# Chapter 12 : Heron's Formula - Ncert Solutions for Class 9 Maths CBSE

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## Chapter 12 - Heron's Formula Excercise Ex. 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 ## Chapter 12 - Heron's Formula Excercise Ex. 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  This quadrilateral is a rectangle.
Area = l b = (6.5 1) cm2 = 6.5 cm2

This quadrilateral is a trapezium.
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

## CBSE Class 9 Maths Homework Help

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