PAPER · REASONING · PRACTICE

Pythagoras converse: conditions and combined problems

Test actual side data and combine the converse with length, midpoint and perpendicularity reasoning.

40 minutes · Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13

Textbook reference: Year 8, lower volume · Chapter 17: Pythagorean theorem · lessons 10–13

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the lower-volume plan

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Teacher reference Arrange in My Classroom

Lesson routePhase

4′Check prerequisites
5′Observe and predict
8′Worked example and guided attempt
15′Independent practice
5′Discuss and correct
3′Exit questions

Observe · explain

A question before practising

3 min: Predict which length cards make right triangles: (5,12,13), (5,12,14), (2,3,5). Do not rely on appearance. 8 min: On paper, first check positive lengths and the strict triangle inequality. For each valid triangle identify its longest side, then compare its square with the sum of the other two squares. The existing right-triangle engine cannot test three arbitrary sides for this converse. 5 min: Explain why the first is right, the second is not right, and the third is degenerate. Distinguish triangle existence from the square equality. 2 min: Individually test (8,15,17), and name the angle opposite the longest side in the conclusion.

Organise the key idea

The converse determines a right angle only after positive side lengths form a triangle and the longest-square equality is established.

Open the original chapter

Use the lesson question here for paper discussion; the original activity opens separately.

Ordinary route: use the first fixed case for the five-minute demonstration. Advanced route: follow the full 18-minute investigation; record predictions before checking.

Worked example1

Can sides 13, 5, 12 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 13; the shorter sum is 17.
$5^{2}+12^{2}=169$ and $13^{2}=169$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Guided attempt2

Can sides 8, 15, 17 cm form a right triangle? Check existence and identify the longest side first.

Existence and longest side: $8+15>\square$ ; longest side = ____.
$8^{2}+15^{2}=\square$ ; $17^{2}=\square$ .
The two square values are ____; therefore, by ____, the angle opposite 17 cm is ____.

17 is the longest side; first check that 8+15 exceeds it.
Compare $64+225$ with $289$ .
The square values agree and the triangle exists. Use the converse to identify the angle opposite the 17 cm side.

The longest side is 17; the shorter sum is 23.
$8^{2}+15^{2}=289$ and $17^{2}=289$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Prerequisite3

Can sides 3, 4, 5 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 5; the shorter sum is 7.
$3^{2}+4^{2}=25$ and $5^{2}=25$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Prerequisite4

Can sides 2, 3, 5 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 5; the shorter sum is 5.
$2^{2}+3^{2}=13$ and $5^{2}=25$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice5

Can sides 5, 12, 14 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 14; the shorter sum is 17.
$5^{2}+12^{2}=169$ and $14^{2}=196$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice6

Can sides 6, 8, 10 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 10; the shorter sum is 14.
$6^{2}+8^{2}=100$ and $10^{2}=100$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice7

A rectangular panel measures 9 cm by 12 cm. A straight brace joins opposite corners. Find the brace length.

Geometry diagram; not drawn to scale except for construction tasks.

Locate the right angle and the opposite hypotenuse.
Write the square relation before substituting.
A length is positive; retain the stated unit.

The right angle is given; add the two leg squares to find the hypotenuse square.
The required square is $9^{2}+12^{2}$ ; take the positive square root.

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice8

AB=AC=5 cm and BC=6 cm. M is the midpoint of BC. Find AM and the triangle area, stating why AM is perpendicular to BC.

Geometry diagram; not drawn to scale except for construction tasks.

Use the isosceles condition before using Pythagoras.
Half the base is 3 cm.
Find the height, then use half base times height.

BM=3 cm; the apex median is perpendicular to BC.
AM²=25−9=16, so AM=4. Area= $\frac{1}{2}\times6\times4=12$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Exit question9

Can sides 4, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 6; the shorter sum is 9.
$4^{2}+5^{2}=41$ and $6^{2}=36$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Exit question10

Can sides 25, 7, 24 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 25; the shorter sum is 31.
$7^{2}+24^{2}=625$ and $25^{2}=625$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice11

Can sides 1.5, 2, 2.5 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 2.5; the shorter sum is 3.5.
$1.5^{2}+2^{2}=6.25$ and $2.5^{2}=6.25$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice12

Can sides 5, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 6; the shorter sum is 10.
$5^{2}+5^{2}=50$ and $6^{2}=36$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice13

A rectangle has diagonal 5 cm and width 3 cm. Find its length.

Locate the right angle and the opposite hypotenuse.
Write the square relation before substituting.
A length is positive; retain the stated unit.

The hypotenuse is known; subtract the square of the known leg from its square.
The required square is $5^{2}-3^{2}$ ; take the positive square root.

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice14

Can sides 29, 20, 21 cm form a right triangle? Check existence and identify the longest side first.

