RD Sharma Class 10 Solutions Chapter 4 - Triangles

Welcome to your complete guide for RD Sharma Class 10 Chapter 4, Triangles. This chapter is fundamental to your geometry curriculum, covering key concepts like the properties of similar triangles, the Pythagorean theorem, and proportionality theorems. To help you master this topic, our RD Sharma Class 10 Solutions offer clear and precise explanations for every problem in the textbook. These solutions are expertly crafted to simplify even the most complex geometric concepts.

Our class 10 RD Sharma triangles solutions are designed to be the perfect companion to your NCERT solutions for class 10 Maths, providing additional practice and detailed, step-by-step guidance. By working through the triangle class 10 RD Sharma exercises with our guide, you will build confidence and a deeper understanding of the material. Each solution for the class 10 RD Sharma triangles problems explains the underlying logic, helping you prepare effectively for your board exams.

If you find any topic in RD Sharma Class 10 Chapter 4 challenging, our resource breaks it down into easy-to-follow steps. Read the full article to access and download the complete class 10 RD Sharma triangles solutions PDF.

RD Sharma Class 10 Chapter 4 Triangles PDF with Solutions

RD Sharma Class 10 Solutions 2025-26 PDF for Chapter 4 helps students learn all the concepts of triangles in a easy and step-by-step way. This chapter explains the rules and properties of similar triangles, their types, and important theorems like Pythagoras and more.

The RD Sharma Class 10 Triangles PDF gives complete solutions for each question. If you want to study offline, you can go for RD Sharma Class 10 Triangles PDF Download, which is very helpful before exams or while doing homework. The solved answers match the latest CBSE Board pattern. Along with NCERT Solutions for Class 10, using this PDF can make your learning more stronger. Many students find it easy to understand with RD Sharma’s clear steps.

Important Questions for RD Sharma Class 10 – Chapter 4: Triangles with Solutions 

Below are important questions from Chapter 4 (Triangles) of RD Sharma Class 10, covering all key concepts such as similarity criteria, proportionality theorems, the Pythagorean theorem, applications, and problem-solving.

Ques: Fill in the blanks (Basics of Similarity)

Fill in the blanks using the correct word given in brackets:

(i) All circles are ________ (congruent, similar).

(ii) All squares are ________ (similar, congruent).

(iii) All ________ triangles are similar (isosceles, equilateral).

(iv) Two triangles are similar, if their corresponding angles are ________ (proportional, equal).

(v) Two triangles are similar, if their corresponding sides are ________ (proportional, equal).

Solution:

(i) similar

(ii) similar

(iii) equilateral

(iv) equal

(v) proportional

Ques: Proportionality in Similar Triangles (Numerical Application)

In ΔABC, D and E are points on AB and AC such that DE || BC. If AD = 6cm, DB = 9cm, and AE = 8cm, find AC.

Solution: By Thales’ theorem,

AD/DB = AE/EC

6/9 = 8/x ⟹ 6x = 72 ⟹ x = 12

So, AC = AE + EC = 8 + 12 = 20cm

Ques: Finding Length Using Ratios

If AD/DB = 3/4 and AC = 15cm, find AE (with D, E as previous).

Solution: Let AE = x, then EC = 15 – x.

3/4 = x/(15–x) ⟹ 45–3x = 4x ⟹ 7x = 45 ⟹ x = 45/7 ≈ 6.43cm1

Ques: Variable-Based Proportionality

If AD = 4cm, AE = 8cm, DB = x–4cm, EC = 3x–19cm and DE || BC, find x.

Solution: 4/(x–4) = 8/(3x–19) ⟹ 4(3x–19) = 8(x–4) ⟹ 12x–76 = 8x–32 ⟹ 4x=44 ⟹ x=11

Ques: Using the Converse of Thales’ Theorem

In ΔABC, D and E on AB and AC such that AD/DB = AE/EC.

Given AB = 12cm, AD = 8cm, AE = 12cm, and AC = 18cm. Prove DE || BC.

Solution: BD = AB–AD = 4cm, EC = AC–AE = 6cm

AD/BD = 8/4 = 2, AE/EC = 12/6 = 2

Since ratios equal, DE || BC (by converse of Thales’ Theorem)

Ques: Angle Bisector Theorem Application

In ΔABC, AD bisects ∠A, meeting BC at D. If BD = 2.5cm, AB = 5cm, AC = 4.2cm, find DC.

Solution: By angle bisector theorem:

AB/AC = BD/DC ⟹ 5/4.2 = 2.5/DC ⟹ 5DC = 2.5×4.2 ⟹ DC = 2.1cm

Ques: Area of Similar Triangles

Triangles ABC and DEF are similar. If area(ΔABC)=16cm², area(ΔDEF)=25cm², and BC=2.3cm, find EF.

Solution: (Area(ABC)/Area(DEF)) = (BC/EF)² ⟹ 16/25 = (2.3/EF)²

EF = 2.3 × (5/4) = 2.875cm

Ques: Finding Ratio of Areas

In similar triangles, the ratio of areas is 81:49. What is the ratio of their corresponding heights?

Solution: √(81/49) = 9/7 (ratio of heights)

Ques: Identify Right Triangle from Sides

If triangle sides are 7cm, 24cm, 25cm, determine if it is a right-angled triangle.

