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

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.

## 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) âˆ†ABC in which AB = 2 cm, BC = 3.5 cm and âˆ C = 800 and, âˆ†DEF in which DE = 2cm, DF = 3.5 cm and âˆ D = 800.

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 = 1800 â€“ (400 + 300)

By further calculation

= 1800 – 700

So we get

= 1100

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 âˆ† ABC,

It is given that

AB = 2cm, BC = 3.5 cm, âˆ C = 800

In âˆ† DEF,

It is given that

DE = 2 cm, DF = 3.5 cm and âˆ D = 800

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:

âˆ† ABD â‰…Â âˆ† ACD

Solution:

In âˆ† ABD and âˆ† ACD

AD = AD is common

It is given that

AB = AC and BD = DC

Here âˆ† ABD â‰…Â âˆ† ACD (SSS Axiom)

Therefore, it is proved.

3. Prove that:

(i) âˆ† ABC â‰…Â âˆ† ADC

(ii) âˆ B = âˆ D

(iii) AC bisects angle DCB.

Solution:

In the figure

AB = AD and CB = CD

In âˆ† ABC andÂ âˆ† ADC

AC = AC is common

It is given that

AB = AD and CB = CD

Here âˆ† ABC â‰…Â âˆ† ADC (SSS Axiom)

âˆ B = âˆ D (c. p. c. t)

So we get

âˆ BCA = âˆ DCA

Therefore, AC bisects âˆ DCB.

4. Prove that:

(i) âˆ†ABDÂ  â‰¡ âˆ†ACD

(ii) âˆ B = âˆ C

(iii) âˆ ADB = âˆ ADC

(iv) âˆ ADB = 90Â°

Solution:

From the figure

AD = AC and BD = CD

In âˆ†ABDÂ and âˆ†ACD

AD = AD is common

(i) âˆ†ABD â‰¡ âˆ†ACD (SSS Axiom)

(ii) âˆ B = âˆ C (c. p. c. t)

(iii) âˆ ADB = âˆ ADC (c. p. c. t)

(iv) We know that

âˆ ADB + âˆ ADC = 1800 is a linear pair

Here âˆ ADB = âˆ ADC

So we get

âˆ ADB = 1800/2

âˆ ADB = 90Â°

5. In the given figure, prove that:

(i) âˆ† ACB â‰… âˆ† ECD

(ii) AB = ED

Solution:

(i) In âˆ† ACB and âˆ† ECD
It is given that AC = CE and BC = CD

âˆ ACB = âˆ DCE are vertically opposite angles

Hence, âˆ† ACB â‰… âˆ† ECD (SAS Axiom)

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

Therefore, it is proved.

6. Prove that

(i) âˆ† ABC â‰… âˆ† ADC

(ii) âˆ B = âˆ D

Solution:

(i) In âˆ† ABC and âˆ† ADC

It is given that

AB = DC and BC = AD

AC = AC is common

Hence, âˆ† ABC â‰… âˆ† ADC (SSS Axiom)

(ii) Here âˆ B = âˆ D (c. p. c. t)

Therefore, it is proved.

7. In the given figure, prove that:

BD = BC.

Solution:

In right âˆ† ABD and âˆ† ABC

AB = AB is common

It is given that

AD = AC

Hence, âˆ† ABD â‰… âˆ† ABC (RHS Axiom)

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

Therefore, it is proved.

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

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

Solution:

In âˆ† ADB and âˆ† ADC

It is given that

AB = AC and âˆ 1 = âˆ 2

AD = AD is common

Hence, âˆ† ADB â‰… âˆ† ADC (SAS Axiom)

(i) âˆ B = âˆ C (c. p. c. t)

(ii) BD = DC (c. p. c. t)
(iii) âˆ ADB = âˆ ADC (c. p. c. t)

We know that

âˆ ADB + âˆ ADC = 1800 is a linear pair

So we get

âˆ ADB = âˆ ADC = 900

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 âˆ† PQR and âˆ† PSR

PR = PR is common

It is given that

âˆ PRQ = âˆ RPS and âˆ PQR = âˆ PSR

âˆ† PQR â‰… âˆ† 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) âˆ† XYZ â‰… âˆ† XPZ
(ii) YZ = PZ
(iii) âˆ YXZ = âˆ PXZ

Solution:

In âˆ† XYZ and âˆ† XPZ

It is given that

XY = XP

XZ = XZ is common

(i) âˆ† XYZ â‰… âˆ† XPZ (RHS Axiom)

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

(iii) âˆ YXZ = âˆ PXZ (c. p. c. t)

Therefore, it is proved.

11. In the given figure, prove that:

(i) âˆ† ABC â‰… âˆ† DCB

(ii) AC = DB

Solution:

In âˆ† ABC and âˆ† DCB

CB = CB is common

âˆ ABC = âˆ BCD = 900

It is given that

AB = CD

(i) âˆ† ABC â‰… âˆ† DCB (SAS Axiom)

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

Therefore, it is proved.

12. In the given figure, prove that:

(i) âˆ† AOD â‰… âˆ† BOC

(ii) AD = BC

(iii) âˆ ADB = âˆ ACB

(iv) âˆ†ADB â‰… âˆ†BCA

Solution:

In âˆ† AOD and âˆ† BOC

It is given that

OA = OB and OD = OC

âˆ AOD = âˆ BOC are vertically opposite angles

(i) âˆ† AOD â‰… âˆ† BOC (SAS Axiom)

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

(iii) âˆ ADB = âˆ ACB (c. p. c. t)

(iv) âˆ†ADB â‰… âˆ†BCA

It is given that

âˆ†ADB = âˆ†BCA

AB = AB is common

Here âˆ† AOB â‰… âˆ† 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 âˆ† ABC

AB = BC = CA

We know that

AD is perpendicular to BC and BE is perpendicular to AC

In âˆ† ADC and âˆ† BEC

âˆ ADC = âˆ BEC = 900

âˆ ACD = âˆ BCE is common

AC = BC are the sides of an equilateral triangle

âˆ† ADC â‰… âˆ† 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

âˆ ABD = 500, âˆ ADB = y – 70

âˆ CBD = x + 50, âˆ CDB = 380

In âˆ† ABD and âˆ† CBD

BD = BD is common

It is given that

AB = BC and AD = CD

Here âˆ† ABD â‰… âˆ† CBD (SSS Axiom)

âˆ ABD = âˆ CBD

So we get

50 = x + 5

x = 50 â€“ 5 = 450

âˆ ADB = âˆ CDB

y â€“ 7 = 38

y = 38 + 7 = 450

Therefore, x = 450 and y = 450.

15. The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that:
(i) âˆ† ABD â‰… âˆ† ACD
(ii) AB = AC
(iii) âˆ B = âˆ C

Solution:

(i) In âˆ† ABC

AD is perpendicular to BC

BD = CD

In âˆ† ABD and âˆ† ACD

AD = AD is common

âˆ ADB = âˆ ADC = 900

It is given that

BD = CD

âˆ† ABD â‰… âˆ† CAD (SAS Axiom)

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

(iii) âˆ B = âˆ C as âˆ† ADB â‰… âˆ† ADC

Therefore, it is proved.