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RD Sharma Solution for Class 9 Mathematics Chapter 17 - Heron's Formula

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 17 - Heron's Formula.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

RD Sharma Solution for Class 9 Mathematics Chapter 17 - Heron's Formula Page/Excercise 17.1

Solution 1

Solution 2

Solution 3

Solution 4



Solution 5

The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
  Perimeter of this triangle = 540 m
             25x + 17x + 12x = 540 m
                         54x = 540 m
                            x = 10 m
  Sides of triangle will be 250 m, 170 m, and 120 m.   Semi-perimeter (s) =    By Heron's formula:     So, area of the triangle is 9000 m2.

Solution 6

Solution 7



Solution 8

Solution 9

Solution 10

Solution 11

RD Sharma Solution for Class 9 Mathematics Chapter 17 - Heron's Formula Page/Excercise 17.2

Solution 1

For ABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, ABC is a right angle triangle, right angled at point B.
Area of ABC For ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
            s = 7 cm
By Heron's formula
Area of triangle Area of ABCD = Area of ABC + Area of ACD
    = (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

Solution 2



Solution 3



Solution 4

Let us join BD.
In BCD applying Pythagoras theorem
BD2 = BC2 + CD2
       = (12)2 + (5)2
       = 144 + 25
BD2 = 169
  BD = 13 m Area of BCD                   For ABD                     By Heron's formula Area of triangle                                 Area of park = Area of ABD + Area of BCD
                         = 35.496 + 30 m2                          = 65.496 m2                          = 65. 5 m2 (approximately)

Solution 5



Solution 6



Solution 7

Solution 8



Solution 09



Solution 10

Solution 11



Solution 12



Solution 13

Solution 14

RD Sharma Solution for Class 9 Mathematics Chapter 17 - Heron's Formula Page/Excercise 17.24

Solution 1

begin mathsize 11px style Let space straight a equals 16 space cm comma space straight b equals 30 space cm comma space straight c equals space 34 space cm
Semi minus perimeter space of space straight a space triangle equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction equals fraction numerator 16 plus 30 plus 34 over denominator 2 end fraction equals 40
Now comma space straight s minus straight a space equals space 24 space cm comma space straight s minus straight b equals 10 space cm space and space straight s minus straight c equals 6 space cm
By space Heron apostrophe straight s space formula comma space
Area space of space straight a space triangle equals square root of straight s space open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 40 cross times 24 cross times 10 cross times 6 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 4 cross times 10 cross times 4 cross times 6 cross times 10 cross times 6 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 4 squared cross times 10 squared cross times 6 squared end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 cross times 10 cross times 6
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 240 space cm squared
Note colon space Correct space option space not space given end style

Solution 2

begin mathsize 12px style Let space ABC space be space an space isosceles space increment comma space right space angled space at space straight A.
rightwards double arrow AB space equals space AC space and space angle CAB equals 90 degree
rightwards double arrow open parentheses AB close parentheses squared plus open parentheses AC close parentheses squared equals open parentheses BC close parentheses squared space equals open parentheses 30 close parentheses squared
rightwards double arrow 2 open parentheses AB close parentheses squared equals open parentheses 30 close parentheses squared
rightwards double arrow left parenthesis AB right parenthesis squared space equals space 900 over 2 equals space AC
Area space of space increment ABC equals 1 half cross times AB cross times AC space
space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half cross times open parentheses AB close parentheses squared space space space space space space space space open parentheses because space AB space equals AC close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 900 over 2
space space space space space space space space space space space space space space space space space space space space space space space space space space equals 225 space cm squared
Hence comma space correct space option space is space open parentheses straight a close parentheses. end style

Solution 3

begin mathsize 12px style Let space straight a equals space 7 space cm comma space straight b equals 9 space cm comma space space space straight c equals 14 space cm
Semi minus perimeter equals straight s equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction equals fraction numerator 7 plus 9 plus 14 over denominator 2 end fraction equals 15 space cm
straight s minus straight a equals 15 minus 7 equals 8 space cm comma space straight s minus straight b space equals space 15 minus 9 equals 6 space cm space and space straight s space minus straight c space equals space 15 minus 14 equals 1 space cm
Area space of space straight a space triangle space equals space square root of straight s open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 15 cross times 8 cross times 6 cross times 1 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 5 cross times 3 cross times 4 cross times 2 cross times 3 cross times 2 end root
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 12 square root of 5 space cm squared
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 4

begin mathsize 12px style straight a equals 325 space straight m comma space space straight b equals 300 space straight m comma space space straight c equals 125 space straight m
straight s equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction equals fraction numerator 325 plus 300 plus 125 over denominator 2 end fraction equals 375 space straight m
straight s minus straight a equals 50 space straight m comma space space space straight s minus straight b equals 75 space straight m comma space space straight s minus straight c equals 250 space straight m
Area space equals space square root of straight s space open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
space space space space space space space space space equals space square root of 375 cross times 50 cross times 75 cross times 250 end root
space space space space space space space space space equals space square root of 15 cross times 25 cross times 25 cross times 2 cross times 3 cross times 25 cross times 25 cross times 10 end root
space space space space space space space space space equals square root of bottom enclose 25 cross times 25 end enclose cross times bottom enclose 25 cross times 25 end enclose cross times bottom enclose 30 cross times 30 end enclose end root
space space space space space space space space space equals 25 cross times 25 cross times 30 space
space space space space space space space space equals 18750 space straight m squared
Hence comma space correct space option space is space open parentheses straight a close parentheses. end style

