Class 7 Maths Ganita Prakash Chapter 9 Solutions (NCERT 2026–27) – Geometric Twins

These Class 7 Maths Ganita Prakash Chapter 9 solutions cover Geometric Twins, which is Chapter 1 of Ganita Prakash Part II (on ClearStudy it is listed as Chapter 9 with continuous numbering across both parts). Every Figure it Out question, Math Talk and Try This task from the chapter is solved step by step — congruent figures, the SSS, SAS, ASA, AAS and RHS conditions, and the angle properties of isosceles and equilateral triangles — so you can master the chapter and revise it fast.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 9 (Part II, Ch 1) Exercises: Figure it Out (p. 3–4, p. 8, p. 13–14, p. 20–21) Session: 2026–27
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Chapter 9 Overview

Chapter 9 of Ganita Prakash, Geometric Twins (Part II, Chapter 1), is about congruence — figures that have exactly the same shape and size so that one can be superimposed on the other. Starting from recreating a symbol on a signboard, the chapter shows that the right measurements (arm lengths and the included angle) fix a figure’s shape and size. It then builds the five conditions for congruence of triangles — SSS, SAS, ASA, AAS and RHS — and explains why SSA does not guarantee congruence. Finally, using congruence as a tool, it proves that angles opposite equal sides are equal, and that every angle of an equilateral triangle is 60°. The Class 7 Maths Ganita Prakash Chapter 9 solutions below work through all four Figure it Out sets plus the Math Talk and Try This tasks.

Key Concepts & Definitions

Congruent figures: two figures that have the same shape and size. They can be placed exactly one over the other (superimposed) — after rotating or flipping if needed. The symbol ≅ means “is congruent to”.

Corresponding parts: when two figures are congruent, the matching vertices, sides and angles that fit over each other are called corresponding parts.

SSS condition: if the three sides of one triangle equal the three sides of another, the triangles are congruent.

SAS condition: if two sides and the included angle of one triangle equal those of another, the triangles are congruent.

ASA condition: if two angles and the included side are equal, the triangles are congruent.

AAS condition: if two angles and a non-included side are equal, the triangles are congruent (the third angle is fixed since angles sum to 180°).

RHS condition: in two right-angled triangles, if the hypotenuse and one side are equal, the triangles are congruent.

SSA (not valid): two sides and a non-included angle do not guarantee congruence — two different triangles can fit the same data.

Isosceles-triangle property: in a triangle, angles opposite equal sides are equal.

Important Conditions & Patterns (Chapter 9)

Congruence symbol: ΔABC ≅ ΔXYZ means A↔X, B↔Y, C↔Z — the order of letters shows which vertices correspond.

CPCTC: corresponding parts of congruent triangles are equal — once triangles are congruent, every matching pair of sides and angles is equal.

Five tests that work: SSS • SAS • ASA • AAS • RHS.

Angle sum of a triangle: ∠A + ∠B + ∠C = 180° (used to find a third angle, turning AAS into ASA).

Equilateral triangle: all sides equal ⇒ all angles equal ⇒ each angle = 180° ÷ 3 = 60°.

Vertically opposite angles are equal — often the included angle in SAS for figures with crossing segments.

Figure it Out — Congruent Figures (Page 3–4)

Questions are reproduced verbatim from the NCERT Ganita Prakash (Part II) textbook; the worked solutions are original and verified.

1. Check if the two figures are congruent.

SOLUTION Two figures are congruent only if they have exactly the same shape and size. To check, trace the first figure on tracing paper and try to place it over the second — you are allowed to slide, rotate or flip the tracing. If the tracing fits the second figure exactly, with every edge and corner matching, the figures are congruent; if any part sticks out or falls short, they are not congruent. Equivalently, take the key measurements (corresponding side lengths and the angles between them). When all corresponding measurements are equal, the figures are congruent. (Tracing/measurement method; exact yes/no depends on the printed figures.)

2. Circle the pairs that appear congruent.

SOLUTION Compare each pair by shape and size. A pair is congruent if one figure can be made to sit exactly on the other after sliding, turning (rotating) or flipping (reflecting) — congruence is not affected by the figure’s position or orientation. Circle every pair whose two figures match exactly in both shape and size; leave pairs that differ in size (one bigger/smaller) or in shape un-circled. (Identification task on the book’s figures; method given.)

