CBSE Important Questions Class 7 Maths Chapter 9 Geometric Twins

CBSE Important Questions Class 7 Maths Chapter 9 – Geometric Twins

Class 7 Maths Chapter 9 Geometric Twins is the first chapter of Part 2 of the new NCERT (National Council of Educational Research and Training) textbook Ganita Prakash. Geometric twins are congruent figures, that is, figures of exactly the same shape and size. Students learn the congruence rules for triangles (SSS, SAS, ASA and RHS), how to match corresponding parts, and how to use CPCT (corresponding parts of congruent triangles) to find equal sides and angles.

These CBSE important questions for Class 7 Maths Chapter 9 include figure-based questions on isosceles triangles, altitudes, right-angled triangles and the RHS, SAS and ASA rules, along with multiple choice questions on congruence statements. Every question has a clear step-by-step answer in simple English, so students can practise on their own, parents can check the working, and teachers can use them in class. A free printable PDF of these important questions is also available on this page.

CBSE Important Questions Class 7 Maths Chapter 9 Geometric Twins

Geometric Twins (New NCERT)
Medium
Q.
Check if the two figures are congruent.
Geometric Twins (New NCERT)
Medium
Q.
Circle the pairs that appear congruent.
Geometric Twins (New NCERT)
Medium
Q.
What measurements would you take to create a figure congruent to a given:
(a) Circle
(b) Rectangle
Using this, state how you would check if two
(a) Are circles congruent?
(b) Rectangles are congruent?
Geometric Twins (New NCERT)
Medium
Q.
How would we check if two figures like the one below are congruent?
Use this to identify whether each of the following pairs is congruent.
Geometric Twins (New NCERT)
Medium
Q.
Suppose ∆HEN is congruent to ∆BIG. List all the other correct ways of expressing this congruence.
Geometric Twins (New NCERT)
Medium
Q.
Determine whether the triangles are congruent. If yes, express the congruence.
Geometric Twins (New NCERT)
Medium
Q.
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.
Geometric Twins (New NCERT)
Medium
Q.
In the figure below, are ∆DFE and ∆GED congruent to each other? It is given that DF = DG and FE = GE.
Geometric Twins (New NCERT)
Medium
Q.
Identify whether the triangles below are congruent. What conditions did you use to establish their congruence? Express the congruence.
Geometric Twins (New NCERT)
Medium
Q.
​Given that ∠ABC = ∠DBC and ∠ACB = ∠DCB, show that ∠BAC = ∠BDC. Are the two triangles congruent?
Geometric Twins (New NCERT)
Medium
Q.
∆AIR ≅ ∆FLY. Identify the corresponding vertices, sides, and angles.
Geometric Twins (New NCERT)
Medium
Q.
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
Geometric Twins (New NCERT)
Medium
Q.
​It is given that OB = OC, and OA = OD. Show that AB is parallel to CD.
Geometric Twins (New NCERT)
Medium
Q.
ABCD is a square. Show that ∆ABC ≅ ∆ADC. Is ∆ABC also congruent to ∆CDA?
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?
Geometric Twins (New NCERT)
Medium
Q.
Find ∠B and ∠C, if A is the centre of the circle.
Geometric Twins (New NCERT)
Medium
Q.
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.

CBSE Important Questions Class 7 Maths Chapter 9 Geometric Twins

Q1. ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.
(i) State the three pairs of equal parts in ΔADB and ΔADC.
(ii) Is ΔADB ≅ ΔADC? Why or why not?
(iii) Is ∠B = ∠C? Why or why not?
(iv) Is BD = CD? Why or why not?

Isosceles triangle ABC with altitude AD

Answer: (i) In ΔADB and ΔADC, the three pairs of equal parts are:
AB = AC (isosceles triangle)
∠BDA = ∠CDA (= 90°, as AD is an altitude)
AD = AD (common side)
(ii) Yes, ΔADB ≅ ΔADC by the RHS congruence rule (right angle, hypotenuse AB = AC, side AD common).
(iii) Yes, ∠B = ∠C, because they are corresponding parts of congruent triangles (CPCT).
(iv) Yes, BD = CD, because they are corresponding parts of congruent triangles (CPCT).

Q2. In the figure, BD and CE are altitudes of ΔABC such that BD = CE.
(i) State the three pairs of equal parts in ΔCBD and ΔBCE.
(ii) Is ΔCBD ≅ ΔBCE? Why or why not?
(iii) Is ∠DCB = ∠EBC? Why or why not?

Triangle ABC with altitudes BD and CE

Answer: (i) In ΔCBD and ΔBCE, the three pairs of equal parts are:
∠BEC = ∠CDB (= 90°, as BD and CE are altitudes)
BD = CE (given)
BC = BC (common side)
(ii) Yes, ΔCBD ≅ ΔBCE by the RHS congruence rule (right angle, common hypotenuse BC, side BD = CE).
(iii) Yes, ∠DCB = ∠EBC, because they are corresponding parts of congruent triangles (CPCT).

Q3. In the figure, the two triangles are congruent. The corresponding parts are marked. Can we write ΔRAT ≅ ΔWON? Give reasons also.

Triangles RAT and WON with equal parts marked

Answer: In ΔRAT and ΔWON, the marked equal parts are:
AR = OW (given)
∠RAT = ∠WON (given)
AT = ON (given)
Two sides and the angle between them are equal, so ΔRAT ≅ ΔWON by the SAS congruence rule. The correspondence R ↔ W, A ↔ O, T ↔ N matches the marked parts, so yes, we can write ΔRAT ≅ ΔWON.

