Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 9 · Chapter companion part
Identify the hypotenuse, justify three proof methods and use the converse without assuming the result.
Class 8 English Version:part A of the chapter9 companion. Eight teaching sections and30 original written tasks. Selected companion coverage,not complete textbook coverage; use both parts and review full proofs,drawings and exact surds with a guardian or teacher.
Open this topic in Skills Lab →
A right triangle has one90° angle. The two sides meeting there are its legs; the opposite, longest side is the hypotenuse. If C is right, AB is the hypotenuse even when the triangle is rotated. We consistently use c for the hypotenuse and a, b for the legs in the formula c²=a²+b². The source changes letter roles between proofs; read the angle and side labels anew rather than assuming b or c always has one role.
For legs6 cm and8 cm, c²=36+64=100cm², so c=10 cm, the positive square root. Adding legs6+8=14 does not find the hypotenuse. With c=13 and one leg5, the other leg is√(169−25)=12. The hypotenuse must exceed each positive leg. The formula applies directly only when the two named legs are perpendicular.
Place B, C, D collinearly with BC=a, CD=b. Erect BA=b and DE=a on the same side, join AC, CE, AE. Triangles ABC and CDE are congruent by SAS, so AC=CE=c. Their acute-angle complements give ACE=90°. The trapezium ABDE area is(a+b)²/2. Its three triangle areas total ab/2+ab/2+c²/2. Equating and multiplying by2 cancels2ab, giving c²=a²+b². This proof does not assume the theorem to establish ACE. If a=b, the outer region is a rectangle; the same parallel-side area formula remains valid.
In right ABC with C=90°, drop CH perpendicular to AB=c; write AH=d, HB=e, BC=a, CA=b. Angle-angle similarity gives a/c=e/a and b/c=d/b, hence a²=ce, b²=cd. Add to obtain a²+b²=c(d+e)=c². The final source line omits an equals sign between c(d+e) and c×c; the intended equality is written here. Similar triangles need correct corresponding sides, not merely a shared shape.
Place four congruent right triangles with legs a, b in the corners of a square of side a+b. Every remaining inner edge is a hypotenuse c. At each inner vertex the two acute triangle angles add to90°, so the remaining angle is90°; equal sides alone would only establish a rhombus. Thus(a+b)²=4(ab/2)+c². Expand and subtract2ab to derive c²=a²+b².
For a≥b>0, four congruent right triangles can lie inside a tilted outer square of side c, leaving an inner square of side a−b. Then c²=4(ab/2)+(a−b)²=2ab+a²−2ab+b²=a²+b². Verify the outer right angles from the complementary acute angles. If a=b, the central gap has area0; it is not a positive-side square.
For positive lengths, first check a+b>c with c the largest. If a²+b²=c², the opposite angle is90°. To prove it, construct another right triangle with legs a, b. The already proved theorem gives its hypotenuse√(a²+b²)=c. SSS makes it congruent to the original triangle, so the corresponding angle is right. A drawing that looks right is not this proof.
A surd is an exact root that cannot be written as a terminating or recurring rational decimal, such as√2. For equal legs5, c=5√2; do not mark7.07 as an exact answer. A finite check can ask c²=50 instead. The two acute angles sum to90°: if one is twice the other, the smaller is30°; if their difference is12°, the smaller is(90−12)/2=39°. State whether a requested approximation is rounded and to what precision.
For right ABC, C=90°, BC=3, CA=4, AB=5:AH=16/5=3.2, HB=9/5=1.8, andCH=12/5=2.4 cm. Similarity and the two area expressions agree.
This companion consistently reservesc for a general hypotenuse, but identifies named sides individually in each task.
A measured3–4–5 sketch supports a check. The geometric/algebraic derivations establish the theorem for all positive right-triangle leg lengths.
RightABC has legs6 cm,8 cm and hypotenusec.
c²=6²+8²=100 cm² c=√100=10 cm
If a leg is missing, subtract its known companion square from the hypotenuse square.
Prove right angles and piece congruence before using their area expressions.
| Construction | Area equality | After simplification |
|---|---|---|
| Trapezium | (a+b)²/2=ab+c²/2 | c²=a²+b² |
| Outer(a+b)-square | (a+b)²=4(ab/2)+c² | c²=a²+b² |
| Outerc-square | c²=4(ab/2)+(a−b)² | c²=a²+b² |
For equal legsa=b, the difference-square gap has zero area. The trapezium becomes a rectangle under the exactly-one-pair naming convention, but the parallel-side area formula still applies.
One fixed7 cm outer square and four congruent3-by4 triangles. No stage plays automatically.
Manual stages, static alternatives and replay are available. No autoplay; reduced motion disables timed playback and leaving the page pauses it.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Show your working on paper. Check your answer, then choose Next.
10
√(6²+8²)=√100=10.