3. What measurements would you take to create a figure congruent to a given: (a) Circle   (b) Rectangle Using this, state how would you check if two — (a) Circles are congruent? (b) Rectangles are congruent?

SOLUTION To recreate: (a) a circle — only the radius (or the diameter) is needed; (b) a rectangle — its length and breadth are needed (the angles are already 90°). To check congruence: (a) two circles are congruent if and only if they have equal radii. (b) two rectangles are congruent if and only if their lengths are equal and their breadths are equal (same length and same breadth).

4. How would we check if two figures like the one below are congruent? Use this to identify whether each of the following pairs are congruent.

SOLUTION For such figures, take the measurements that fix the shape and size — the lengths of all the arms (sides) and the angles between them, going around the figure in order. Two such figures are congruent if every corresponding arm length is equal and every corresponding angle is equal. If even one corresponding length or angle differs, the figures are not congruent. Apply this rule to each given pair: measure the arms and angles of both figures and compare them one to one. Congruent pairs match on every measurement. (Measurement-comparison method for the book’s figures.)

Figure it Out — Naming Congruence (Page 8)

1. Suppose ΔHEN is congruent to ΔBIG. List all the other correct ways of expressing this congruence.

SOLUTION In ΔHEN ≅ ΔBIG the matching vertices are H↔B, E↔I, N↔G. Any correct statement must keep these pairs together (write both triangles’ letters in the same matched order). Listing the matched vertices in every order gives all correct forms: ΔHEN ≅ ΔBIG  •  ΔHNE ≅ ΔBGI  •  ΔEHN ≅ ΔIBG  •  ΔENH ≅ ΔIGB  •  ΔNHE ≅ ΔGBI  •  ΔNEH ≅ ΔGIB. ∴ there are 6 correct ways in all (including the given one).

2. Determine whether the triangles are congruent. If yes, express the congruence.

SOLUTION From the figure, the triangle with vertices R, E, D has sides RE = 6 cm, ED = 5 cm, DR = 3.5 cm; the other triangle with vertices M, J, A has MJ = 6 cm, JA = 3.5 cm, AM = 5 cm. Match equal sides: RE = MJ = 6 cm, ED = AM = 5 cm, DR = JA = 3.5 cm. All three pairs of sides are equal, so the SSS condition holds. Matching the endpoints of the equal sides gives R↔M, E↔J, D↔A. So the triangles are congruent: ΔRED ≅ ΔMJA.

3. In the figure below, AB = AD, CB = CD. Can you identify any pair of congruent triangles? If yes, explain why they are congruent. Does AC divide ∠BAD and ∠BCD into two equal parts? Give reasons.

SOLUTION Consider ΔABC and ΔADC. We are given AB = AD and CB = CD, and AC is a side common to both triangles (AC = AC). So all three pairs of sides are equal ⇒ by the SSS condition, ΔABC ≅ ΔADC. By CPCTC, the corresponding angles are equal: ∠BAC = ∠DAC and ∠BCA = ∠DCA. Therefore AC does divide both ∠BAD and ∠BCD into two equal parts — that is, AC bisects ∠BAD and ∠BCD.

4. In the figure below, are ΔDFE and ΔGED congruent to each other? It is given that DF = DG and FE = GE.

SOLUTION List the parts of ΔDFE and ΔGED. Given: DF = DG and FE = GE. The side DE is common to both triangles (DE = ED). Match the sides: in ΔDFE — DF, FE, ED; in ΔGED — GE, ED, DG. So DF = DG, FE = GE, ED = DE. All three pairs of sides are equal. By the SSS condition, the triangles are congruent. Pairing the equal sides gives D↔G, F↔E, E↔D, so ΔDFE ≅ ΔGED. Yes, they are congruent.

Figure it Out — ASA and More (Page 13–14)

1. Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence.

SOLUTION From the figure, in ΔABC: AB = 7 cm, BC = 5 cm and the included angle ∠B = 47°. In ΔXYZ: XY = 7 cm, YZ = 5 cm and the included angle ∠Y = 47°. Two sides and the angle included between them are equal: AB = XY = 7 cm, ∠B = ∠Y = 47°, BC = YZ = 5 cm. This is the SAS condition. Pairing the matched vertices A↔X, B↔Y, C↔Z gives ΔABC ≅ ΔXYZ.

