RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles - Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 4 – Triangles are designed by expert teachers to enhance students’ performance in the board exams. Mathematics is often considered one of the toughest subjects, which is why at Infinity Learn, we have created comprehensive RD Sharma Solutions with the help of our experienced team. These solutions are aimed at helping students grasp the concepts clearly and effectively.

Students can easily refer to and download the RD Sharma Solutions for Class 10 Chapter 4 from the link provided here. Our subject experts have carefully formulated these exercises to assist students in their exam preparation, ensuring they achieve excellent marks in Maths. By practicing these solutions, students will strengthen their understanding of the concepts and build the confidence essential for performing well in exams.

In this chapter, we explore triangles, recognized as the strongest polygon, and delve into their unique characteristics and properties. Unlike previous lessons on congruent figures, this chapter focuses on the similarity of geometric figures, especially triangles. Let’s take a quick look at the main concepts covered in this chapter:

• Similar polygons
• Similar triangles and their properties
• Basic results on proportionality
• Basic Proportionality Theorem
• Criteria for the similarity of triangles
• Areas of two similar triangles

Download RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles PDF Here

RD Sharma Class 10 Chapter 4 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 10 Chapter 4 PDF.

Access Answers to RD Sharma Solutions for Class 10 Maths Chapter 4- Triangles

1. In a Δ ABC, D and E are points on the sides AB and AC, respectively, such that DE || BC.

i) If AD/DB = 3/4 and AC = 15 cm, Find AE.

Solution:

Given: AD/BD = 3/4 and AC = 15 cm [As DE ∥ BC]

Required to find AE.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

Let, AE = x, then CE = 15-x.

⇒ 3/4 = x/ (15–x)

45 – 3x = 4x

-3x – 4x = – 45

7x = 45

x = 45/7

x = 6.43 cm

∴ AE= 6.43cm

ii) If AD/DB = 2/3 and AC = 18 cm, Find AE.

Solution:

Given: AD/BD = 2/3 and AC = 18 cm

Required to find AE.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

Let, AE = x and CE = 18 – x

⇒ 23 = x/ (18–x)

3x = 36 – 2x

5x = 36 cm

x = 36/5 cm

x = 7.2 cm

∴ AE = 7.2 cm

iii) If AD = 6 cm, DB = 9 cm and AE = 8 cm, Find AC.

Solution:

Given: Δ ABC, DE ∥ BC, AD = 6 cm, DB = 9 cm and AE = 8 cm.

Required to find AC.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

Let CE = x

So then,

6/9 = 8/x

6x = 72 cm

x = 72/6 cm

x = 12 cm

∴ AC = AE + CE = 12 + 8 = 20.

iv) If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.

Solution:

Given: AD = 4 cm, DB = 4.5 cm, AE = 8 cm

Required to find AC.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

4/4.5 = 8/AC

AC = (4.5 × 8)/4 cm

∴ AC = 9 cm

v) If AD = x cm, DB = x – 2 cm, AE = x + 2 cm, and EC = x – 1 cm, find the value of x.

Solution:

Given: AD = x, DB = x – 2, AE = x + 2 and EC = x – 1

Required to find the value of x.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

So, x/ (x–2) = (x+2)/ (x–1)

x(x – 1) = (x – 2)(x + 2)

x² – x – x² + 4 = 0

x = 4

vi) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1, and CE = 5x – 3, find the value of x.

Solution:

Given: AD = 4x – 3, BD = 3x – 1, AE = 8x – 7 and EC = 5x – 3

Required to find x.

By using Thales Theorem, [As DE ∥ BC]

AD/BD = AE/CE

So, (4x–3)/ (3x–1) = (8x–7)/ (5x–3)

(4x – 3)(5x – 3) = (3x – 1)(8x – 7)

4x(5x – 3) -3(5x – 3) = 3x(8x – 7) -1(8x – 7)

20x² – 12x – 15x + 9 = 24x² – 29x + 7

20x² -27x + 9 = 242 -29x + 7

⇒ -4x² + 2x + 2 = 0

4x² – 2x – 2 = 0

4x² – 4x + 2x – 2 = 0

4x(x – 1) + 2(x – 1) = 0

(4x + 2)(x – 1) = 0

⇒ x = 1 or x = -2/4

We know that the side of a triangle can never be negative. Therefore, we take the positive value.

∴ x = 1