Q4. It is to be established by the RHS congruence rule that ΔABC ≅ ΔPQR. What additional information is needed, if it is given that ∠B = ∠Q = 90° and AC = PR?

Right-angled triangles PQR and ABC

Answer: For the RHS rule we need a right angle, an equal hypotenuse and one more pair of equal sides. The right angles (∠B = ∠Q = 90°) and the hypotenuses (AC = PR) are already given. So the additional information needed is one pair of equal sides: either BC = QR or AB = PQ.

Q5. In the following figure, DA ⊥ AB, CB ⊥ AB and AC = BD. State the three pairs of equal parts in ΔABC and ΔDAB. Which of the following statements is meaningful?
(i) ΔABC ≅ ΔBAD (ii) ΔABC ≅ ΔABD

Figure with DA perpendicular to AB, CB perpendicular to AB and diagonals AC and BD

Answer: The three pairs of equal parts are:
∠ABC = ∠BAD (= 90°)
AC = BD (given)
AB = BA (common side)
So ΔABC ≅ ΔBAD by the RHS congruence rule. Statement (i) is true and meaningful.
Statement (ii) ΔABC ≅ ΔABD is not meaningful, because the vertices do not correspond correctly (A ↔ A, B ↔ B, C ↔ D does not match the equal parts).

Q6. In ΔPQR, PQ = 5 cm, QR = 4 cm and ∠PQR = 90°. If ΔPQR ≅ ΔMNO, then the length of MO is _______.
(a) 2√41 cm
(b) √43 cm
(c) √41 cm
(d) 2√13 cm

Answer: (c) √41 cm
In right-angled ΔPQR, PR is the hypotenuse. By the Pythagoras property, PR² = PQ² + QR² = 5² + 4² = 25 + 16 = 41, so PR = √41 cm.
Since ΔPQR ≅ ΔMNO, side PR corresponds to side MO. So MO = PR = √41 cm.

Q7. The given triangles ABC and PQR are congruent by the rule _______.
(a) RHS
(b) ASA
(c) SAS
(d) SSS

Right-angled triangles ABC and PQR with sides 3 cm and 6.5 cm

Answer: (a) RHS
Both triangles are right-angled (∠B = ∠Q = 90°). The hypotenuses are equal (AC = PR = 6.5 cm) and one pair of sides is equal (BC = PQ = 3 cm). So the triangles are congruent by the RHS (Right angle – Hypotenuse – Side) rule.

Q8. Which of the following conditions are required to prove ΔADB and ΔADC congruent?
(a) AB = AC, AD = DA and ∠BDA = ∠CDA = 90°
(b) AD = AC, AB = AD and ∠BAD = ∠CAD
(c) BD = CD, AD = AC and ∠ABD = ∠CDA
(d) AB = CD, AD = DA and ∠BDA = ∠CDA = 90°

Triangle ABC with AD perpendicular to BC and AB = AC

Answer: (a) AB = AC, AD = DA and ∠BDA = ∠CDA = 90°
With these conditions, both triangles are right-angled at D, their hypotenuses AB and AC are equal, and side AD is common. This is exactly the RHS congruence rule, so ΔADB ≅ ΔADC.

Q9. In ΔABC and ΔRPQ, ∠B = ∠P = 90° and AB = RP. Which of the following pairs of sides is required to show the triangles ΔABC and ΔRPQ congruent by the RHS congruence rule?
(a) AC = PQ
(b) AB = PQ
(c) BC = RQ
(d) AC = RQ

Answer: (d) AC = RQ
For the RHS rule we need equal hypotenuses. In ΔABC the hypotenuse is AC (opposite the right angle at B) and in ΔRPQ the hypotenuse is RQ (opposite the right angle at P). So we need AC = RQ.

Q10. If ΔABC ≅ ΔDEF, then the value of DF will be _____.
(a) 13 cm
(b) 5 cm
(c) 3.5 cm
(d) 2 cm

Triangle ABC with sides 2 cm, 5 cm, 3.5 cm and triangle DEF

Answer: (c) 3.5 cm
In ΔABC ≅ ΔDEF, the vertices correspond as A ↔ D, B ↔ E, C ↔ F. So side AC corresponds to side DF. Since AC = 3.5 cm, DF = 3.5 cm.

Q11. If ΔABC and ΔPQR are congruent by the ASA criterion, the pair of corresponding equal sides is ______.
(a) BC = PR
(b) BC = RQ
(c) RQ = AB
(d) RQ = AC

Triangles ABC and PQR with equal angles marked

Answer: (b) BC = RQ
In the figure, ∠B = ∠Q (right angles) and ∠C = ∠R (double arcs). For the ASA rule, the side between the two equal angles must be equal. In ΔABC that side is BC and in ΔPQR it is QR. So BC = RQ.

Q12. What can be said for the following statement: 'Two triangles with equal corresponding angles need not be congruent.'?
(a) It is always true.
(b) It is always false.
(c) It is partially true.
(d) It cannot be determined.

Answer: (c) It is partially true.
Triangles with all corresponding angles equal have the same shape (they are similar), but their sizes can be different. They are congruent only when the corresponding sides are also equal. So equal angles alone do not guarantee congruence, which makes the statement partially true.

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