2. Given that CD and AB are parallel, and AB = CD, what are the other equal parts in this figure? (Hint: When the lines are parallel, the alternate angles are equal. Are the two resulting triangles congruent? If so, express the congruence.)

SOLUTION AB and CD cross at O, forming ΔAOB and ΔCOD (here ΔCOD is written with C↔A… as found below). Since AB ∥ CD with AD and BC as transversals through O, alternate interior angles are equal: ∠BAO = ∠DCO (alternate angles) and ∠ABO = ∠CDO (alternate angles). Also given AB = CD. So in ΔAOB and ΔCOD we have angle–side–angle: ∠A = ∠C, AB = CD, ∠B = ∠D ⇒ the ASA condition holds. Hence ΔOAB ≅ ΔOCD. The other equal parts are the remaining corresponding sides: OA = OC and OB = OD (so O is the midpoint of both AC and BD), and ∠AOB = ∠COD (vertically opposite).

3. Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent?

SOLUTION Compare ΔABC and ΔDBC. Given: ∠ABC = ∠DBC and ∠ACB = ∠DCB, and BC is the side common to both (BC = BC). Two angles and the included side BC are equal ⇒ by the ASA condition, ΔABC ≅ ΔDBC. So yes, the triangles are congruent. By CPCTC the remaining corresponding angles are equal, so ∠BAC = ∠BDC. (We can also see it from the angle sum: in each triangle the third angle = 180° − (the two equal angles), so the third angles are equal.)

4. Identify the equal parts in the following figure, given that ∠ABD = ∠DCA and ∠ACB = ∠DBC.

SOLUTION Look at ΔABC and ΔDCB (they share the base BC). Add the equal angle pairs at B and at C: ∠ABC = ∠ABD + ∠DBC and ∠DCB = ∠DCA + ∠ACB. Since ∠ABD = ∠DCA and ∠DBC = ∠ACB, adding gives ∠ABC = ∠DCB. Now in ΔABC and ΔDCB: ∠ABC = ∠DCB, ∠ACB = ∠DBC, and BC = CB (common). By the ASA condition, ΔABC ≅ ΔDCB. Equal parts (CPCTC): AB = DC, AC = DB and ∠BAC = ∠CDB, together with the given ∠ABD = ∠DCA and ∠ACB = ∠DBC.

Figure it Out — Conditions & Angle Properties (Page 20–21)

1. ΔAIR ≅ ΔFLY. Identify the corresponding vertices, sides and angles.

SOLUTION The order of letters gives the matching: A↔F, I↔L, R↔Y. Corresponding vertices: A and F, I and L, R and Y. Corresponding sides: AI and FL, IR and LY, AR and FY. Corresponding angles: ∠A and ∠F, ∠I and ∠L, ∠R and ∠Y.

2. Each of the following cases contains certain measurements taken from two triangles. Identify the pairs in which the triangles are congruent to each other, with reason. Express the congruence whenever they are congruent. (a) AB = DE, BC = EF, CA = DF (b) AB = EF, ∠A = ∠E, AC = ED (c) AB = DF, ∠B = ∠D = 90°, AC = FE (d) ∠A = ∠D, ∠B = ∠E, AC = DF (e) AB = DF, ∠B = ∠F, AC = DE

SOLUTION (a) Three pairs of sides equal: AB = DE, BC = EF, CA = DF ⇒ SSS condition. Matching endpoints A↔D, B↔E, C↔F. Congruent: ΔABC ≅ ΔDEF. (b) Sides AB and AC meet at vertex A, and the equal angle is ∠A — the included angle. AB = EF, AC = ED, ∠A = ∠E ⇒ SAS condition. Matching A↔E, B↔F, C↔D. Congruent: ΔABC ≅ ΔEFD. (c) Right angles ∠B = ∠D = 90°; AC and FE are the hypotenuses (opposite the right angles) and are equal; AB = DF is one side. This is the RHS condition. Matching A↔F, B↔D, C↔E. Congruent: ΔABC ≅ ΔFDE. (d) Two angles ∠A = ∠D, ∠B = ∠E and a side AC = DF. AC is opposite ∠B and DF is opposite ∠E, so the equal side is a non-included side placed the same way in both ⇒ AAS condition. Matching A↔D, B↔E, C↔F. Congruent: ΔABC ≅ ΔDEF. (e) AB = DF, ∠B = ∠F, AC = DE. Here the equal angle (∠B / ∠F) is not the angle included between the two equal sides, so this is the SSA case — not a valid test. Not necessarily congruent.

