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# NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula

NCERT Textbook Solutions are considered extremely helpful when preparing for your CBSE Class 9 Mathematics exams. TopperLearning study resources infuse profound knowledge, and our Textbook Solutions compiled by our subject experts are no different. Here you will find all the answers to the NCERT textbook questions of Chapter 12 - Heron's Formula.

All our solutions for Chapter 12 - Heron's Formula are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

Exercise/Page

## NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula Page/Excercise 12.1

Solution 1

Side of traffic signal board = a
Perimeter of traffic signal board = 3  a By Heron's formula Perimeter of traffic signal board = 180 cm
Side of traffic signal board Using equation (1), area of traffic of signal board

Solution 2

We may observe that sides of triangle a, b, c are of 122 m, 22 m, and 120 m respectively
Perimeter of triangle = (122 + 22 + 120) m
2s = 264 m
s = 132 m
By Heron's formula
Rent of 1 m2 area per year = Rs.5000
Rent of 1 m2 area per month = Rs     Rent of 1320 m2 area for 3 months
= Rs.(5000  330) = Rs.1650000
So, company had to pay Rs.1650000.

Solution 3

We may observe that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m.
Perimeter of such triangle = (11 + 6 + 15) m
2 s = 32 m
s = 16 m
By Heron's formula     So, the area painted in colour is .

Solution 4

Let third side of triangle be x.
Perimeter of given triangle = 42 cm
18 cm + 10 cm + x = 42
x = 14 cm By Heron's formula

Solution 5

Let the common ratio between the sides of given triangle be x.
So, side of triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
Sides of triangle will be 120 cm, 170 cm, and 250 cm. By Heron's formula     So, area of this triangle will be 9000 cm2.

Solution 6

Let third side of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm
x = 6 cm
By Heron's formula

## NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula Page/Excercise 12.2

Solution 1

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 2

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 DAC
Perimeter = 2s = DA + AC + CD = (5 + 5 + 4) 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 3

For triangle I
This triangle is a isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm            For quadrilateral II
Area = l  b = (6.5  1) cm2 = 6.5 cm2

Perpendicular height of parallelogram                                                       Area = Area of parallelogram + Area of equilateral triangle
= 0.866 + 0.433 = 1.299 cm2 Area of triangle (iv) = Area of triangle in (v) Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5  2
= 19.287 cm2

Solution 4

For triangle
Perimeter of triangle = (26 + 28 + 30) cm = 84 cm
2s = 84 cm
s = 42 cm
By Heron's formula
Area of triangle Area of triangle                        Let height of parallelogram be h
Area of parallelogram = Area of triangle
h  28 cm = 336 cm2
h = 12 cm
So, height of the parallelogram is 12 cm.

Solution 5

Let ABCD be a rhombus shaped field.

For BCD
Semi perimeter, Area of triangle Therefore area of BCD                                  Area of field = 2  Area of BCD
= (2  432) m2 = 864 m2
Area for grazing for 1 cow= = 48 m2
Each cow will be getting 48 m2 area of grass field.

Solution 6

For each triangular piece  Semi perimeter
By Heron's formula Area of triangle

Since, there are 5 triangular pieces made of two different colours cloth.

So, area of each cloth required

Solution 7

We know that
Area of square  (diagonal)2 Area of given kite    Area of 1st shade = Area of 2nd shade

So, area of paper required in each shape = 256 cm2.

For IIIrd triangle Semi perimeter     By Heron's formula   Area of triangle
Area of IIIrd triangle                                     Area of paper required for IIIrd shade  = 17.92 cm2

Solution 8

We may observe that
Semi perimeter of each triangular shaped tile           By Heron's formula   Area of triangle     Area of each tile                                                   = (36  2.45) cm2
=88.2 cm2
Area of 16 tiles = (16  88.2) cm2= 1411.2 cm2

Cost of polishing per cm2 area = 50 p
Cost of polishing 1411.2 cm2 area = Rs. (1411.2  0.50) = Rs.705.60
So, it will cost Rs.705.60 while polishing all the tiles.

Solution 9

Draw a line BE parallel to AD and draw a perpendicular BF on CD.

Now we may observe that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 - ED = 15 m
For BEC

Semi perimeter     By Heron's formula   Area of triangle   Area of BEC                       m2 = 84 m2               Area of  BEC
Area of ABED = BF  DE = 11.2  10
= 112 m2
Area of field = 84 + 112
= 196 m2

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CBSE IX - Mathematics

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