Heron's Formula

Given:
Perimeter of Signal board = 180 cm

Length of the side of the equilateral triangle = a

The perimeter of the signal board = perimeter of the triangle

3 a = 180 cm

a = 60 cm

Semi perimeter of the signal board (s) = Perimeter of board /2 = 180/2 = 90 cm

Using Heron’s formula,

Area of the signal board =
where, s = semi perimeter of signal board = 90 cm
a = b = c = 60 cm
Putting the value we get,

Method 2:
we know that,

where a = side of the equilateral triangle.
putting the value of a = 60 cm we get,

Area = 900√3 cm²

As shown in the figure the sides of the triangular side walls of flyover are 122 m, 22 m, and 120 m.

Perimeter of the triangle = 122 + 22 + 120 = 264 m

Semi perimeter of triangle(s) =

= 132 m

Now,

Using Heron’s formula

Area of the advertisement =

=

=

=

= 132 x 10 m²

= 1320 m²

Rent of advertising per year = Rs. 5000 per m²

Rent of one wall for 3 months =

Rent of one wall for 3 months = Rs. 1650000

Given: The sides of the triangle are 15 m, 11 m and 6 m.
To find: Area painted in color
Explanation:

Perimeter of the triangular walls = sum of 3 sides

Perimeter = 15 + 11 + 6 = 32 m

Semi perimeter of triangular walls(s) =

= 16 m

Using Heron’s formula,

Area of the advertisement = Area of triangle =

where, s = semi perimeter of the triangle

a, b and c are the sides of the triangle

Therefore for the given triangle,

Area =

=

=

= √800 m²

= √(20 × 20 × 2)

(Taking 20 outside the square root, as a pair is present in square root, we get,)

Thus the area of painted color is 20√2 m².

To find : Area of the triangle
Given:
Two sides of the triangle are 18 cm and 10 cm (Given)

Perimeter of the triangle = 42 cm

Third side of triangle, x = 42 – (18 + 10)

= 14 cm

Semi perimeter of triangle(s) =

= 21cm

Using Heron’s formula,

Area of the triangle =

=

=

=

= cm²

Let the common ratio of the sides of the triangle be x and sides are 12x, 17x and 25x

The perimeter of the triangle = 540 cm
Sum of sides of the triangle = 540 cm

12x + 17x + 25x = 540 cm
54 x = 540

x = 10

Sides of triangle:

12 x = 12×10 = 120 cm

17 x = 17×10 = 170 cm

25 x = 25×10 = 250 cm

Semi perimeter of triangle(s) =

= 270cm

Now,

Using Heron’s formula,

Area of the triangle =
where, s = semiperimeter of the triangle and a, b and c are the sides of the triangle

=

=

= 9000 cm²
Area of triangle = 9000 cm²

Length of equal sides of the triangle = 12cm

Perimeter of the triangle = 30 cm

Third side of triangle = 30 – (12 + 12)

= 6cm

Semi perimeter of triangle(s) =

= 15cm

Now,

Using Heron’s formula,

Area of the triangle =

=

=

= cm²

Hence, the area of triangle is cm²

Given: The figure is given below:

The dimensions of the park as follows:

Angle C = 90°,

AB = 9m,

BC = 12m,

CD = 5m

And, AD = 8m

Now, BD is joined.
Thus quadrilateral ABCD can now be divided into two triangles ABD and BCD
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In ΔBCD by Pythagoras theorem,

BD² = BC² + CD²

BD² = 12² + 5²

BD² = 169

BD = 13m

As triangle BCD is right angled triangle,

Now, Semi perimeter of triangle (ABD) = (8 + 9 + 13)/2

= 30/2 = 15m

Using Heron’s formula,
ar(ABD) =
where, s = semiperimeter of the triangle and a, b, and c are the sides of the triangle.

=

=

=

= 6√35 m²

= 6 × 5.91
= 35.46 m²

Area of quadrilateral ABCD = 35.46 + 30

Area of quadrilateral ABCD = 65.46 m²

Thus, the park acquires an area of 65.5 m²(approx).

Given: AB = 3 cm, BC = 4 cm, CD = 4 cm, DA =5 cm and AC = 5 cm

To find: Area of quadrilateral ABCD.
Now we can divide the quadrilateral into two separate triangles and then calculate their area.
Now, in Δ ABC,

Using Pythagoras theorem,

AC² = BC² + AB²

5² = 3² + 4²

25= 16 + 9

25 = 25

Thus,

Triangle is right angled triangle and since AB and BC are smaller sides so they work as legs.