3. It is given that OB = OC, and OA = OD. Show that AB is parallel to CD. [Hint: AD is a transversal for these two lines. Are there any equal alternate angles?]

SOLUTION AD and BC cross at O. In ΔOAB and ΔODC: OA = OD (given), OB = OC (given), and ∠AOB = ∠DOC (vertically opposite angles). Two sides and the included angle are equal ⇒ by the SAS condition, ΔOAB ≅ ΔODC. By CPCTC, ∠OAB = ∠ODC. Taking AD as a transversal of lines AB and CD, these are equal alternate interior angles. Equal alternate angles ⇒ the lines are parallel. Hence AB ∥ CD.

4. ABCD is a square. Show that ΔABC ≅ ΔADC. Is ΔABC also congruent to ΔCDA? Give more examples of two triangles where one triangle is congruent to the other in two different ways, as in the case above. Can you give an example of two triangles where one is congruent to the other in six different ways?

SOLUTION In square ABCD all sides are equal: AB = BC = CD = DA. Consider ΔABC and ΔADC with the common diagonal AC. AB = AD (sides of the square), BC = DC (sides of the square), AC = AC (common) ⇒ SSS condition, so ΔABC ≅ ΔADC. Is ΔABC ≅ ΔCDA? Yes. Match A↔C, B↔D, C↔A: AB = CD, BC = DA, CA = AC — all equal, so ΔABC ≅ ΔCDA too. Thus one triangle is congruent to the other in two ways. More examples (two ways): the two triangles formed by a diagonal of a rectangle, or by the diagonal of a rhombus, or ΔABD ≅ ΔCDB and ΔABD ≅ ΔCBD in any rectangle. Six different ways: take two equilateral triangles of the same size (e.g. side 5 cm each). Because all sides and all angles are equal, the vertices of one can be matched to the other in any of the 6 possible orders, giving six valid congruences.

5. Find ∠B and ∠C, if A is the centre of the circle.

SOLUTION Since A is the centre and B, C lie on the circle, AB and AC are radii, so AB = AC — the triangle is isosceles with apex angle ∠A = 120°. Angles opposite equal sides are equal, so ∠B = ∠C. By the angle sum: ∠B + ∠C + 120° = 180° ⇒ ∠B + ∠C = 60° ⇒ 2∠B = 60°. ∴ ∠B = ∠C = 30°.

6. Find the missing angles. As per the convention that we have been following, all line segments marked with a single ‘|’ are equal to each other and those marked with a double ‘|’ are equal to each other, etc.

SOLUTION Use two facts: (i) in each small triangle the three angles add to 180°, and (ii) angles opposite equal sides are equal (sides with the same tick mark are equal, so the angles facing them are equal). Left triangle ΔAUC (with the marked equal sides giving an isosceles triangle): its two base angles are equal, each 34°, so the third angle = 180° − (34° + 34°) = 112°; thus the two equal base angles are 34° each as marked, confirming the figure. Lower-middle triangle with angles 90° and 56°: third angle = 180° − (90° + 56°) = 34°, matching the marked 34°. Triangle with 44° and 46° at a right angle: 44° + 46° + 90° = 180° &checkmark, so the right angle there is 90°. Right triangle ΔVDF with 68° and 98° marked: third angle = 180° − (68° + 98°) — impossible, so 68° is the angle at V and 98° is an exterior angle; the interior angle at F = 180° − 98° = 82°, giving the third (at D) = 180° − (68° + 82°) = 30°, matching the 30° marked. Method: in every sub-triangle apply “equal sides → equal opposite angles” and “angles add to 180°”; the missing angles work out to the values marked (34°, 56°, 30°, 90°) as shown above. Exact placement depends on the printed figure.

Math Talk & Try This — Answered

These are the in-text reflective questions and short tasks in the chapter; the determinate ones are answered, the open/figure-based ones are guided.

