. A displayed approximation is not an exact equality.

Solve six right-triangle tasks and check exactness

Six original tasks. Every row states a right triangle or a rectangle whose diagonal creates one. All supplied lengths are in centimetres. Cover the final two columns, draw a labelled schematic, choose addition or subtraction and solve. Use exact values unless a rounding instruction is given; in row 6 also give three decimal places.

Task	Given lengths in cm	Unknown	Squared calculation	Result
1	Legs 6 and 8	Hypotenuse	36+64=100	10 cm
2	Hypotenuse 13; leg 5	Other leg	169−25=144	12 cm
3	Legs 9 and 12	Hypotenuse	81+144=225	15 cm
4	Hypotenuse 10; leg 6	Other leg	100−36=64	8 cm
5	Rectangle sides 7 and 24	Diagonal	49+576=625	25 cm
6	Legs 1 and 1	Hypotenuse	1+1=2	√2 cm ≈ 1.414 cm

For row 3, the exact check is 9²+12²=81+144=225=15². It is also a scale factor of 1.5 applied to the 6–8–10 triple. This second route checks both the arithmetic and the correspondence between legs and hypotenuse. In row 4, 6²+8²=100 confirms the missing leg 8, rather than the invalid sum √136.

For row 5, 25 is longer than 24 and 7, but shorter than their sum 31. For row 6, √2 is an exact irrational length. The rounded 1.414 squares to 1.999396, close to but not exactly 2. That is expected after rounding, not a reason to replace the exact value. State ≈ rather than = when switching to the three-decimal approximation. The hypotenuse is longer than either unit leg.

Unfamiliar transfer. A right triangle has hypotenuse 17 mm and one leg 8 mm. The other leg has square 289−64=225 mm² and length 15 mm. The exact triple check gives 8²+15²=64+225=289. In a new coordinate problem with points (−2, 1) and (3, 13), horizontal and vertical changes are 5 and 12, giving distance 13 coordinate units. Both require identifying perpendicular components before applying the theorem.

Compare methods. A labelled squared equation is general and makes the hypotenuse assignment visible. Recognising a scaled triple may be faster for integer data, but it still needs verification and does not help with every irrational root. A second solution method should independently test the relationship, rather than simply retyping the same calculator sequence.

Retain all six solutions with equations, positive lengths, units, longest-side checks and one corrected first error. Use an estimate to identify gross errors, then an exact substitution where possible. A table of correct final answers without the side assignment does not show that the theorem was used in a valid situation. Revisit one problem after a real delay and record your actual response; no delayed mastery or practical measuring skill is assumed from completing this lesson.

Interactive Tasks

Quiz: Test Your Knowledge

Which side is the hypotenuse? (The side opposite the right angle) (!The shortest side) (!The vertical side in every drawing) (!The sum of the legs)

Legs 6 cm and 8 cm: hypotenuse? (10 cm) (!14 cm) (!100 cm) (!2 cm)

Hypotenuse 13 cm and leg 5 cm: other leg? (12 cm) (!18 cm) (!√194 cm) (!8 cm)

Which condition is required by this theorem here? (A right-angled triangle) (!Any three lengths) (!A triangle that looks nearly right) (!Two equal sides)

Legs 9 and 12: hypotenuse? (15) (!21) (!225) (!3)

Hypotenuse 10 and leg 6: other leg? (8) (!4) (!16) (!√136)

Rectangle sides 7 and 24: diagonal? (25) (!31) (!17) (!625)

Unit perpendicular legs: exact diagonal? (√2) (!2) (!1.414 exactly) (!0)

Which is the squared unit of centimetres? (cm²) (!cm) (!cm³) (!mm)

Points 1, 2 and 7, 10: straight-line distance? (10 coordinate units) (!14 coordinate units) (!26 coordinate units) (!100 coordinate units)

Memory Game

Hypotenuse	Side opposite the right angle
Legs	Two sides meeting at the right angle
Squared length	Area of the square built on that side
Positive root	Root selected for a nonzero physical length
Diagonal	Straight segment between opposite vertices
Approximation	Rounded value marked as not exact

