Selina Solutions Concise Maths Class 7 Chapter 19 Congruency: Congruent Triangles provides students with a clear idea about the fundamentals of congruency. The solutions contain stepwise explanations in simple language to boost interest among students towards the subject. The answers are prepared in an interactive manner, which covers all the important shortcut tricks in solving problems in less time. Selina Solutions Concise Maths Class 7 Chapter 19 Congruency: Congruent Triangles, PDF links are given here with a free download option.

Chapter 19 contains important topics like congruency in triangles, corresponding sides and angles and the conditions of congruency. The solutions PDF help students clear their doubts by themselves and solve tricky problems with ease.

## Selina Solutions Concise Maths Class 7 Chapter 19: Congruency: Congruent Triangles Download PDF

## Access Selina Solutions Concise Maths Class 7 Chapter 19: Congruency: Congruent Triangles

Exercise 19 page: 213

**1. State, whether the pairs of triangles given in the following figures are congruent or not:**

**(vii) ∆ABC in which AB = 2 cm, BC = 3.5 cm and ∠C = 80 ^{0} and, ∆DEF in which DE = 2cm, DF = 3.5 cm and ∠D = 80^{0}.**

**Solution:**

**(i) In the given figure, corresponding sides of the triangles are not equal.**

**Therefore, the given triangles are not congruent.**

**(ii) In the first triangle**

**Third angle = 180 ^{0} – (40^{0} + 30^{0})**

**By further calculation**

**= 180 ^{0} – 70^{0}**

**So we get**

**= 110 ^{0}**

**In the two triangles, the sides and included angle of one are equal to the corresponding sides and included angle.**

**Therefore, the given triangles are congruent. (SAS axiom)**

**(iii) In the given figure, corresponding two sides are equal and the included angles are not equal.**

**Therefore, the given triangles are not congruent.**

**(iv) In the given figure, the corresponding three sides are equal.**

**Therefore, the given triangles are congruent. (SSS Axiom)**

**(v) In the right triangles, one side and diagonal of one triangle are equal to the corresponding side and diagonal of the other.**

**Therefore, the given triangles are congruent. (RHS Axiom)**

**(vi) In the given figure, two sides and one angle of one triangle are equal to the corresponding sides and one angle of the other.**

**Therefore, the given triangles are congruent. (SSA Axiom)**

**(vii) In ∆ ABC, **

**It is given that**

**AB = 2cm, BC = 3.5 cm, ∠C = 80 ^{0}**

**In ∆ DEF,**

**It is given that**

**DE = 2 cm, DF = 3.5 cm and ∠D = 80 ^{0}**

**We get to know that two corresponding sides are equal but the included angles are not equal.**

**Therefore, the triangles are not congruent.**

**2. In the given figure, prove that:**

**∆ ABD ≅∆ ACD**

**Solution:**

**In ∆ ABD and ∆ ACD**

**AD = AD is common**

**It is given that**

**AB = AC and BD = DC**

**Here ∆ ABD ≅∆ ACD (SSS Axiom)**

**Therefore, it is proved.**

**3. Prove that:**

**(i) ∆ ABC ≅∆ ADC**

**(ii) ∠B = ∠D**

**(iii) AC bisects angle DCB.**

**Solution:**

**In the figure **

**AB = AD and CB = CD**

**In ∆ ABC and∆ ADC**

**AC = AC is common**

**It is given that**

**AB = AD and CB = CD**

**Here ∆ ABC ≅∆ ADC (SSS Axiom)**

**∠B = ∠D (c. p. c. t)**

**So we get**

**∠BCA = ∠DCA**

**Therefore, AC bisects ∠DCB.**

**4. Prove that:**

**(i) ∆ABD ≡ ∆ACD**

**(ii) ∠B = ∠C**

**(iii) ∠ADB = ∠ADC**

**(iv) ∠ADB = 90°**

**Solution:**

**From the figure**

**AD = AC and BD = CD**

**In ∆ABDand ∆ACD**

**AD = AD is common**

**(i) ∆ABD ≡ ∆ACD (SSS Axiom)**

**(ii) ∠B = ∠C (c. p. c. t)**

**(iii) ∠ADB = ∠ADC (c. p. c. t)**

**(iv) We know that**

**∠ADB + ∠ADC = 180 ^{0} is a linear pair**

**Here ∠ADB = ∠ADC**

**So we get**

**∠ADB = 180 ^{0}/2**

**∠ADB = 90°**

**5. In the given figure, prove that:**

**(i) ∆ ACB ≅ ∆ ECD**

**(ii) AB = ED**

**Solution:**

**(i) In ∆ ACB and ∆ ECDIt is given that AC = CE and BC = CD**

**∠ACB = ∠DCE are vertically opposite angles**

**Hence, ∆ ACB ≅ ∆ ECD (SAS Axiom)**

**(ii) Here AB = ED (c. p. c. t)**

**Therefore, it is proved.**

**6. Prove that **

**(i) ∆ ABC ≅ ∆ ADC**

**(ii) ∠B = ∠D**

**Solution:**

**(i) In ∆ ABC and ∆ ADC**

It is given that

AB = DC and BC = AD

AC = AC is common

Hence, **∆ ABC ≅ ∆ ADC (SSS Axiom)**

**(ii) Here ∠B = ∠D (c. p. c. t)**

**Therefore, it is proved.**

**7. In the given figure, prove that: **

**BD = BC.**

**Solution:**

In right **∆ ABD and ∆ ABC**

**AB = AB is common**

**It is given that**

**AD = AC **

**Hence, ∆ ABD ≅ ∆ ABC (RHS Axiom)**

**Here BD = BC (c. p. c. t)**

**Therefore, it is proved.**

**8. In the given figure, ∠1 = ∠2 and AB = AC. **

**Prove that: (i) ∠B = ∠ C (ii) BD = DC (iii) AD is perpendicular to BC.**

**Solution:**

**In ∆ ADB and ∆ ADC**

**It is given that**

**AB = AC and ∠1 = ∠2**

**AD = AD is common**

**Hence, ∆ ADB ≅ ∆ ADC (SAS Axiom)**

**(i) ∠B = ∠C (c. p. c. t)**

(ii) BD = DC (c. p. c. t)

