All Courses
Get the App Signup / Login
Sign in Open App
Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  RD Sharma Solutions: Heron's Formula

Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
Page 2


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
Page 3


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
Page 4


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 70° + 70° = 180° 
? ?A + 140°= 180° 
??A = 180°- 140° = 40° 
Question 6. 
The vertical angle of an isosceles triangle is 100°. Find its base angles. 
Solution: 
In ?ABC, AB = AC and ?A = 100° 
But AB = AC (In isosceles triangle) 
 
? ?C = ?B (Angles opposite to equal sides) 
?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? 100° + ?B + ?B = 180° ( ? ?C = ?B) 
? 2 ?B = 180° – 100° = 80° 
? ?C = ?B = 40° 
Hence ?B = 40°, ?C = 40° 
Question 7. 
In the figure, AB = AC and ?ACD = 105°, find ?BAC. 
 
Solution: 
In ?ABC, AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?ACB + ?ACD = 180° (Linear pair) 
? ?ACB + 105°= 180° 
? ?ACB = 180°-105° = 75° 
Page 5


RD Sharma Class 9 Solution Chapter 12 Heron’s Formula 
Question 1. 
In the figure, the sides BA and CA have been produced such that BA = AD and CA = 
AE. Prove the segment DE || BC.
Solution: 
Given : Sides BA and CA of ?ABC are produced such that BA = AD are CA = AE. ED is 
joined. 
To prove : DE || BC 
Proof: In ?ABC and ?DAE AB=AD (Given) 
AC = AE (Given) 
?BAC = ?DAE (Vertically opposite angles) 
? ?ABC ? ?DAE (SAS axiom) 
? ?ABC = ?ADE (c.p.c.t.) 
But there are alternate angles 
? DE || BC 
Question 2. 
In a ?PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP 
respectively. Prove that LN = MN. 
Solution: 
Given : In ?PQR, PQ = QR 
L, M and N are the mid points of the sides PQ, QR and PR respectively 
 
To prove : LM = MN 
Proof : In ?LPN and ?MRH 
PN = RN ( ? M is mid point of PR) 
LP = MR (Half of equal sides) 
?P = ?R (Angles opposite to equal sides) 
? ALPN ? AMRH (SAS axiom) 
? LN = MN (c.p.c.t.) 
Question 3. 
Prove that the medians of an equilateral triangle are equal. 
Solution: 
Given : In ?ABC, AD, BE and CF are the medians of triangle and AB = BC = CA 
 
To prove : AD = BE = CF 
Proof : In ?BCE and ?BCF, 
BC = BC (Common side) 
CE = BF (Half of equal sides) 
?C = ?B (Angles opposite to equal sides) 
? ABCE ? ABCF (SAS axiom) 
? BE = CF (c.p.c.t.) …(i) 
Similarly, we can prove that 
? ?CAD ? ?CAF 
? AD = CF …(ii) 
From (i) and (ii) 
BE = CF = AD 
? AD = BE = CF 
Question 4. 
In a ?ABC, if ?A = 120° and AB = AC. Find ?B and ?C. 
Solution: 
In ?ABC, ?A = 120° and AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
 
? 120° + ?B + ?B = 180° 
? 2 ?B = 180° – 120° = 60° 
? ?B = 60 °2 = 30° 
and ?C = ?B = 30° 
Hence ?B = 30° and ?C = 30° 
Question 5. 
In a ?ABC, if AB = AC and ?B = 70°, find ?A. 
Solution: 
In ?ABC, ?B = 70° 
AB =AC 
? ?B = ?C (Angles opposite to equal sides) 
 
But ?B = 70° 
? ?C = 70° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 70° + 70° = 180° 
? ?A + 140°= 180° 
??A = 180°- 140° = 40° 
Question 6. 
The vertical angle of an isosceles triangle is 100°. Find its base angles. 
Solution: 
In ?ABC, AB = AC and ?A = 100° 
But AB = AC (In isosceles triangle) 
 
? ?C = ?B (Angles opposite to equal sides) 
?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? 100° + ?B + ?B = 180° ( ? ?C = ?B) 
? 2 ?B = 180° – 100° = 80° 
? ?C = ?B = 40° 
Hence ?B = 40°, ?C = 40° 
Question 7. 
In the figure, AB = AC and ?ACD = 105°, find ?BAC. 
 
