RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B

In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15B for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15B Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 7
SubjectMaths
ChapterChapter 15
Chapter NameProperties of Triangles
Exercise15B

RS Aggarwal Solutions for Class 7 Chapter 15 Properties of Triangles Ex 15B Download PDF

Question 1.
Solution:
In ∆ABC,
∠A = 75°, ∠B = 45°
side BC is produced to D
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 1
forming exterior ∠ ACD
Exterior ∠ACD = ∠A + ∠B (Exterior angle is equal to sum of its interior opposite angles)
= 75° + 45° = 120°

Question 2.
Solution:
In ∆ABC, BC is produced to D forming an exterior angle ACD
∠ B = 68°, ∠ A = x°, ∠ ACB = y° and ∠ACD = 130°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 2
In triangle,
Exterior angles is equal to sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 130° = x + 68°
⇒ x = 130° – 68° = 62°
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
Hence x = 62° and y = 50°

Question 3.
Solution:
In ∆ABC, side BC is produced to D forming exterior angle ACD.
∠ACD = 65°, ∠A = 32°
∠B = x, ∠ACB = y
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 3
In a triangle, the exterior angles is equal to the sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 65° = 32° + x
⇒ x = 65° – 32° = 33°
But ∠ ACD + ∠ ACB = 180° (Linear pair)
⇒ 65° + y = 180°
⇒ y = 180°- 65° = 115°
x = 33° and y = 115°

Question 4.
Solution:
In ∆ABC, side BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 4
∠ACD = 110°, and ∠A : ∠B = 2 : 3
In a triangle, exterior angles is equal to the sum of its interior opposite angles
⇒ ∠ACD = ∠A + ∠B
⇒ ∠A + ∠B = 110°
But ∠A : ∠B = 2 : 3
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 5
But ∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
⇒ 44° + 66° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
Hence ∠ A = 44°, ∠ B = 66° and ∠ C = 70°

Question 5.
Solution:
In ∆ABC, side BC is produced to forming exterior angle ACD.
∠ACD = 100° and ∠A = ∠B
Exterior angle of a triangle is equal to the sum of its interior opposite angles.
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 6
∠ACD = ∠A + ∠B But ∠A = ∠B
∠A + ∠A = ∠ACD = 100°
⇒ 2 ∠A = 100°
⇒ ∠A = 50°
∠B = ∠A = 50°
But ∠A + ∠B + ∠ ACB = 180° (sum of angles of a triangle)
⇒ 50° + 50° + ∠ ACB = 180°
⇒ 100° + ∠ ACB = 180°
⇒ ∠ ACB = 180° – 100° = 80°
Hence ∠ A = 50°, ∠ B = 50° and ∠ C = 80°

Question 6.
Solution:
In ∆ABC, side BC is produced to D From D, draw a line meeting AC at E so that ∠D = 40°
∠A = 25°, ∠B = 45°
In ∆ABC,
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 7
Exterior ∠ACD = ∠A + ∠B = 25° + 45° = 70°
Again, in ∆CDE,
Exterior ∠ AED = ∠ ECD + ∠ D = ∠ACD + ∠D = 70° + 40° = 110°
Hence ∠ACD = 70° and ∠AED = 110°

Question 7.
Solution:
In ∆ABC, sides BC is produced to D and BA to E
∠CAD = 50°, ∠B = 40° and ∠ACB = 100°
∠ ACB + ∠ ACD = 180° (Linear pair)
⇒ 100° + ∠ ACD = 180°
⇒ ∠ ACD = 180° – 100° = 80°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 8
In ∆ACD,
∠ CAD + ∠ ACD + ∠ ADC = 180° (sum of angles of a triangle)
⇒ 50° + 80° + ∠ ADC = 180°
⇒ 130° + ∠ ADC = 180°
⇒ ∠ ADC = 180° – 130° = 50°
Now, in ∆ABD, BA is produced to E
Exterior ∠DAE = ∠ACD + ∠ADC = 80° + 50° = 130°
Hence ∠ ACD = 80°, ∠ ADC = 50° and ∠DAE = 130°

Question 8.
Solution:
In ∆ABC, BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 9
∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.
x : y = 2 : 3
Now, in ∆ABC,
Exterior ∠ACD = ∠A + ∠B
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 10
But ∠A + ∠B + ∠ACB = 180° (sum of angles of a triangle)
⇒ 78° + 52° + ∠ACB = 180°
⇒ 130° + ∠ACB = 180°
⇒ ∠ACB = 180° – 130°
⇒ ∠ACB = 50°
⇒ ∠ = 50°
Hence x = 52°, y = 78° and z = 50°

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