Solution 5

begin mathsize 12px style The space smallest space altitude space is space perpendicular drawn space to space the space largest space side space of space straight a space triangle space from space opposite space point.
straight i. straight e. space BD
Area space of space triangle equals 1 half cross times Ac cross times BD space equals 1 half cross times 112 space straight x space BD space equals 56 cross times BD
straight s equals fraction numerator 50 plus 78 plus 112 over denominator 2 end fraction equals 120 space cm
straight s minus AB equals 70 space cm comma space space straight s minus BC space equals space 42 space cm comma space space straight s minus AC equals 8 space cm
Area equals square root of straight s open parentheses straight s minus AB close parentheses minus open parentheses straight s minus BC close parentheses open parentheses straight s minus AC close parentheses end root equals square root of 120 cross times 70 cross times 42 cross times 8 end root equals 1680 space cm squared
Now comma space space 56 cross times BD equals 1680 space cm squared
rightwards double arrow BD equals 1680 over 56 equals 30 space cm
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

RD Sharma Solution for Class 9 Mathematics Chapter 17 - Heron's Formula Page/Excercise 17.25

Solution 6

begin mathsize 12px style Area space of space triangle equals 1 half space Base space cross times Height
The space smallest space side space is space 11 space straight m
rightwards double arrow Area space equals space 1 half cross times 11 cross times Height space space space space space space.... open parentheses 1 close parentheses
Area space by space Heron apostrophe straight s space Formula space equals square root of straight s open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
straight s space fraction numerator 11 plus 60 plus 61 over denominator 2 end fraction equals 66 space straight m
rightwards double arrow Area space equals space square root of 66 cross times 55 cross times 6 cross times 5 end root space equals 330 space straight m squared
From space eq space open parentheses 1 close parentheses space
330 space equals 1 half cross times 11 cross times height space
rightwards double arrow Height equals fraction numerator 2 cross times 330 over denominator 11 end fraction equals 60 space straight m
Hence comma space correct space option space is space left parenthesis straight d right parenthesis. end style

Solution 7

begin mathsize 12px style straight s equals fraction numerator 11 plus 15 plus 16 over denominator 2 end fraction equals 21 space cm
Area space of space triangle equals square root of straight s open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root space equals space square root of 21 cross times 10 cross times 6 cross times 5 end root equals 30 square root of 7 space cm squared
Also space if space we space choose space largest space side space and space its space Altitude comma space the space area space would space be
straight A equals 1 half cross times largest space side space cross times straight h
rightwards double arrow 1 half cross times 16 cross times straight h space equals space 30 square root of 7
rightwards double arrow straight h equals fraction numerator 30 square root of 7 over denominator 8 end fraction equals 15 over 4 square root of 7 space cm
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 10

begin mathsize 12px style Let space each space of space the space two space equal space sides space of space an space isosceles space right space triangle space be space straight a space cm.
Then comma space third space side equals straight a square root of 2 space cm
space Area space of triangle equals 1 half cross times straight a cross times straight a space
rightwards double arrow 8 equals straight a squared over 2
rightwards double arrow straight a squared equals 16
rightwards double arrow straight a equals space 4 space cm
rightwards double arrow Perimeter space rightwards double arrow straight a plus straight a plus straight a square root of 2 space equals space 4 plus 4 plus 4 square root of 2 equals 8 space plus space 4 square root of 2 space cm
Hence comma space correct space option space is space left parenthesis straight b right parenthesis. end style

Solution 11

begin mathsize 12px style Let space the space sides space of triangle ABC space be space straight n comma space straight n plus 1 comma space straight n space plus space 2. space
rightwards double arrow Perimeter space equals space straight n plus straight n plus 1 plus straight n plus 2
rightwards double arrow open parentheses 9 plus 9 plus 9 close parentheses space equals space 3 straight n space plus space 3
rightwards double arrow 27 space equals space 3 straight n space plus space 3
rightwards double arrow 3 straight n space equals space 24
rightwards double arrow straight n space equals space 8 space cm
Thus comma space the space shortest space side space is space 8 space cm.
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Solution 12

