RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4C

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4C Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4C Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4C Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameTriangles
Exercise4C
CategoryRS Aggarwal Solutions

RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4C

Question 1.
Solution:
Given : Area of ∆ABC = 64 cm²
and area of ∆DEF =121 cm²
EF = 15.4 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 1.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 1.2

Question 2.
Solution:
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 2.1

Question 3.
Solution:
∆ABC ~ ∆PQR
ar (∆ABC) = 4ar (∆PQR),
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 3.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 3.2

Question 4.
Solution:
Areas of two similar triangles are 169 cm² and 121 cm²
Longest side of largest triangle = 26 cm
Let longest side of smallest triangle = x
∆s are similar
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 4.1

Question 5.
Solution:
Area of ∆ABC = 100 cm²
and area of ∆DEF = 49 cm²
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 5.1

Question 6.
Solution:
Given : Corresponding altitudes of two similar triangles are 6 cm and 9 cm
We know that the areas of two similar triangles are in the ratio of the squares of their corresponding altitudes.
Ratio in the areas of two similar triangles = (6)² : (9)² = 36 : 81 = 4 : 9 (Dividing by 9)

Question 7.
Solution:
The areas of two similar triangles are 81 cm² and 49 cm²
Altitude of the first triangle = 6.3 cm
Let altitude of second triangle = x cm
The areas of two similar triangles are in the ratio of the squares on their corresponding altitude,
Let area of ∆ABC = 81 cm²
and area of ∆DEF = 49cm²
Altitude AL = 6 – 3 cm
Let altitude DM = x cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 7.1

Question 8.
Solution:
Areas of two similar triangles are 100 cm² and 64 cm²
Let area of ∆ABC = 100 cm²
and area of ∆DEF = 64 cm²
Median DM of ∆DEF = 5.6 cm
Let median AL of ∆ABC = x
The areas of two similar triangles is proportional to the squares of their corresponding median.
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 8.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 8.2

Question 9.
Solution:
Given : In ∆ABC, PQ is a line which meets AB in P and AC in Q.
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm QC = 4.5 cm
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 9.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 9.2

Question 10.
Solution:
In ∆ABC,
DE || BC
DE = 3 cm, BC = 6 cm
area (∆ADE) = 15 cm²
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 10.1

Question 11.
Solution:
Given : In right ∆ABC, ∠A = 90°
AD ⊥ BC
BC = 13 cm, AC = 5 cm
To find : Ratio in area of ∆ABC and ∆ADC
In ∆ABC and ∆ADC
∠C = ∠C (common)
∠BAC = ∠ADC (each 90°)
∆ABC ~ ∆ADC (AA axiom)
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 11.1

Question 12.
Solution:
In the given figure ∆ABC,
DE || BC and DE : BC = 3 : 5.
In ∆ABC and ∆ADE,
DE || BC
∆ABC ~ ∆ADE
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 12.1

Question 13.
Solution:
In ∆ABC, D and E are the midpoints of sides AB and AC respectively.
DE || BC and DE = 12 BC
∆ADE ~ ∆ABC
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 13.1
RS Aggarwal Solutions Class 10 Chapter 4 Triangles 4C 13.2

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