Solution: 
In ?ABC, AB = AC 
? ?B = ?C (Angles opposite to equal sides) 
But ?ACB + ?ACD = 180° (Linear pair) 
? ?ACB + 105°= 180° 
? ?ACB = 180°-105° = 75° 
? ?ABC = ?ACB = 75° 
But ?A + ?B + ?C = 180° (Sum of angles of a triangle) 
? ?A + 75° + 75° = 180° 
? ?A + 150°= 180° 
? ?A= 180°- 150° = 30° 
? ?BAC = 30° 
Question 8. 
Find the measure of each exterior angle of an equilateral triangle. 
Solution: 
In an equilateral triangle, each interior angle is 60° 
 
But interior angle + exterior angle at each vertex = 180° 
? Each exterior angle = 180° – 60° = 120° 
Question 9. 
If the base of an isosceles triangle is produced on both sides, prove that the exterior 
angles so formed are equal to each other. 
Solution: 
Given : In an isosceles ?ABC, AB = AC 
and base BC is produced both ways 
 
To prove : ?ACD = ?ABE
Read More
44 videos|412 docs|54 tests
Join Course for Free

Top Courses for Class 9

FAQs on Heron's Formula RD Sharma Solutions - Mathematics (Maths) Class 9

1. What is Heron's formula?
Ans. Heron's formula is a mathematical formula used to find the area of a triangle when the lengths of its sides are known. It is named after Hero of Alexandria, a Greek mathematician who derived this formula.
2. How is Heron's formula derived?
Ans. Heron's formula is derived using the concept of semiperimeter of a triangle. The semiperimeter is calculated by adding the lengths of all three sides of the triangle and dividing it by 2. Then, using the semiperimeter, the area of the triangle is calculated using the formula A = √(s(s-a)(s-b)(s-c)), where A is the area, s is the semiperimeter, and a, b, and c are the lengths of the sides of the triangle.
3. Can Heron's formula be used for all types of triangles?
Ans. Yes, Heron's formula can be used to find the area of any type of triangle, whether it is equilateral, scalene, or isosceles. It is a general formula that works for all triangles, as long as the lengths of the sides are known.
4. How is Heron's formula helpful in real-life applications?
Ans. Heron's formula is used in various real-life applications, especially in fields like architecture, engineering, and surveying. It helps in calculating the area of irregularly shaped land or fields, determining the amount of materials required for construction, and in designing structures like bridges and buildings.
5. Are there any limitations or special cases when using Heron's formula?
Ans. Yes, there are a few limitations and special cases to consider when using Heron's formula. One limitation is that the lengths of the sides must form a valid triangle, i.e., the sum of any two sides must be greater than the third side. Additionally, when the lengths of the sides are very large, the calculations involved in the formula may result in rounding errors. Finally, if the triangle is degenerate (having collinear vertices), Heron's formula will not yield a valid area.
Related Exams
About this Document
4.90/5 Rating
Dec 23, 2024 Last updated
Document Description: RD Sharma Solutions: Heron's Formula for Class 9 2024 is part of Mathematics (Maths) Class 9 preparation. The notes and questions for RD Sharma Solutions: Heron's Formula have been prepared according to the Class 9 exam syllabus. Information about RD Sharma Solutions: Heron's Formula covers topics like and RD Sharma Solutions: Heron's Formula Example, for Class 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Heron's Formula.
Introduction of RD Sharma Solutions: Heron's Formula in English is available as part of our Mathematics (Maths) Class 9 for Class 9 & RD Sharma Solutions: Heron's Formula in Hindi for Mathematics (Maths) Class 9 course. Download more important topics related with notes, lectures and mock test series for Class 9 Exam by signing up for free. Class 9: Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9
Description
Full syllabus notes, lecture & questions for Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9 - Class 9 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 9 | Best notes, free PDF download
Information about RD Sharma Solutions: Heron's Formula
In this doc you can find the meaning of RD Sharma Solutions: Heron's Formula defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions: Heron's Formula theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Heron's Formula tests, examples and also practice Class 9 tests
44 videos|412 docs|54 tests
Join Course for Free
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9

,

Important questions

,

Objective type Questions

,

ppt

,

practice quizzes

,

Extra Questions

,

Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9

,

Heron's Formula RD Sharma Solutions | Mathematics (Maths) Class 9

,

mock tests for examination

,

video lectures

,

Free

,

pdf

,

Semester Notes

,

Exam

,

study material

,

Previous Year Questions with Solutions

,

past year papers

,

shortcuts and tricks

,

Viva Questions

,

Summary

,

Sample Paper

,

MCQs

;
Additional Information about RD Sharma Solutions: Heron's Formula for Class 9 Preparation

RD Sharma Solutions: Heron's Formula Free PDF Download

The RD Sharma Solutions: Heron's Formula is an invaluable resource that delves deep into the core of the Class 9 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Heron's Formula now and kickstart your journey towards success in the Class 9 exam.

Importance of RD Sharma Solutions: Heron's Formula

The importance of RD Sharma Solutions: Heron's Formula cannot be overstated, especially for Class 9 aspirants. This document holds the key to success in the Class 9 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions: Heron's Formula Notes

RD Sharma Solutions: Heron's Formula Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions: Heron's Formula. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Heron's Formula Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions: Heron's Formula Class 9 Questions

The "RD Sharma Solutions: Heron's Formula Class 9 Questions" guide is a valuable resource for all aspiring students preparing for the Class 9 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions: Heron's Formula on the App

Students of Class 9 can study RD Sharma Solutions: Heron's Formula alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Heron's Formula, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions: Heron's Formula is prepared as per the latest Class 9 syllabus.
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup