CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition
Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 10 Solutions: Congruency of Triangles (त्रिभुजों की सर्वांगसमता) in Bilingual Format (English & Hindi Medium).
Q1
In a right angled triangle Δ ABC C…
सिद्ध कीजिए कि समांतर चतुर्भुज के विकर्ण द्वारा…
📝 Question:
In a right angled triangle Δ ABC, ∠ C = 90°, M is the midpoint of hypotenuse AB. C is joined to M and extended to D such that DM = CM. Point D is joined to B. Show that: 1. Δ AMC ≅ Δ BMD 2. CM = 1/2 AB 3. Δ DBC ≅ Δ ACB 4. ∠ DBC = 90°: Δ ABC ∠ C = 90°, M AB C M D DM = CM D B : 1. Δ AMC ≅ Δ BMD 2. CM = 1/2 AB 3. Δ DBC ≅ Δ ACB 4. ∠ DBC = 90° Solution / : 1. Proof of Δ AMC ≅ Δ BMD: In Δ AMC and Δ BMD: AM = BM (M is the midpoint of AB) ∠ AMC = ∠ BMD (Vertically opposite angles) CM = DM (Given) By SAS congruence rule, Δ AMC ≅ Δ BMD. 2. Proof of ∠ DBC = 90°: Since Δ AMC ≅ Δ BMD, ∠ MAC = ∠ MBD (CPCT). Since these alternate interior angles are equal, DB ∥ AC. Thus, ∠ ACB + ∠ DBC = 180°(Co-interior angles). 90°+ ∠ DBC = 180° ⇒ {∠ DBC = 90°. 3. Proof of Δ DBC ≅ Δ ACB: In Δ DBC and Δ ACB: DB = AC (CPCT from Δ AMC ≅ Δ BMD) ∠ DBC = ∠ ACB = 90° BC = CB (Common side) By SAS congruence rule, Δ DBC ≅ Δ ACB. 4. Proof of CM = 1/2 AB: Since Δ DBC ≅ Δ ACB ⇒ DC = AB (CPCT). * CM = 1/2 DC ⇒ CM = 1/2 AB. Prove that the two triangles formed by the diagonal of a parallelogram are always congruent to each other
💡 Solution:
Let ABCD be a parallelogram with diagonal AC.
- In Δ ABC and Δ CDA
- AB = CD (Opposite sides of parallelogram)
- BC = DA (Opposite sides of parallelogram)
- AC = CA (Common side)
- By SSS congruence rule, Δ ABC ≅ Δ CDA.
Q2
Try This Page 216 Q1 In the given…
Try This Page 216 Q1 In the given…
📝 Question:
Try This (Page 216) / * Q.1 In the given isosceles triangles, find the value of x / x
💡 Solution:
(i) Isosceles triangle with sides 3x – 2 and 16:**
- (ii) Isosceles triangle with base angles 65°and vertex angle x°:
- (iii) Isosceles triangle with vertex angle 70°and base angle x°:
- Solution / :
- 140°+ ∠ 2 = 180° ⇒ {∠ 2 = 40°.
- Since BC = AC, the opposite angles are equal: {∠ 4 = ∠ 2 = 40°.
- In Δ ABC: ∠ 2 + ∠ 3 + ∠ 4 = 180° ⇒ 40°+ ∠ 3 + 40°= 180° ⇒ {∠ 3 = 100°.
- Solution / :
- Given parallel line at top with angle 62°. Alternate interior angle to the base of isosceles triangle is 62°.
- Since the triangle is isosceles, both base angles are 62°.
- The angle x°on the straight top line = 180°- (62°+ 88°) ⇒ {x = 30°.
Q3
Try This Page 226 Q1 From the following…
Try This Page 226 Q1 From the following…
📝 Question:
Try This (Page 226) / Q.1 From the following given measurements, which can form a triangle? / – ?* (i) 10 cm, 5 cm, 4 cm (ii) 8 cm, 6 cm, 3 cm (iii) 5 cm, 8 cm, 3 cm (iv) 14 cm, 6 cm, 7 cm
💡 Solution:
(i) R.H.S.** (Right angle-Hypotenuse-Side)
- (ii) SAS (Side-Angle-Side)
- (iii) ASA (Angle-Side-Angle)
- (iv) AAS (Angle-Angle-Side)
- (v) SSS (Side-Side-Side)
- Δ ABD ≅ Δ BAC
- BD = AC
- ∠ ABD = ∠ BAC
- Δ ABD ≅ Δ BAC
- BD = AC
- ∠ ABD = ∠ BAC
- Solution / :
- In Δ ABD and Δ BAC:
- AD = BC (Given)
- ∠ DAB = ∠ CBA (Given)
- AB = BA (Common side)
- By SAS congruence rule, Δ ABD ≅ Δ BAC.
- Since Δ ABD ≅ Δ BAC ⇒ BD = AC (CPCT).
- Since Δ ABD ≅ Δ BAC ⇒ ∠ ABD = ∠ BAC (CPCT). Hence proved.
- Solution / :
- Given BE = FC.
- Add EC to both sides: BE + EC = FC + EC ⇒ BC = FE.
- In Δ ABC and Δ DFE:
- AB = DF (Given)
- AC = DE (Given)
- BC = FE (Proved above)
- By SSS congruence rule, Δ ABC ≅ Δ DFE. Hence proved.
- Solution / :
- In Δ AOB: ∠ B < ∠ A ⇒ AO < BO ------ (1) (Side opposite to smaller angle is smaller)
- In Δ COD: ∠ C < ∠ D ⇒ OD < OC ------ (2)
- Adding equations (1) and (2):
- AO + OD < BO + OC ⇒ AD < BC. Hence proved.
- Solution / :
- Interior angle at M: ∠ PMN = 180°- x.
- Interior angle at N: ∠ PNM = 180°- y.
- Since x > y ⇒ -x < -y ⇒ 180°- x < 180°- y ⇒ ∠ PMN < ∠ PNM.
- Side opposite to larger angle ∠ PNM is MP, and side opposite to smaller angle ∠ PMN is NP.
- Therefore, MP > NP. Hence proved.
- Solution / :
- In Δ PQR: PQ > PR ⇒ ∠ PRQ > ∠ PQR.
- Dividing by 2: 1/2∠ PRQ > 1/2∠ PQR ⇒ ∠ SRQ > ∠ SQR.
- In Δ SQR, side opposite to larger angle ∠ SRQ is SQ, and side opposite to smaller angle ∠ SQR is SR.
- Therefore, SQ > SR. Hence proved.
- Solution / :
- In Δ PQR: PQ = PR ⇒ ∠ PQR = ∠ PRQ.
- Exterior angle of Δ PQS at Q: ∠ PQR = ∠ S + ∠ QPS > ∠ S.
- Thus, ∠ PRQ = ∠ PQR > ∠ S.
- In Δ PRS, since ∠ PRQ > ∠ S ⇒ PS > PR.
- Since PR = PQ, we get PS > PQ. Hence proved.