Chapter 17 : Heron's Formula - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 17 - Heron's Formula Excercise Ex. 17.1

Question 1

Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4



Question 5
The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.
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.
Question 6

The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.

Solution 6

Question 7

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Solution 7



Question 8

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

Solution 8

Question 9

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.

Solution 9

Question 10

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.

Solution 10

Question 11

Find the area of the shaded region in fig.12.12

Solution 11

Chapter 17 - Heron's Formula Excercise Ex. 17.2

Question 1
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
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)

Question 2

Solution 2



Question 3

Solution 3



Question 4
A park, in the shape of a quadrilateral ABCD, has  = 90o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
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)

Question 5

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.

Solution 5



Question 6

A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m2. Find the cost of painting.

Solution 6



Question 7

Solution 7

Question 8

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.

Solution 8



Question 9

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Solution 9



Question 10

Find the area of the blades of the magnetic compass shown in fig.

(Take √11 = 3.32)

 

 

Solution 10

Question 11

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.

Solution 11



Question 12

Solution 12



Question 13

Solution 13

Question 14

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Solution 14

Chapter 17 - Heron's Formula Excercise 17.24

Question 1

The side of triangle are 16 cm, 30 cm, 34 cm, its area is

 

begin mathsize 11px style open parentheses a close parentheses 225 space c m squared end style

begin mathsize 11px style open parentheses b close parentheses 225 square root of 3 c m squared end style

begin mathsize 12px style open parentheses c close parentheses space 225 square root of 2 c m squared end style

begin mathsize 12px style open parentheses d close parentheses 450 space c m squared end style

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

Question 2

The base of an isosceles right triangle is 30 cm. Its area is

begin mathsize 12px style open parentheses a close parentheses 225 space c m squared
open parentheses b close parentheses 225 square root of 3 space c m squared
open parentheses c close parentheses 225 square root of 2 space c m squared
open parentheses d close parentheses 450 space c m squared space
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

Question 3

The sides of a triangle are 7 cm, 9 cm, and 14 cm. Its area is

begin mathsize 12px style open parentheses a close parentheses 12 square root of 5 c m squared
open parentheses b close parentheses 12 square root of 3 c m squared
open parentheses c close parentheses 24 square root of 5 c m squared
open parentheses d close parentheses 63 space c m squared 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

Question 4

The sides of a triangle are 325 m, 300 m and 125 m. Its area is

(a) 18750 m2

(b) 37500 m2

(c) 97500 m2

(d) 48750 m2

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

Question 5

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is 

(a) 20 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

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

Chapter 17 - Heron's Formula Excercise 17.25

Question 1

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

(a) 11 m

(b) 66 m

(c) 50 m

(d) 60 m

Solution 1

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

Question 2

begin mathsize 12px style The space sides space of space triangle space are space 11 space cm comma space 15 space cm space and space 16 space cm. space The space altitude space to space the space largest space side space is end style

begin mathsize 12px style open parentheses straight a close parentheses space 30 square root of 7 cm
open parentheses straight b close parentheses space fraction numerator 15 square root of 7 over denominator 2 end fraction space cm
open parentheses straight c close parentheses space fraction numerator 15 square root of 7 over denominator 4 end fraction space cm
open parentheses straight d close parentheses space 30 space cm
end style

Solution 2

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

Question 3

begin mathsize 12px style If space the space area space of space an space isosceles space right space triangle space is space 8 space cm squared comma space what space is space the space perimeter space of space the space triangle ? end style

begin mathsize 12px style open parentheses straight a close parentheses space space 8 plus square root of 2 cm
open parentheses straight b close parentheses space space 8 plus 4 square root of 2 cm
open parentheses straight c close parentheses space space 4 plus 8 square root of 2 space cm
open parentheses straight d close parentheses space 12 square root of 2 cm end style

Solution 3

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

Question 4

begin mathsize 12px style The space lengths space of space the space sides space of space triangle ABC space are space space consecutive space integers.
If space triangle ABC space has space the space same space perimeter space as space an space equilateral space triangle space with space straight a space side space of space length space 9 space cm comma space
what space is space the space length space of space the space shortest space side space of space triangle ABC ? end style

(a) 4

(b) 6

(c) 8

(d) 10

Solution 4

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

Question 5

begin mathsize 12px style In space figure comma space the space ratio space of space AD space to space DC space is space 3 space to space 2. space If space the space area space of space triangle ABC space is space 40 space cm squared comma space
what space is space the space area space of space triangle ABC ?
open parentheses straight a space close parentheses 16 space cm squared
open parentheses straight b close parentheses space 24 space cm squared
open parentheses straight c close parentheses space 30 space cm squared
open parentheses straight d close parentheses space 36 space cm squared end style

Solution 5

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

Question 6

begin mathsize 12px style If space every space side space of space straight a space triangle space is space doubled comma space then space increase space in space the space area space of space the space triangle space is
open parentheses straight a close parentheses space 100 square root of 2 percent sign
open parentheses straight b close parentheses space 200 percent sign
open parentheses straight c close parentheses space 300 percent sign
open parentheses straight d close parentheses space 400 percent sign end style

Solution 6

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

Question 7

The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. it is area is:

(a) 25 cm2

(b) 28 cm2

(c) 30 cm2

(d) 40 cm2

Solution 7

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

Question 8

begin mathsize 12px style The space length space of space each space side space of space an space equilateral space triangle space of space area space 4 square root of 3 space end root space cm squared comma space is
open parentheses straight a close parentheses space space 4 space cm
open parentheses straight b space close parentheses space fraction numerator 4 over denominator square root of 3 end fraction space cm
open parentheses straight c close parentheses space fraction numerator square root of 3 over denominator 4 end fraction cm
open parentheses straight d close parentheses space space 3 space cm end style

Solution 8

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

Question 9

begin mathsize 12px style If space the space length space of space the space median space of space an space equilateral space triangle space is space straight x space cm comma space then space itis space area space is
open parentheses straight a close parentheses space straight x squared
open parentheses straight b close parentheses space fraction numerator square root of 3 over denominator 2 end fraction straight x squared
open parentheses straight c close parentheses fraction numerator straight x squared over denominator square root of 3 end fraction
open parentheses straight d close parentheses space straight x squared over 2 end style

Solution 9

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

Question 10

begin mathsize 12px style straight A space square space and space an space equilateral space triangle space have space equal space perimeters. space If space diagonal space of space the space square
is space 12 square root of 2 cm comma space then space the space area space of space the space triangle space is
open parentheses straight a close parentheses space 24 square root of 2 cm squared
open parentheses straight b close parentheses space space 24 square root of 3 cm squared
open parentheses straight c close parentheses space space 48 square root of 3 cm squared
open parentheses straight d close parentheses space space 64 square root of 3 cm squared end style

Solution 10

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

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