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Heron's Formula

Class 9th Mathematics CBSE Solution

Exercise 12.1
Question 1.

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?ṇ


Answer:


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

(Perimeter of Equilateral Triangle = 3 x Side)

3 a = 180 cm

a = 60 cm

Therefore, each side of triangle = 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


Question 2.

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?


Answer:

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 =

where, s = semi perimeter of triangle and a, b, and c are the sides of the triangle.

=

=

=

= 132 x 10 m2

= 1320 m2

Rent of advertising per year = Rs. 5000 per m2


Rent of one wall for 1 year = 1320 x 5000

Rent of one wall for 3 months =


Rent of one wall for 3 months = Rs. 1650000


Question 3.

There is a slide in a park. One of its side walls has been painted in some color with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in color.



Answer:

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.


Question 4.

Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.


Answer:

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

Perimeter of Triangle = Sum of sides of triangle
Let the third side of triangle be x, so
x + 18 + 10 = 42

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


Question 5.

Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.


Answer:

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


=


=

= √(81000000) cm2

= 9000 cm2
Area of triangle = 9000 cm2


Question 6.

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.


Answer:


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 =

where, s = semi perimeter of the triangle
a, b and c are the sides of the triangle

=

=

= cm2


Hence, the area of triangle is cm2



Exercise 12.2
Question 1.

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


Answer:

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).


Question 2.

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.


Answer:


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 =

where, s = semi perimeter of the triangle and a, b, and c are the sides of the triangle

=

=

= cm2

Take √21 = 4.58

= 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)


Question 3.

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.


Answer:

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.


Question 4.

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?


Answer:

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


Question 5.

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the umbrella?


Answer:

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 =


where, s is the semiperimeter of the triangle and a, b and c are the sides of the triangle.

=


=


= cm2

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

=1000√6 cm2


Question 6.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?


Answer:

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 =

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

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


Question 7.

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm2



Answer:



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

Total Area of 16 tiles = 576 x 2.45 cm2 [√6 = 2.45]
Total Area of 16 tiles = 1411.2 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


Question 8.

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.


Answer:

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 =


where, s = semi perimeter, a, b and c are the sides of the triangle.

=

=

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

For CEB, CM will be the height and 15 cm will be base of the triangle.
And we calculated the area from above. So,

×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