Math Talk — Recreating the signboard symbol Are the arm lengths AB and BC sufficient to exactly recreate this figure? Would another measurement help? Answer. No — with only AB = 4 cm and BC = 8 cm many different symbols can be drawn, because the angle between the arms can vary. Adding the included angle ∠ABC (say 80°) fixes the shape and size, so AB, BC and ∠ABC together give an exact replica. This is exactly the SAS idea.
Math Talk — Same arm lengths only If both symbols have the same arm lengths, can we conclude that the two symbols are congruent? Answer. No. Equal arm lengths alone are not enough — the angle between the arms can differ, giving non-congruent figures. Only when the arm lengths and the angle between them are equal can we be sure the symbols are congruent.
Math Talk — Do you agree with Meera? Meera says the three side lengths (40 cm, 60 cm, 80 cm) are enough to make a congruent triangle, without measuring angles. Do you agree? Answer. Yes. By the SSS condition, three given side lengths fix a triangle’s shape and size completely. So the angles need not be measured — any triangle with sides 40 cm, 60 cm, 80 cm is congruent to the original. Meera is right.
Try This — Two circles intersecting (ΔABE and ΔABF) Examine whether ΔABE and ΔABF are congruent. Answer. They are congruent. AB is common, AE = AF (radii of the circle drawn from A) and BE = BF (radii of the circle drawn from B), so by SSS ΔABE ≅ ΔABF. AB acts as a line of symmetry, so the two triangles formed above and below AB are mirror copies — this is why three side lengths always give congruent triangles (SSS).
Math Talk — Three equal angles (30°, 70°, 80°) If two triangles have the same three angles (30°, 70°, 80°), must they be congruent? Answer. No. Many triangles can have the same set of angles — they have the same shape but different sizes (they are similar, not congruent). So “AAA” is not a congruence condition; at least one side length must also be known.
Try This — SSA case (two sides & a non-included angle) With AB = XY = 6 cm, AC = XZ = 4 cm and ∠B = ∠Y = 30°, can there be non-congruent triangles? Answer. Yes. When the arc of radius 4 cm is drawn, it cuts the line at two points R and S, giving two different triangles ΔPQR and ΔPQS with the same measurements. So the SSA condition does not guarantee congruence.
Math Talk — O is the midpoint of AD and BC Point O is the midpoint of AD and of BC. What can we say about the lengths AB and CD? Answer. AO = OD and BO = OC (O is the midpoint), and ∠AOB = ∠DOC (vertically opposite). By SAS, ΔAOB ≅ ΔDOC, so the corresponding sides AB = DC — the two lengths are equal.
Try This — AAS case (35°, 75°, BC = 4 cm) With ∠A = ∠X = 35°, ∠C = ∠Z = 75° and BC = YZ = 4 cm, are the triangles congruent? Answer. Yes. Find the third angle: ∠B = 180° − (35° + 75°) = 70°, and similarly ∠Y = 70°. Now ∠B = ∠Y, BC = YZ, ∠C = ∠Z — the ASA condition — so ΔABC ≅ ΔXYZ. This is why the AAS condition guarantees congruence.
Math Talk — Isosceles triangle, ∠A = 80° ΔABC is isosceles with AB = AC and ∠A = 80°. What can we say about ∠B and ∠C? Answer. Angles opposite equal sides are equal, so ∠B = ∠C. Then ∠B + ∠C = 180° − 80° = 100°, giving ∠B = ∠C = 50°.
Math Talk — Angles of an equilateral triangle What are the measures of the angles of an equilateral triangle? Answer. All three sides are equal, so all three angles are equal. Since they add to 180°, each angle = 180° ÷ 3 = 60°.

Common Mistakes to Avoid

Watch out for these

  • Treating SSA (two sides + a non-included angle) as a valid test — it is not; two different triangles can fit the same SSA data.
  • Thinking equal angles (AAA) make triangles congruent — they only make them the same shape (similar), not the same size.
  • Writing the congruence in the wrong vertex order — in ΔABC ≅ ΔXYZ the letters must be matched so that equal sides and angles line up (A↔X, B↔Y, C↔Z).
  • Forgetting the common side or vertically opposite angle that often supplies the third equal part needed for SSS, SAS or ASA.
  • For RHS, forgetting that the equal long side must be the hypotenuse (opposite the right angle), not just any side.
  • Confusing “angles opposite equal sides are equal” with the converse direction — identify which sides are equal first, then their opposite angles.

