GANITA PRAKASH · GRADE 7 · PART II · CHAPTER 1
Chapter Summary
Think first. Use AI purposefully. Verify always.
Concept Map
Criteria Bank
- SSS: three equal corresponding sides.
- SAS: two equal corresponding sides and their included angle.
- ASA: two equal corresponding angles and their included side.
- AAS: two equal corresponding angles and a corresponding non-included side.
- RHS: both triangles right, equal hypotenuse and one equal corresponding leg.
AAA does not fix scale. SSA is not sufficient in general. Every criterion requires a consistent correspondence and valid, non-degenerate triangles.
Visual Recall
Match vertices before writing the order
Correct order: A ↔ X, B ↔ Y, C ↔ Z.
Correspondence Guide
A ↔ X, B ↔ Y, C ↔ Z. Thus AB ↔ XY, BC ↔ YZ, CA ↔ ZX and the corresponding angles match. Changing the first order requires the same permutation in the second.
Important Connections
Rigid movements preserve lengths and angles. The included-angle hinge explains SAS; circle intersections explain both SSS uniqueness up to reflection and SSA ambiguity.
AAS becomes ASA using the angle sum. RHS proves the isosceles base-angle property; repeated equal-side reasoning gives 60° in an equilateral triangle.
Common Mistakes
- Orientation does not affect congruence. Same shape with different size is not enough.
- Two given sides alone leave the hinge free to move.
- Two circle intersections do not by themselves mean two non-congruent triangles.
- Naming the same two triangles with an arbitrary order can state a false correspondence.
- The included angle between AB and AC is ∠A, not ∠B. AAA is not a congruence criterion.
- Do not say SSA never works. Some data have one solution; the right-triangle special case gives RHS.
- AAS is valid even though the side is not between the two originally stated angles.
- The longest-looking side is not proof of a hypotenuse. Locate the established right angle.
- Do not mark a midpoint or angle bisector as given when it still needs to be proved.
- Equal area is not sufficient for congruence.
- Unmarked visual similarities are not proof.
- Every proof step needs a given fact, a construction or an established property.
Mini Examples
MATCH — Whether the circles are congruent
Given
Two circles of radius 3 cm.
To Find
Whether the circles are congruent
Property
A circle is fixed by its radius.
Solution
- Move one centre onto the other.
- Both boundaries are exactly 3 cm from that centre.
Result
The circles are congruent.
FIX — Measurements needed for an exact copy
Given
A two-arm symbol with AB = 4 cm, BC = 8 cm and ∠ABC = 80°.
To Find
Measurements needed for an exact copy
Property
The included angle is at the shared endpoint B.
Solution
- Draw BC of length 8 cm.
- At B construct an 80° ray and mark A at distance 4 cm.
- The opposite-side construction is a reflected copy.
Result
The arm lengths and included angle determine a congruent copy.
TRIPLE — △ABE ≅ △ABF
Given
AB = 6 cm; AE = AF = 4 cm; BE = BF = 8 cm.
To Prove
△ABE ≅ △ABF
Property
SSS; the construction circles fix the three side pairs.
Proof
- AB = AB is common.
- AE = AF and BE = BF are radii of the respective construction circles.
- Match A ↔ A, B ↔ B, E ↔ F.
Result
△ABE ≅ △ABF; reflection in AB superimposes them.
ORDER — Equivalent ordered statements for the same correspondence
Given
△HEN ≅ △BIG.
To Find
Equivalent ordered statements for the same correspondence
Property
H ↔ B, E ↔ I, N ↔ G must remain paired.
Solution
- Permute both lists in the same way.
- HNE ↔ BGI; EHN ↔ IBG; ENH ↔ IGB; NHE ↔ GBI; NEH ↔ GIB.
Result
The five other orders preserve the original vertex mapping.
HINGE — △ABC ≅ △XYZ
Given
AB = XY = 6 cm, AC = XZ = 5 cm, ∠A = ∠X = 30°.
To Prove
△ABC ≅ △XYZ
Property
SAS: ∠A and ∠X lie between the given side pairs.
Proof
- AB ↔ XY, AC ↔ XZ fix A ↔ X.
- The included angles agree. Match B ↔ Y and C ↔ Z.
Result
△ABC ≅ △XYZ by SAS.
AMBIGUOUS — Whether △PQR and △PQS must be congruent
Given
PQ = 6 cm; QR = QS = 4 cm; R and S lie on the same 30° ray from P.
To Find
Whether △PQR and △PQS must be congruent
Property
The two circle–ray intersections satisfy the same SSA data.
Solution
- Both share PQ and the angle at P.
- Both Q-to-intersection lengths are 4 cm.
- The intersections are different distances along the ray, so PR ≠ PS.
Result
The two triangles are non-congruent. SSA is insufficient in general.
ANGLE — △ABC ≅ △XYZ
Given
∠A = ∠X = 35°, ∠C = ∠Z = 75°, BC = YZ = 4 cm.
To Prove
△ABC ≅ △XYZ
Property
Angle sum followed by ASA, equivalently AAS.
Proof
- ∠B = ∠Y = 180° − 35° − 75° = 70°.
- The endpoint angles B, C match Y, Z and the included sides BC, YZ are equal.
Result
△ABC ≅ △XYZ.
RIGHT — △ABC ≅ △XYZ
Given
∠B = ∠Y = 90°, AC = XZ = 5 cm and BC = YZ = 4 cm.
To Prove
△ABC ≅ △XYZ
Property
RHS: AC and XZ are opposite the right angles.
Proof
- B ↔ Y are right-angle vertices.
- Hypotenuse AC matches XZ and leg BC matches YZ.
- Thus C ↔ Z and A ↔ X.
Result
△ABC ≅ △XYZ by RHS.
DEDUCE — The measures of ∠ABC and ∠ACB
Given
AB = AC and ∠BAC = 80°.
To Find
The measures of ∠ABC and ∠ACB
Property
Construction: draw AD perpendicular to BC. Use RHS, then the triangle angle sum.
Solution
- AB = AC, AD = AD and ∠ADB = ∠ADC = 90°.
- △ADB ≅ △ADC by RHS, so ∠ABC = ∠ACB.
- Their sum is 180° − 80° = 100°; each is 50°.
Result
∠ABC = ∠ACB = 50°.
5-Question Quick Check
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