GANITA PRAKASH · GRADE 7 · PART II · CHAPTER 1
Source Investigations
Think first. Use AI purposefully. Verify always.
Selected source investigations retain their printed page connections. The seven Think+ tasks are available separately. The final angle network includes an explicit source-consistency caution.
Circles and rectangles
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 3
Which measurements establish an exact copy?
Read the complete reasoning
Given
A circle or a rectangle to be copied.
To Find
Sufficient measurements for congruence
Property
A radius fixes a circle; two adjacent side lengths fix a rectangle.
Solution
- For circles, compare radii.
- For rectangles, compare the unordered pair of length and width; rotation may swap their orientation.
- Area alone is insufficient: 2 × 6 and 3 × 4 have the same area but different dimensions.
Result
Use radius for a circle, and length and width for a rectangle.
A three-arm symbol
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 4
What must be fixed in the source’s branched line figure?
Read the complete reasoning
Given
Three arms from one junction, with two of the arms lying on the same straight line.
To Find
Measurements that fix the branched symbol
Property
Lengths from the junction and the angle of the third arm.
Solution
- Measure each arm from the junction to its endpoint.
- Record the straight-line relationship of the opposite arms.
- Measure the angle between the third arm and one specified arm.
Result
These data fix the figure up to a rigid movement or reflection; appearance alone does not verify a pair.
RED and JAM
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 8
RE = JA = 3.5 cm, ED = AM = 5 cm, RD = JM = 6 cm. Write the correspondence.
Read the complete reasoning
Given
RE = JA, ED = AM and RD = JM.
To Find
The congruence order
Property
SSS.
Solution
- R is where the 3.5 cm and 6 cm sides meet; it matches J.
- E matches A, and D matches M.
Result
△RED ≅ △JAM.
A diagonal that bisects angles
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 9
AB = AD and CB = CD. Does AC bisect ∠BAD and ∠BCD?
Read the complete reasoning
Given
AB = AD, CB = CD and common side AC.
To Prove
∠BAC = ∠CAD and ∠BCA = ∠ACD
Property
SSS then corresponding angles.
Proof
- AB = AD, BC = DC and AC = AC give △ABC ≅ △ADC.
- A ↔ A, B ↔ D, C ↔ C.
- The two angles adjacent to AC at A match; so do those at C.
Result
AC bisects both specified angles.
A deliberate order trap
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 9
DF = DG and FE = GE. The source asks about triangles DFE and GED. Check the order before writing a congruence statement.
Read the complete reasoning
Given
DF = DG, FE = GE and common side DE.
To Find
The correctly ordered congruence
Property
SSS with D ↔ D, F ↔ G and E ↔ E.
Solution
- DF corresponds to DG; FE corresponds to GE; DE corresponds to DE.
- The correct statement is △DFE ≅ △DGE.
- △DFE ≅ △GED would wrongly pair DF with GE, which is not guaranteed.
Result
The triangles are congruent, but △DFE ≅ △GED is not a justified order.
The included 47° angle
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 13
AB = XZ = 7 cm, BC = ZY = 5 cm and ∠B = ∠Z = 47°. Establish the order.
Read the complete reasoning
Given
Two corresponding sides with the included angles B and Z.
To Find
The congruence order
Property
SAS.
Solution
- B ↔ Z lies between the 7 cm and 5 cm sides.
- A ↔ X and C ↔ Y.
Result
△ABC ≅ △XZY.
Parallel sides and a crossing
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 14
In the source arrangement A,O,C and B,O,D are collinear, AB ∥ CD and AB = CD. Identify the other equal parts.
Read the complete reasoning
Given
A,O,C and B,O,D are straight lines; AB ∥ CD and AB = CD.
To Find
Corresponding equal parts of the crossing triangles
Property
Alternate interior angles and ASA.
Solution
- ∠BAO = ∠DCO; ∠ABO = ∠CDO.
- AB = CD, so △ABO ≅ △CDO by ASA.
- The correspondence is A ↔ C, B ↔ D, O ↔ O.
Result
AO = CO, BO = DO and ∠AOB = ∠COD.
Two triangles sharing BC
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 14
∠ABC = ∠DBC and ∠ACB = ∠DCB. Prove the remaining angles equal and establish congruence.
Read the complete reasoning
Given
The two angle pairs and common side BC.
To Prove
∠BAC = ∠BDC and △ABC ≅ △DBC
Property
Angle sum and ASA.
Proof
- The third angles are each 180° minus the same two base angles.
- The known angles have common included side BC.
Result
△ABC ≅ △DBC; the third angles are equal.