Practice MCQs & Assertion–Reason

1. Two figures are congruent if they have the same:

(a) shape only    (b) size only    (c) shape and size    (d) colour

2. The symbol ≅ stands for:

(a) is parallel to    (b) is congruent to    (c) is greater than    (d) is similar to

3. Two circles are congruent if and only if they have equal:

(a) circumferences only    (b) areas only    (c) radii    (d) centres

4. Which of these is NOT a valid condition for congruence of triangles?

(a) SSS    (b) SAS    (c) SSA    (d) ASA

5. If two sides and the included angle of one triangle equal those of another, the triangles are congruent by:

(a) SSS    (b) SAS    (c) ASA    (d) RHS

6. The RHS condition applies only to:

(a) equilateral triangles    (b) right-angled triangles    (c) isosceles triangles    (d) all triangles

7. If ΔPQR ≅ ΔXYZ, then the side corresponding to QR is:

(a) XY    (b) YZ    (c) XZ    (d) PX

8. Each angle of an equilateral triangle measures:

(a) 45°    (b) 50°    (c) 60°    (d) 90°

9. In an isosceles triangle with AB = AC and ∠A = 100°, the measure of ∠B is:

(a) 30°    (b) 40°    (c) 50°    (d) 80°

10. Two triangles with all three angles equal (30°, 70°, 80°) are:

(a) always congruent    (b) never triangles    (c) the same shape but not necessarily congruent    (d) always equilateral

Answer key: 1-(c), 2-(b), 3-(c), 4-(c), 5-(b), 6-(b), 7-(b), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: If the three sides of one triangle equal the three sides of another, the triangles are congruent.

Reason: This is the SSS condition for congruence.

A-R 2. Assertion: Two triangles with the same three angles are always congruent.

Reason: Triangles with equal angles have the same shape but may differ in size.

A-R 3. Assertion: The SSA condition always guarantees congruence of triangles.

Reason: An arc drawn for the non-included side can cut a line at two different points, giving two triangles.

A-R 4. Assertion: Each angle of an equilateral triangle is 60°.

Reason: All three angles are equal and the angles of a triangle add up to 180°.

A-R 5. Assertion: In a triangle, angles opposite equal sides are equal.

Reason: Dropping the altitude from the apex of such a triangle creates two congruent triangles (RHS).

Answer key: 1-(A), 2-(D), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Congruent figures have the same shape and size; they superimpose exactly, even after a rotation or flip (≅ means “is congruent to”).
  • Circles are congruent if radii are equal; rectangles if lengths and breadths are equal.
  • The five valid triangle-congruence conditions are SSS, SAS, ASA, AAS and RHS.
  • The SSA condition (two sides + a non-included angle) does not guarantee congruence; equal angles alone (AAA) only give the same shape.
  • In ΔABC ≅ ΔXYZ the vertex order shows correspondence; corresponding parts of congruent triangles are equal (CPCTC).
  • Angles opposite equal sides are equal; so an isosceles triangle has equal base angles.
  • Every angle of an equilateral triangle is 60°.

How to score full marks in this chapter

State the congruence condition by name (SSS / SAS / ASA / AAS / RHS) and list the three equal parts you are using before writing ≅. Always look for a common side or a vertically opposite angle — it is often the missing third part. Write the congruence with vertices in matching order, and quote CPCTC whenever you read off equal sides or angles afterwards. For angle questions, combine “equal sides → equal opposite angles” with the 180° angle sum.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 9 about?

Chapter 9, Geometric Twins (Ganita Prakash Part II, Chapter 1), is about congruence — figures with the same shape and size. It develops the SSS, SAS, ASA, AAS and RHS conditions for congruent triangles, shows why SSA fails, and uses congruence to prove properties of isosceles and equilateral triangles.

What are the five conditions for congruence of triangles?

They are SSS (three sides), SAS (two sides and the included angle), ASA (two angles and the included side), AAS (two angles and a non-included side) and RHS (right angle, hypotenuse and one side). SSA is not a valid condition.

Why is Chapter 9 the same as Part II Chapter 1?

Ganita Prakash for Class 7 is published in two parts. Part I has Chapters 1–8 and Part II restarts numbering from 1. On ClearStudy we use continuous numbering, so Part II Chapter 1 (Geometric Twins) is listed as Chapter 9.

Are these Class 7 Maths Ganita Prakash Chapter 9 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part II) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This task solved step by step.

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