12
√(13²−5²)=√144=12.
90
AB is opposite C.
24.5
(3+4)²/2=49/2=24.5.
12.5
Two outer triangles contribute6 each;24.5−12=12.5.
3.2
CA²=AB×AH gives16/5=3.2.
1.8
BC²=AB×HB gives9/5=1.8.
49
(3+4)²=49.
24
4×3×4/2=24.
25
49−24=25.
1
(4−3)²=1.
90
8²+15²=64+225=289=17², so the converse applies.
50
c²=25+25=50; the exact length is5√2 cm.
30
x+2x=90.
39
x+(x+12)=90 givesx=39.
Write in your own words. A guardian or teacher should review this response.
AB is opposite C. Rotation changes orientation but preserves right-angle position and lengths. It remains the longest side.
Use opposite the right angle, not the sloping-looking side.
Write in your own words. A guardian or teacher should review this response.
c²=6²+8²=100cm²; taking the positive square root givesc=10 cm.14 cm would incorrectly add the two leg lengths.
Keep square and linear units distinct.
Write in your own words. A guardian or teacher should review this response.
x²=81−25=56, so x=√56=2√14 cm. A rounded decimal is an approximation only.
Select the positive root and retain the exact expression.
Write in your own words. A guardian or teacher should review this response.
Construct perpendicular BA=b, DE=a on baseline BD with BC=a, CD=b. SAS gives AC=CE=c. Acute-angle complementarity proves ACE=90°. Equate(a+b)²/2=ab/2+ab/2+c²/2, expand and cancel to obtainc²=a²+b².
Justify the central right angle before assigning its area.
Write in your own words. A guardian or teacher should review this response.
AA gives a/c=e/a and b/c=d/b. Hencea²=ce, b²=cd. Add:a²+b²=c(e+d)=c², sinced+e=c.
State both similarity correspondences and the segment sum.
Write in your own words. A guardian or teacher should review this response.
The projections are16/5=3.2 and9/5=1.8 cm, summing5. Area is3×4/2=6cm², also5h/2, soh=2.4 cm. Their product3.2×1.8=5.76 equals2.4².
All lengths refer to this explicitly supplied triangle.
Write in your own words. A guardian or teacher should review this response.
Four equal sides would permit a non-right rhombus. The adjacent acute triangle angles are complementary, so the interior gap angle is180−90=90°. Thus it is a square of areac².
Do not use appearance or equal sides alone.
Write in your own words. A guardian or teacher should review this response.
(a+b)²=4(ab/2)+c². Expanda²+2ab+b²=2ab+c², then subtract2ab from both sides.
Explain the four triangles and remaining square before the algebra.
Write in your own words. A guardian or teacher should review this response.
Arrange four right triangles in an outer c-square around a central(a−b)-square. Then c²=2ab+(a−b)²=a²+b². For a=b, the central area is0 and the triangles fill the outer square.
Verify the arrangement and handle the zero gap without claiming a positive-side square.
Write in your own words. A guardian or teacher should review this response.
Construct a separate right triangle with the two smaller lengths a, b. Its hypotenuse is c by the proved theorem and given equality. SSS makes it congruent to the original, so the corresponding angle is90°.
Keep the constructed triangle separate from the original assumption.
Write in your own words. A guardian or teacher should review this response.
They form a triangle since6+8>11, but36+64=100 differs from121. Therefore it is not right-angled; a successful triangle inequality alone does not imply rightness.
Check existence first, then compare squares.
Write in your own words. A guardian or teacher should review this response.
The triangle inequality fails:1+2<4. No nondegenerate triangle has those lengths. The equality-of-squares test does not replace the existence check.
Identify the failed prerequisite.
Write in your own words. A guardian or teacher should review this response.
After scaling by positive k, the sides arek, k, k√2. Their squares satisfyk²+k²=(k√2)², so the converse gives90° opposite the largest side. The other two angles are45°.
A ratio describes proportional lengths, not the number of angles.
Write in your own words. A guardian or teacher should review this response.
x+4x=90 givesx=18°, and the other angle is72°. Both are acute and total90°.
State the equation and check the result.
Write in your own words. A guardian or teacher should review this response.
Locate the right angle in each diagram, identify its opposite side, and rewrite the equation with that side squared alone. Our general notationc²=a²+b² is a convention, not a rule that a printed letterc must be the hypotenuse.
Label a fresh drawing before applying the formula.
Build two dissection proofs and one converse check.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 9, printed pages 152–158. It does not replace the complete chapter or its practical teaching.
Official book catalogue and source →
Teacher-led lessons, schedules and prices for this outline are upcoming. You can use any starter practice above now without booking.
Solo practice works without an account. Sign in to save a practice record to your learner profile.
Exact lesson links and related activities are labelled separately. Return-to-course links retain your selected learning path.
Play this exact lesson in Skills Lab →