Combine adjacent angles
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 14
∠ABD = ∠DCA and ∠ACB = ∠DBC in the source crossing figure. Identify the equal triangle parts.
Read the complete reasoning
Given
B and C are the common base endpoints; the two given angle equalities.
To Find
The correspondence between △ABC and △DCB
Property
Angle addition followed by ASA.
Solution
- ∠ABC = ∠ABD + ∠DBC; ∠DCB = ∠DCA + ∠ACB. Hence ∠ABC = ∠DCB.
- ∠ACB = ∠DBC and BC = CB.
- Thus △ABC ≅ △DCB.
Result
AB = DC, AC = DB and ∠BAC = ∠CDB.
End exercise Q2 — criterion and order
Textbook connection: Ganita Prakash · Grade 7 · Part II · Chapter 1 — Geometric Twins · printed page 20
Case (a): AB = DE, BC = EF, CA = DF
SSS
The endpoints of AB and AC give A ↔ D. Then B ↔ E and C ↔ F.
△ABC ≅ △DEF
Case (b): AB = EF, ∠A = ∠E, AC = ED
SAS
The included angles A and E lie between the given side pairs; A ↔ E, B ↔ F, C ↔ D.
△ABC ≅ △EFD
Case (c): AB = DF, ∠B = ∠D = 90°, AC = FE
RHS
AC and FE are hypotenuses; the right-angle vertices B and D correspond. AB ↔ FD gives A ↔ F and C ↔ E.
△ABC ≅ △FDE
Case (d): ∠A = ∠D, ∠B = ∠E, AC = DF
AAS
The third angles C and F agree by the angle sum, so AAS applies.
△ABC ≅ △DEF
Case (e): AB = DF, ∠B = ∠F, AC = DE
SSA — insufficient in general
The angle is not included between AB and AC, or between DF and DE. The two-intersection construction is a counterexample.
No congruence guaranteed
Source figure audit — the final angle network
Printed page 21, Figure it Out Q6. M, N, O and P below name the source’s four unnamed interior points. The diagram is a schematic transcription, not an exact metric construction.
Read the source critically
The full collection of marks and printed angles is not jointly exact. Equal single-marked segments give CD = 3u and CA = 2u. The right corners make the outer figure a rectangle, so its diagonal AD makes approximately 33.69° with AB, not exactly the printed 34°. Do not treat all marks and integer angle labels as an exact construction specification.
The table records local angle-chasing deductions from the stated subsets of information. It does not certify global consistency or establish congruence from appearance. A corrected source figure is needed before this network can be used as an exact scored construction task.
Single, double and triple ticks preserve the source’s distinct equality groups. The 90° label at O is the printed angle NOP. The outer corner and altitude squares are the source’s right-angle marks.
| Triangle | Angles | Reason |
|---|---|---|
| CUR | 90°, 45°, 45° | CU = CR; angle C is right. |
| AUN | 56°, 68°, 56° | AU = UN; use the angle sum. |
| UMN | 34°, 73°, 73° | UM = UN; the two base angles share 146°. |
| RVO | 56°, 68°, 56° | RV = VO; the two base angles share 112°. |
| VOD | 112°, 34°, 34° | Straight line RVD gives 112° at V; VO = VD. |
| DOF | 56°, 26°, 98° | Corner D is 90°; subtract 34°, then use the triangle sum. |
| PFB | 60°, 60°, 60° | PF = FB = BP are double-marked. |
| KLP | 30°, 90°, 60° | KL = LB and common perpendicular LP give △KLP ≅ △BLP by SAS; K,P,F are collinear. |
| PLB | 60°, 90°, 30° | The equilateral angle at B leaves 30° in the right corner. |
| NOP | 34°, 90°, 56° | Use the two given angles and the angle sum. |
| ANK | 34°, 44°, 102° | Two printed angles give the third. |
| NKP | 102°, 48°, 30° | AN and NO are opposite rays, so ∠KNO = 136°; subtract ∠PNO = 34°. |
| PFO | 94°, 22°, 64° | ∠OFB = 82° and ∠PFB = 60°; then use ∠NPO = 56° and ∠NPK = 30° with K,P,F collinear. |
| MNO | 83°, 51°, 46° | At N: 180° − 56° − 73° = 51°; at O the printed angle is 46°. |
| MRO | 102°, 34°, 44° | Two printed angles give the third. |
| URM | 33°, 45°, 102° | At R on CRV: 180° − 45° − 34° − 56° = 45°; at U on CUA: 180° − 45° − 34° − 68° = 33°. |