Solution: 7² + 24² = 49+576 = 625 = 25²

Hence, right-angled triangle by Pythagoras’ theorem

Ques: Basic Proportionality Use (Variable)

In ΔABC, D and E are midpoints of AB and AC. Find the ratio of areas ΔADE and ΔABC.

Solution: DE || BC and DE = ½ BC

∴ area ratio = (AD/AB)² = (1/2)² = 1/4

Ques: Finding Distance Using Pythagoras’ Theorem

A man goes 15m west and then 8m north. How far is he from the starting point?

Solution: Distance = √(15² + 8²) = √(225+64) = √289 = 17m

Ques: Length from Proportionality (Composite Problem)

A vertical stick 10cm casts a shadow 8cm long. At the same time, a tower casts a shadow 30m long. Find the height of the tower.

Solution: Let h = height of tower.

10/8 = h/3000 ⟹ h = (10/8) × 3000 = 3750cm = 37.5m

Ques: Altitude in Isosceles Triangle

In isosceles ΔABC, AB = AC = 25cm, BC = 14cm. Calculate the altitude from A on BC.

Solution: Let D be midpoint of BC: BD = 7cm

AD² + 7² = 25² ⟹ AD² = 625 – 49 = 576 ⟹ AD = 24cm

Ques: Proving AA Similarity

If ΔABC and ΔDEF are similar by AA criterion, prove all corresponding sides are proportional.

Solution: By AA similarity, ∠A=∠D, ∠B=∠E

So, AB/DE = BC/EF = AC/DF (by definition of similar triangles)

Ques: Finding Side in Similar Triangles

ΔACB ∼ ΔAPQ. If BC=8cm, PQ=4cm, BA=6.5cm, AP=2.8cm, find CA and AQ.

Solution: BA/AQ = BC/PQ ⟹ 6.5/AQ = 8/4 ⟹ AQ=3.25cm

CA/AP = BC/PQ ⟹ CA/2.8 = 8/4 ⟹ CA=5.6cm

Ques: Application: Ladder Distance

A ladder 17m long reaches a window 15m above the ground. Find distance of the foot from the building.

Solution: Let the distance = x.

x² + 15² = 17² ⟹ x² = 289–225=64 ⟹ x=8m

Ques: Finding Median in Similar Triangles

Areas of two similar triangles are 121cm² and 64cm². If the median of the first is 12.1cm, find the median of the other.

Solution: (121/64) = (12.1/m)² ⟹ 11/8 = 12.1/m ⟹ m = 8.8cm1

Ques: Basic Proportionality in Trapezium

Diagonals AC, BD of trapezium ABCD (AB || DC) meet at O. Show OA/OC = OB/OD.

Solution: ΔAOB ~ ΔCOD (Vertically opposite & alternate angles)

Thus, OA/OC = OB/OD

Ques: Ratio of Area with Altitudes

Corresponding altitudes of two similar triangles are 6cm and 9cm. Find the ratio of their areas.

Solution: Area ratio = (6/9)² = 4/9

Ques: Theoretical Proof: ab = cx

In a right triangle with sides a, b, hypotenuse c, and the altitude to hypotenuse x, prove ab = cx.

Solution: By similarity: a/x = c/b ⟹ ab = cx

RD Sharma Solutions Class 10 Chapter 4 – Triangles PDF Overview

The RD Sharma Class 10 Chapter 4 PDF makes learning triangles very simple and easy for students. This chapter, named Triangles, is a big part of geometry and is also included in the CBSE Class 10 syllabus. It helps students learn important triangle concepts like similarity, properties, and theorems that are very useful in solving exam questions.

Main topics in Chapter 4:

• Understanding triangle shapes and their types
• Similar triangles and rules of similarity (AA, SAS, SSS)
• Basic Proportionality Theorem (Thales Theorem)
• Pythagoras Theorem and its converse
• Areas of similar triangles
• Applications of triangle theorems in problems

The RD Sharma Class 10 Solutions Chapter 4 PDF explains every question with easy steps and helpful tricks. Many students find triangle problems hard, but this PDF makes it simple and clears all doubts. Even if you make some mistake while solving, you can check the steps and try again.

Why use RD Sharma Solutions Class 10 Chapter 4 PDF?

• Clear and simple step-by-step solutions for all problems
• Best for homework, revision, and exam preparation
• All formulas explained with examples
• Can download and study offline from the rd sharma class 10 triangles pdf download link
• Helpful hints even for difficult triangle questions
• Follows the latest CBSE Class 10 syllabus properly

Advantages of Solving RD Sharma Class 10 Chapter 4 – Triangles

• Stronger concepts: Helps you fully understand triangles, similarity, and proofs – all are needed in higher classes too
• Better problem-solving: Regular practice improves logical thinking and confidence
• Exam-focused: Questions are as per the CBSE Class 10 exam pattern
• Speed & accuracy: Practice from RD Sharma Class 10 Triangles PDF increases solving speed
• Full explanations: Each step is clearly shown, no need to get confused
• Competitive exam prep: Triangle concepts help in JEE, NEET, and other entrance tests too

If you want to improve in triangles class 10, then download the full RD Sharma Class 10 Triangles PDF with solutions. It is free, easy to read, and very useful for mastering this chapter.