Heron's Formula Important Questions

Important Questions Class 9 Maths Chapter 12 Heron’s Formula is a very important resource for students preparing for IX Board Examination. Here we have provided Solutions for Important questions of Class 9 Maths Chapter 12 heron’s formula. Students who are preparing for their final exams for the academic sessions can practice these questions to get better marks. All the problems are based on the NCERT book and CBSE syllabus, presented by our Math experts. In this article, we will solve heron's formula extra questions with answers.


Heron's Formula Extra Questions

Here we will solve class 9th heron's formula extra questions with answers

Q1: Find the Area of a Triangle whose two sides are 18 cm and 10 cm respectively and the perimeter is 42cm.

Solution:

Let us consider the third side of the triangle to be “c”.

Now, the three sides of the triangle are a = 18 cm, b = 10 cm, and “c” cm 

It is given that the perimeter of the triangle = 42cm

So, 

18 + 10 + c = 42

c = 42 – (18 + 10) cm = 14 cm

∴ The semi perimeter of triangle (s) = 42/2 = 21 cm

Using Heron’s formula,

Area of the triangle A = \[\sqrt{s(s-a)(s-b)(s-c)}\]

= \[\sqrt{21(21 – 18) (21 – 10) (21 – 14)}\]

=  \[\sqrt{21 × 3 × 11 × 7}\]

= \[\sqrt{21 × 21 × 11}\]

= \[21\sqrt{11}\]cm2

Q2: Sides of a Triangle are in the ratio of 14 : 20: 25 and its perimeter is 590cm. Find its area.

Solution:

The ratio of the sides of the triangle is given as 14: 20: 25

Let us consider the common ratio between the sides of the triangle be “a”

∴ The sides are 14a, 20a and 25a

It is also given that the perimeter of the triangle = 590 cm

12a + 17a + 25a = 590 

=> 59a = 590

So, a = 10

Now, the sides of the triangle are 140 cm, 200 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 590/2 = 295 cm

Using Heron’s formula for Area of the triangle

= \[\sqrt{s(s-a)(s-b)(s-c)}\]

= \[\sqrt{295(295-140)(295-200)(295-250)}\]

= \[\sqrt{295 × 155 × 95 × 45}\]

= \[\sqrt{195,474,375}\]

= 13981.21cm2

Q3: A field is in the Shape of a Trapezium whose parallel sides are 22 m and 10 m. The non-parallel sides are given as 13 m and 14 m. Find the area of the field.

Solution:

Draw a line segment BE line AD. Then, draw a perpendicular on the line segment CD from point B

[Image will be uploaded soon]

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = DE = 10 m

AD = BE = 13 m

EC = DC– ED 

= 22 – 10 = 12 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+ 14 + 12)/2 

= 39/2 m

=19.5m

By using Heron’s formula,

Area of ΔBEC =

= \[\sqrt{s(s-a)(s-b)(s-c)}\]

= \[\sqrt{19.5(19.5-13)(19.5-14)(19.5-15)}\]

= \[\sqrt{19.5 × 6.5 × 5.5 × 4.5}\]

= \[\sqrt{3137.06 }\]

= 56m2

We also know that the area of ΔBEC = (½) × EC × BF

56 cm2 = (½) × 12 × BF

BF = 56 x 2 /12cm 

     = 9.3 cm

So, the total area of ABED will be BF × DE i.e. 9.3 × 10 = 93 m2

∴ Area of the field = 93 + 56 = 149 m2

Q4: Find the Area of the Triangular field of sides 55 m, 60 m, and 65 m. Find the cost of laying the grass in the triangular field at the rate of Rs 8 per m2.

Solution:

Given that 

Sides of the triangular field are 50 m, 60 m and 65 m.

Cost of laying grass in a triangular field = Rs 8 per m2

Let a = 55, b = 60, c = 65

Semi- Perimeter s = (a + b + c)/2

⇒ s = (55 + 60 + 65)/2

= 180/2

= 90.

By using Heron’s formula,

Area of ΔBEC =

= \[\sqrt{s(s-a)(s-b)(s-c)}\]

= \[\sqrt{90(90-55)(90-60)(90-65)}\]

= \[\sqrt{90×35×30×25}\]

=\[\sqrt{2362500}\]

= 1537m2

Cost of laying grass = Area of triangle × Cost of laying grass per m2

= 1537×8

= Rs.12296

Q5: The Perimeter of an Isosceles triangle is 42 cm. The ratio of the equal side to its base is 3: 4. Find the area of the triangle.

Solution:

Given that,

The perimeter of the isosceles triangle = 42 cm

It is also given that,

Ratio of equal side to base = 3 : 4

Let the equal side = 3x

So, base = 4x

Perimeter of the triangle = 42

⇒ 3x + 3x + 4x = 42

⇒ 10x = 42

⇒ x = 4.2

Equal side = 3x = 3×4.2 = 12.6

Base = 4x = 4×4.2 = 16.8

The sides of the triangle = 12.6cm, 12.6cm and 16.8cm.

Let a = 12.6, b = 12.6, c = 16.8

s = (a + b + c)/2

⇒ s = (12.6 + 12.6 + 16.8)/2

= 42/2

= 21.

By using Heron’s formula,

Area of ΔBEC =

= \[\sqrt{s(s-a)(s-b)(s-c)}\]

= \[\sqrt{21(21-12.6)(21 - 12.6)(21 - 16.8)}\]

=\[\sqrt{21 × 8.4 × 8.4 × 4.2}\]

= \[\sqrt{6223.39}\]

= 78.88cm2


Extra Questions of Heron's Formula 9th class

  1. A parallelogram has sides 30 m and 40 m and one of its diagonals is 20m long. Find the area of the parallelogram.

  2. The sides of a quadrilateral ABCD are 4cm, 8cm, 12cm, and 16cm respectively. Find its area.

  3. A rhombus-shaped sheet with perimeter 50cm and one diagonal 10 cm, is painted on both sides at the rate of Rs. 7 per m2. Find the cost of painting.

  4. From a point in the interior of an equilateral triangle, perpendicular are drawn on the three sides. The lengths of the perpendicular are 12cm, 14cm, and 8cm. Find the area of the triangle.