begin mathsize 12px style AD over DC equals 3 over 2
Let space AD equals 3 straight x space and space DC equals 2 straight x
Area space of space triangle ABC space equals 1 half cross times AC cross times BE space space space space space space space space space space space open parentheses BE equals straight h close parentheses
rightwards double arrow 40 equals 1 half cross times 5 straight x cross times straight h
rightwards double arrow 80 equals 5 xh
rightwards double arrow xh space equals space 16 space cm squared space space space space.... left parenthesis 1 right parenthesis
Now space Area space of space triangle ABD space equals 1 half cross times 3 straight x cross times straight h space equals fraction numerator 3 xh over denominator 2 end fraction equals 3 over 2 cross times 16 equals 24 space cm squared
Area space of space triangle BDC space equals space Area space of space triangle ABC minus space Area space of space triangle ABD equals 40 minus 24 equals 16 space cm squared
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 14

begin mathsize 12px style straight s equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction comma space space space straight A equals square root of straight s open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
Now comma space if space straight a apostrophe equals 2 straight a comma space space straight b apostrophe equals 2 straight b space space and space space straight c apostrophe space equals space 2 straight c
then comma space straight s apostrophe space equals fraction numerator straight a apostrophe plus straight b apostrophe plus straight c apostrophe over denominator 2 end fraction space equals fraction numerator 2 straight a space plus 2 straight b plus 2 straight c over denominator 2 end fraction equals 2 straight s
straight A apostrophe space equals square root of straight s apostrophe open parentheses straight s apostrophe minus straight a apostrophe close parentheses open parentheses straight s apostrophe minus straight b apostrophe close parentheses open parentheses straight s apostrophe minus straight c apostrophe close parentheses end root
space space space space space equals square root of 2 straight s open parentheses 2 straight s minus 2 straight a close parentheses open parentheses 2 straight s minus 2 straight b close parentheses open parentheses 2 straight s space minus space 2 straight c close parentheses end root
space space space space equals space 4 square root of straight s open parentheses straight s minus straight a close parentheses open parentheses straight s minus straight b close parentheses open parentheses straight s minus straight c close parentheses end root
rightwards double arrow straight A apostrophe equals 4 straight A
rightwards double arrow Increase space in space Area space equals space fraction numerator 4 straight A space minus straight A over denominator straight A end fraction cross times 100 space percent sign equals 300 percent sign
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 8

begin mathsize 12px style AB space equals square root of open parentheses 13 close parentheses squared minus open parentheses 5 close parentheses squared end root equals 12 space cm
Area space equals 1 half cross times BC cross times AB space equals space 1 half cross times 5 cross times 12 space equals space 30 space cm squared
Hence comma space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 9

begin mathsize 12px style If space side space of space an space equilateral space triangle space is space apostrophe straight a apostrophe comma space then space its space
Area space equals fraction numerator square root of 3 over denominator 4 end fraction straight a squared
Now space fraction numerator up diagonal strike square root of 3 end strike over denominator 4 end fraction straight a squared equals 4 square root of 3
rightwards double arrow straight a squared space equals space 4 squared
rightwards double arrow straight a space equals space 4 space cm
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Solution 13

begin mathsize 12px style Let space the space side space of space equilateral triangle ABC space be space straight a space cm
The space median space of space equilateral space triangle space is space its space altitude space drawn space from space straight A space to space BC.
left parenthesis straight i. straight e. space the space height space of space triangle over space Base space BC right parenthesis
rightwards double arrow AD space equals space straight a space sin 60 degree
rightwards double arrow straight x equals space fraction numerator straight a square root of 3 over denominator 2 end fraction space space space space left square bracket AD equals space straight x space open parentheses given close parentheses
rightwards double arrow straight a equals fraction numerator 2 straight x over denominator square root of 3 end fraction
Area space of space equilateral space triangle space of space side space straight a space
equals fraction numerator square root of 3 straight a squared over denominator 4 end fraction
equals space fraction numerator square root of 3 over denominator 4 end fraction open parentheses fraction numerator 2 straight x over denominator square root of 3 end fraction close parentheses cubed
equals fraction numerator straight x squared over denominator square root of 3 end fraction space
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Solution 15

begin mathsize 12px style If space side space of space straight a space square space is space straight a space cm
then comma space its space diagonal space equals space square root of 2 straight a space cm
But space diagonal equals 12 square root of 2 space cm
rightwards double arrow square root of 2 straight a equals 12 square root of 2
rightwards double arrow straight a equals 12 space cm
rightwards double arrow Perimeter space of space straight a space square equals 4 straight a equals 4 cross times 12 equals 48 space cm
Now comma space perimeter space of space an space equilateral space triangle space with space side space straight x space equals 3 straight x space cm
But comma space perimeter space of space equilateral space triangle equals Perimeter space of space square
rightwards double arrow 3 straight x space equals space 48
rightwards double arrow straight x space equals space 16 space cm
Now comma space Area space of space equilateral space triangle space equals space fraction numerator square root of 3 straight x squared over denominator 4 end fraction equals fraction numerator square root of 3 over denominator 4 end fraction cross times 16 cross times 16 space equals 64 square root of 3 cm squared
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 17 - Heron's Formula for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 17 - Heron's Formula.

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CBSE IX - Mathematics

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