All lengths must be positive.
Compare the two shorter squares with the longest square.
The converse needs equality, not approximate visual similarity.

The longest side is 29; the shorter sum is 41.
$20^{2}+21^{2}=841$ and $29^{2}=841$ .

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Extension15

A= $(0,0)$ , B= $(6,0)$ , C= $(0,8)$ . Verify the three lengths and use the converse to identify the right angle.

Use the stated conditions and the labelled scales.

The longest side is BC.
Compute the squared coordinate differences for BC.
Name the angle opposite BC.

Horizontal and vertical distances give AB=6 and AC=8. The diagonal length is 10.
6²+8²=10², so the angle opposite BC is a right angle at A.

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Extension16

A parallelogram has adjacent sides 6 cm and 8 cm, and the diagonal joining their free endpoints is 10 cm. Prove it is a rectangle.

Geometry diagram; not drawn to scale except for construction tasks.

Use the triangle formed by those two adjacent sides and the diagonal.
The converse gives one right angle.
Retain the parallelogram premise in the final criterion.

The sides 6,8,10 form a triangle and satisfy 6²+8²=10².
The angle between the two given adjacent sides is therefore right; use the parallelogram premise for the rectangle criterion.

Check your work

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Preparation and assessment notes

Prerequisites

Triangle inequalities, squares and the Pythagorean theorem.

Look for in students’ work

Use the two prerequisite checks to choose support. Project the worked and guided tasks, then collect the eight paper core questions. Follow-ups and extensions are optional; accept other complete valid methods. This page does not save learners’ paper answers.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Print edition

Textbook reference and authorship

Original chapter-core practice after the cited lower-volume textbook lessons. References identify skills; textbook exercises are not reproduced.

大湾区初中数学三维课堂（八年级下册）正文.pdf

Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Pythagoras converse: conditions and combined problems

Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Can sides 13, 5, 12 cm form a right triangle? Check existence and identify the longest side first.

Guided attempt2

Can sides 8, 15, 17 cm form a right triangle? Check existence and identify the longest side first.

Existence and longest side: $8+15>\square$ ; longest side = ____.
$8^{2}+15^{2}=\square$ ; $17^{2}=\square$ .
The two square values are ____; therefore, by ____, the angle opposite 17 cm is ____.

Prerequisite3

Can sides 3, 4, 5 cm form a right triangle? Check existence and identify the longest side first.

Prerequisite4

Can sides 2, 3, 5 cm form a right triangle? Check existence and identify the longest side first.

Pythagoras converse: conditions and combined problems

Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

Can sides 5, 12, 14 cm form a right triangle? Check existence and identify the longest side first.

Independent practice6

Can sides 6, 8, 10 cm form a right triangle? Check existence and identify the longest side first.

Independent practice7

A rectangular panel measures 9 cm by 12 cm. A straight brace joins opposite corners. Find the brace length.

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice8

AB=AC=5 cm and BC=6 cm. M is the midpoint of BC. Find AM and the triangle area, stating why AM is perpendicular to BC.

Geometry diagram; not drawn to scale except for construction tasks.

Exit question9

Can sides 4, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

Exit question10

Can sides 25, 7, 24 cm form a right triangle? Check existence and identify the longest side first.

Pythagoras converse: conditions and combined problems — Further practice

Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13

Further practice11

Can sides 1.5, 2, 2.5 cm form a right triangle? Check existence and identify the longest side first.

Further practice12

Can sides 5, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

Further practice13

A rectangle has diagonal 5 cm and width 3 cm. Find its length.

Further practice14

Can sides 29, 20, 21 cm form a right triangle? Check existence and identify the longest side first.

Extension15

A= $(0,0)$ , B= $(6,0)$ , C= $(0,8)$ . Verify the three lengths and use the converse to identify the right angle.

Use the stated conditions and the labelled scales.

Extension16

A parallelogram has adjacent sides 6 cm and 8 cm, and the diagonal joining their free endpoints is 10 cm. Prove it is a rectangle.

Geometry diagram; not drawn to scale except for construction tasks.

Pythagoras converse: conditions and combined problems — Teacher answers

Year 8, lower volume · text pp. 19–26 · textbook lessons 10–13

Ordinary route: prerequisites 4, short inquiry 5, worked/guided 8, independent 15, correction 5, exit 3 minutes. Advanced route: inquiry 18, organise 4, selected practice 13, exit 5 minutes.

Prerequisites: Triangle inequalities, squares and the Pythagorean theorem.

Ordinary route: use the first fixed case for the five-minute demonstration. Advanced route: complete the full investigation.

3′ Predict which length cards make right triangles: (5,12,13), (5,12,14), (2,3,5). Do not rely on appearance.
8′ On paper, first check positive lengths and the strict triangle inequality. For each valid triangle identify its longest side, then compare its square with the sum of the other two squares. The existing right-triangle engine cannot test three arbitrary sides for this converse.
5′ Explain why the first is right, the second is not right, and the third is degenerate. Distinguish triangle existence from the square equality.
2′ Individually test (8,15,17), and name the angle opposite the longest side in the conclusion.

Worked example1

Can sides 13, 5, 12 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 13; the shorter sum is 17.
$5^{2}+12^{2}=169$ and $13^{2}=169$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Guided attempt2

Can sides 8, 15, 17 cm form a right triangle? Check existence and identify the longest side first.

Existence and longest side: $8+15>\square$ ; longest side = ____.
$8^{2}+15^{2}=\square$ ; $17^{2}=\square$ .
The two square values are ____; therefore, by ____, the angle opposite 17 cm is ____.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 17; the shorter sum is 23.
$8^{2}+15^{2}=289$ and $17^{2}=289$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Prerequisite3

Can sides 3, 4, 5 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 5; the shorter sum is 7.
$3^{2}+4^{2}=25$ and $5^{2}=25$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Prerequisite4

Can sides 2, 3, 5 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

No: the required squared equality fails (or the lengths do not form a triangle).

The longest side is 5; the shorter sum is 5.
$2^{2}+3^{2}=13$ and $5^{2}=25$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice5

Can sides 5, 12, 14 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

No: the required squared equality fails (or the lengths do not form a triangle).

The longest side is 14; the shorter sum is 17.
$5^{2}+12^{2}=169$ and $14^{2}=196$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice6

Can sides 6, 8, 10 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 10; the shorter sum is 14.
$6^{2}+8^{2}=100$ and $10^{2}=100$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice7

A rectangular panel measures 9 cm by 12 cm. A straight brace joins opposite corners. Find the brace length.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

$15$ cm.

15

The right angle is given; add the two leg squares to find the hypotenuse square.
The required square is $9^{2}+12^{2}$ ; take the positive square root.

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Independent practice8

AB=AC=5 cm and BC=6 cm. M is the midpoint of BC. Find AM and the triangle area, stating why AM is perpendicular to BC.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

AM=4 cm; area=12 cm². In an isosceles triangle, the median from the apex is also an altitude.

BM=3 cm; the apex median is perpendicular to BC.
AM²=25−9=16, so AM=4. Area= $\frac{1}{2}\times6\times4=12$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Exit question9

Can sides 4, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

No: the required squared equality fails (or the lengths do not form a triangle).

The longest side is 6; the shorter sum is 9.
$4^{2}+5^{2}=41$ and $6^{2}=36$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Exit question10

Can sides 25, 7, 24 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 25; the shorter sum is 31.
$7^{2}+24^{2}=625$ and $25^{2}=625$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice11

Can sides 1.5, 2, 2.5 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 2.5; the shorter sum is 3.5.
$1.5^{2}+2^{2}=6.25$ and $2.5^{2}=6.25$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice12

Can sides 5, 5, 6 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

No: the required squared equality fails (or the lengths do not form a triangle).

The longest side is 6; the shorter sum is 10.
$5^{2}+5^{2}=50$ and $6^{2}=36$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice13

A rectangle has diagonal 5 cm and width 3 cm. Find its length.

Reference answer

$4$ cm.

4

The hypotenuse is known; subtract the square of the known leg from its square.
The required square is $5^{2}-3^{2}$ ; take the positive square root.

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Further practice14

Can sides 29, 20, 21 cm form a right triangle? Check existence and identify the longest side first.

Reference answer

Yes: the positive lengths satisfy the triangle inequalities and the Pythagorean converse.

The longest side is 29; the shorter sum is 41.
$20^{2}+21^{2}=841$ and $29^{2}=841$ .

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Extension15

A= $(0,0)$ , B= $(6,0)$ , C= $(0,8)$ . Verify the three lengths and use the converse to identify the right angle.

Use the stated conditions and the labelled scales.

Reference answer

AB=6, AC=8, BC=10; ∠BAC is 90°.

Horizontal and vertical distances give AB=6 and AC=8. The diagonal length is 10.
6²+8²=10², so the angle opposite BC is a right angle at A.

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.

Use the stated conditions and the labelled scales.

Extension16

A parallelogram has adjacent sides 6 cm and 8 cm, and the diagonal joining their free endpoints is 10 cm. Prove it is a rectangle.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The included angle is 90° by the Pythagorean converse. A parallelogram with one right angle is a rectangle.

The sides 6,8,10 form a triangle and satisfy 6²+8²=10².
The angle between the two given adjacent sides is therefore right; use the parallelogram premise for the rectangle criterion.

Use the stated conditions and justify the relation before calculating.
Give an exact answer where possible, with units and any domain restrictions.