(iii) ∠ADB = ∠ADC (c. p. c. t)

**We know that**

**∠ADB + ∠ADC = 180 ^{0} is a linear pair**

**So we get**

**∠ADB = ∠ADC = 90 ^{0}**

**Here, AD is perpendicular to BC**

**Therefore, it is proved.**

**9. In the given figure, prove that:(i) PQ = RS(ii) PS = QR**

**Solution:**

In **∆ PQR and ∆ PSR**

**PR = PR is common**

**It is given that**

**∠PRQ = ∠RPS and ∠PQR = ∠PSR**

**∆ PQR ≅ ∆ PSR (AAS Axiom)**

**(i) PQ = RS (c. p. c. t)**

(ii) QR = PS or PS = QR (c. p. c. t)

Therefore, it is proved.

**10. In the given figure, prove that:**

**(i) ∆ XYZ ≅ ∆ XPZ(ii) YZ = PZ(iii) ∠YXZ = ∠PXZ**

**Solution:**

In **∆ XYZ and ∆ XPZ**

**It is given that**

**XY = XP**

**XZ = XZ is common**

**(i) ∆ XYZ ≅ ∆ XPZ (RHS Axiom)**

(ii) **YZ = PZ (c. p. c. t)**

(iii) ∠YXZ = ∠PXZ (c. p. c. t)

**Therefore, it is proved.**

**11. In the given figure, prove that:**

**(i) ∆ ABC ≅ ∆ DCB**

**(ii) AC = DB**

**Solution:**

In **∆ ABC and ∆ DCB**

**CB = CB is common**

**∠ABC = ∠BCD = 90 ^{0}**

**It is given that**

**AB = CD**

**(i) ∆ ABC ≅ ∆ DCB (SAS Axiom)**

(ii) AC = DB (c. p. c. t)

**Therefore, it is proved.**

**12. In the given figure, prove that:**

**(i) ∆ AOD ≅ ∆ BOC**

**(ii) AD = BC**

**(iii) ∠ADB = ∠ACB**

**(iv) ∆ADB ≅ ∆BCA**

**Solution:**

In **∆ AOD and ∆ BOC**

**It is given that**

**OA = OB and OD = OC**

**∠AOD = ∠BOC are vertically opposite angles**

**(i) ∆ AOD ≅ ∆ BOC (SAS Axiom)**

(ii) AD = BC (c. p. c. t)

(iii) ∠ADB = ∠ACB (c. p. c. t)

(iv) ∆ADB ≅ ∆BCA

**It is given that**

**∆ADB = ∆BCA**

AB = AB is common

Here **∆ AOB ≅ ∆ BCA**

**Therefore, it is proved.**

**13. ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:(i) AD = BE(ii)BD = CE**

**Solution:**

In **∆ ABC**

**AB = BC = CA**

**We know that **

**AD is perpendicular to BC and BE is perpendicular to AC**

In **∆ ADC and ∆ BEC**

**∠ADC = ∠BEC = 90 ^{0}**

**∠ACD = ∠BCE is common **

**AC = BC are the sides of an equilateral triangle**

**∆ ADC ≅ ∆ BEC (AAS Axiom)**

**(i) AD = BE (c. p. c. t)**

(ii) BD = CE (c. p. c. t)

Therefore, it is proved.

**14. Use the informations given in the following figure to prove triangles ABD and CBD are congruent.**

**Also, find the values of x and y.**

**Solution:**

In the figure

AB = BC and AD = DC

**∠ABD = 50 ^{0}, ∠ADB = y – 7^{0}**

**∠CBD = x + 5 ^{0}, ∠CDB = 38^{0}**

In **∆ ABD and ∆ CBD**

**BD = BD is common**

**It is given that**

**AB = BC and AD = CD**

**Here ∆ ABD ≅ ∆ CBD (SSS Axiom)**

**∠ABD = ∠CBD**

**So we get**

**50 = x + 5**

**x = 50 – 5 = 45 ^{0}**

**∠ADB = ∠CDB**

**y – 7 = 38**

**y = 38 + 7 = 45 ^{0}**

**Therefore, x = 45 ^{0} and y = 45^{0}.**

**15. The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that: (i) ∆ ABD ≅ ∆ ACD(ii) AB = AC(iii) ∠B = ∠C**

**Solution:**

**(i) In ∆ ABC**

AD is perpendicular to BC

BD = CD

**In ∆ ABD and ∆ ACD**

**AD = AD is common**

**∠ADB = ∠ADC = 90 ^{0}**

**It is given that**

**BD = CD**

**∆ ABD ≅ ∆ CAD (SAS Axiom)**

**(ii) AB = AC (c. p. c. t)**

**(iii) ∠B = ∠C as ∆ ADB ≅ ∆ ADC**

**Therefore, it is proved.**