Drag and Drop

10 cm	Hypotenuse from perpendicular legs 6 and 8
12 cm	Other leg from hypotenuse 13 and leg 5
15 cm	Hypotenuse from perpendicular legs 9 and 12
8 cm	Other leg from hypotenuse 10 and leg 6
25 cm	Diagonal of a 7 by 24 rectangle

Crossword Puzzle

HYPOTENUSE	Longest side opposite the right angle
PYTHAGORAS	Theorem relating squares of right-triangle sides
SQUARE	Second power of a side length
ROOT	Operation reversing squaring on nonnegative values
DIAGONAL	Segment joining opposite rectangle vertices
PERPENDICULAR	Describes lines meeting at a right angle

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Name the side roles: Rotate a schematic and identify the side opposite the right angle.
Solve the first source example: Show the complete 6-and -8 equation, root and longest-side check.
Solve the second source example: Rearrange x²+25=169 before taking the root.
Read the image correctly: Explain why a height-foot right angle does not make the whole acute triangle right-angled.

Standard

Solve all six tasks: Retain equations, positive lengths, units and checks; row 6 requires exact and rounded values.
Correct both errors: Find the first invalid step in adding 13²+5² for a leg and in adding lengths 6+8.
Check a diagonal: Compare the rectangle diagonal with the two-edge route.
Use coordinate differences: Find the distance from 1, 2 to 7, 10 and state the coordinate-unit limit.

Advanced

Transfer to 17 and 8: Find the missing leg in millimetres and check by substitution.
Use new coordinates: Find distance from −2, 1 to 3, 13 through perpendicular differences.
Compare two methods: Use a squared equation and a verified scaled triple, explaining where recognition stops helping.
Explain exactness: Compare √2 with 1.414 and retain an actual later retrieval response.

Learning Check

Six-task answer key: Solve all six and justify the hypotenuse assignment.
Source conditions and proof model: Explain side roles and the square-area connection.
First-error corrections: Correct both supplied flawed solutions.
Independent transfer: Solve the new side and coordinate tasks.
Methods and rounding: Explain checks beyond a repeated calculator entry.
Portfolio and evidence limits: Retain equations, interpretations and actual later work.

Worked checks and assessment guidance

Results 10, 12, 15, 8, 25, √2 cm. Rows 1, 3, 5, 6 use a sum; rows 2, 4 subtract a known leg square from the hypotenuse square. Each complete triangle has its hypotenuse longest. Row 6 rounds to 1.414 cm at three decimal places.
a²+b²=c² applies to perpendicular legs and the opposite hypotenuse. Four right triangles in a square of side a+b leave area a²+b²=c². The acute source image contains right subtriangles; the full base is not automatically one subtriangle leg.
For hypotenuse 13 and leg 5, subtract 169−25=144, giving 12. Adding gives a supposed leg longer than the hypotenuse. For legs 6 and 8, 14 is the two-edge route; 14² does not equal 36+64.
17²−8²=225 gives 15 mm; 64+225=289 checks. Coordinate changes from −2, 1 to 3, 13 are 5 and 12, so distance 13 coordinate units; no physical scale is assumed.
Recognise and verify 6–8–10 scaled by 1.5 for 9–12–15. For √2 the rounded 1.414 squared is 1.999396; retain the exact root and use an approximation sign. Triangle-side bounds help catch gross errors.
Keep all six labelled solutions, first-error corrections and a new transfer with units. Online geometry models do not certify physical measurement skill; only record delayed retrieval when it really occurred.

Learning evidence

Retain six complete right-triangle solutions, two first-error corrections and new side/coordinate transfers with valid conditions, units, exactness and independent checks.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, geometry, measures, statistics, fluency, reasoning and problem solving. Year 8 allocation is editorial within KS3. All numerical cases, original schematic graphs and feedback are newly authored teaching material. Actual Commons creators/licences follow; original-publisher videos supplement the complete written teaching. Reviewed 12 November 2026.

Triangle (acute) bh.PNG, H Padleckas at English Wikibooks, Public domain. Unchanged original; caption and learning activities are original.

2D Cartesian Coordinates.svg, H Padleckas, Public domain. Unchanged original; caption and learning activities are original.