And we know that,

Now,

Perimeter of Δ ACD = 5 + 5 + 4

Semi perimeter of Δ ACD =

=

= 7 cm

Using Heron’s formula,

Area of =

=

=

= cm²

= 9.16 cm² (approx.)

Area of quadrilateral ABCD = Area of Δ ABC+ Area of Δ ACD

= 6 + 9.16

= 15.16 cm²

By rounding off,

= 15.2 cm² ( approx)

Given,

Area of the parallelogram and triangle are equal

Sides of triangle are 26 cm, 28 cm and 30 cm.

Semi perimeter of triangle ,s =

s =

= 42 cm

Using Heron’s formula,

Area of triangle =

=
=

= 336 cm²

Let the height of parallelogram be h,

As parallelogram and triangle having same base and same height have equal areas.

∴ Area of parallelogram = Area of triangle

28×h = 336

h =

h = 12 cm

Hence, the height of the parallelogram is 12 cm.

Diagonals divide the rhombus ABCD into two congruent triangles of equal area.

Semi perimeter of = 54 m

Using Heron’s formula,

Area of =

=

= 432 m²

Area of field = 2 x area of

= 864 m²

Thus,

Area of grass field which each cow will be getting = = 48 m²

To Find: Total Amount of cloth required
Given: Umbrella is made up of 10 triangular pieces of sides 50, 50 and 20 cm
Perimeter of each triangular piece of cloth = 50 + 50 + 20

Semi perimeter of each triangular piece of cloth =

=

= 60 cm

Using Heron’s formula,

Area of one triangular piece =

=

=

= cm²

Area of piece of cloth of one colour = 5 x 200 √6

=1000√6 cm²

For I and II section:
For calculating the area of the square, Let the length of side = x cm

Each Interior Angle of square = 90º
Pythagoras Theorem: Square of Hypotenuse equals to the sum of squares of other two sides
Now by Pythagoras theorem,
x² + x² = 32 x 32
2 x² = 32 x 32
x² = 16 x 32 = 512 cm²
Area of Square = (side)²
And this will be the area of Square.
Area of square = 512 cm²
Area of section I = Area of section II = 1/2 Area of square
Now, we need half of this area,
So half of the area of square = 256 cm²

For the III section

Length of the sides of triangle = 6cm, 6cm and 8cm
Perimeter of triangle = 6 + 6 + 8 = 20 cm
Semi-Perimeter of Triangle, s = 10 cm
Using Heron’s formula,
Area of the III triangular piece =

Area of Triangle =
=
Area of Triangle = 8√5 cm²
= 17.92 cm²

So, we need to find the area of all 16 tiles or we can say that Area of 16 triangles.
Perimeter of each triangular shaped tiles = 28 + 9 + 35

Semi perimeter of each triangular shaped tiles, s =

s =

s = 36cm

Now,

Using Heron’s formula,

Area of each triangular shape=
where s = semi perimeter of triangle and
a, b, c are the sides of the triangle

Area of 1 triangular shape =

=
Area of 1 tile = 36√6 cm²

Total area of 16 tiles = 16 x 36√6 cm² = 576√6 cm²

Rate of polishing tiles = 50 p per cm²

Hence,

The total cost of polishing tiles = 50×1411.2
= 70560 paise
As 1 paisa = 1/100 Rs
= RS. 705.60

Let ABCD be the given trapezium whose

parallel sides are:

AB = 25m and CD = 10m

And,

Non-parallel sides are AD = 13m and BC = 14m

Now,

CM perpendicular to AB and CE ‖ AD

In

BC = 14m, CE = AD = 13m and

BE = AB – AE

= 25- 10

= 15m

Semi perimeter of =

=

= 21cm

Now,

Using Heron’s formula,

Area of =

=

=

= 84m²
Now from Δ CEB, Area of triangle = 1/2 x base x height

×15 ×CM = 84

CM = m

Area of parallelogram AECD = Base ×Altitude

= AE ×CM

=
= 112 m²

Area of trapezium ABCD = Area of AECD + Area of triangle BCE

= (112 + 84) m²

= 196 m²