JKBOSE 9th Class Mathematics Solutions Chapter 12 Heron’s Formula

As we know that region covered by plane figures (two dimensional figures) is called Area.

In the previous classes we have learnt about figures of different shapes such as squares, rectangles (playground, walls or floor of the classroom), triangles and quadrilaterals. We have also calculated perimeters and the areas of some of these figures like rectangle and square.

We also know that unit of measurement for length or breadth is taken as metre (m) or centimetre (cm) etc. and unit of measurement for area of any plane figure is taken as square metre (m²) or square centimetre (cm²) etc.

In this chapter, we shall lay stress over determining the area of triangle when all its sides are given.

1. Area of a triangle when its base and height are known is calculated by using the formula Area of triangle = 1/2 base × height Formula-I

2. Area of an equilateral triangle with side a is √3a²/4.

Heron was a mathematician born in about 10 AD (possibly) in Alexandria in Egypt. He has derived the famous formula for the area of triangle when all its three sides are given.

The formula given by Heron about the area of a triangle, is also known as Hero’s formula.

It is stated as

where a, b and c are the sides of the triangle, and s= semi-perimeter i.e. half the perimeter of the triangle = a + b + c

This formula is helpful where it is not possible to find the height of the triangle easily.

Solution.— Traffic signal board is in the shape of equilateral triangle. Let name it  ΔABC.

∴ Side of equilateral triangle ABC = a units (given)

Perimeter of equilateral triangle i.e ; 2s = a + a + a = 3a

Semiperimeter; s = 3a/2

Therefore using Hero’s formula,

Area of triangle = √s(s – a)(s – a) (s – a)

[Area of triangle = √s(s – AB) (s – BC)(s – AC) Here AB = BC = AC = a]

Solution.— Sides of coloured triangular wall are 15 m, 11 m and 6 m.

∴ Perimeter of this triangular wall;

2s = (15 + 11 + 6)m

Solution.— Perimeter of triangle = 42 cm

⇒ 18 cm + 10 cm + Third side = 42 cm

⇒  Third side = (42-18 10) cm

⇒ Third side = 14 cm

Now Semiperimeter of triangle;

s = 42/2

⇒ s = 21 cm

Now we have three sides of triangle as 18 cm, 10 cm and 14 cm. Area of triangle

= √21(21 – 18) (21 – 10) (21 – 14) cm²

[Area of Δ (by using Hero’s formula) = √s(s – a)(s – b)(s – c) ]

Hence area of triangle is 21 √11 cm²

Solution.— Given that the sides of triangle are in the ratio of 12 : 17 : 25

∴ Let sides of triangle in cm are 12x, 17x and 25x

Perimeter of triangle = 540 cm (given)

⇒ 12x + 17x + 25x = 540 cm

⇒ 54x = 540 cm

⇒ x  = 540/54 m

⇒ x = 10 cm

Now sides of triangle are

12x = 12 × 10 = 120 cm

17x = 17 × 10 = 170 cm

and 25x = 25 × 10 = 250 cm

Semiperimeter; s of triangle = Perimeter/2

= 540/ 2 cm

= 270

Using Hero’s formula ;

Area of triangle = √270(270 – 120)(270 – 170)(270 – 250) cm²

= √270 × 150 × 100 × 20 cm²

= √30 × 3 × 3 × 30 × 5 × 10 × 10 × 2 × 2 × 5 cm²

=  √2 × 2 × 2 × 3 × 3 × 5 × 5 × 10 × 10 × 30 × 30 cm²

= 2 × 3 × 5 × 10 × 30 cm²

= 9000 cm²

Hence area of required triangle is 9000 cm².

Solution.— Perimeter of an isosceles triangle = 30 cm

⇒12 cm + 12 cm + Third side = 30 cm

[∴ in an isosceles triangle two sides are of equal length and here each of the equal sides is 12 cm]

⇒ Third side = (30 – 12 – 12) cm

⇒ Third side = 6 cm

Now semiperimeter; s of isosceles triangle = Perimeter / 2

⇒ s = 30/2

⇒ s = 15 cm

Using Hero’s formula ; area of triangle

= √15(15 – 12)(15 – 12) (15 – 6) cm²

= √15 × 3 × 3 × 9 cm²

= √15 × 3 × 3 × 3 × 3 cm²

= 3 × 3√15 cm²

= 9√15 cm²

Hence required area of an isosceles triangle = 9 √15 cm².

Application of Heron’s Formula in Finding Areas of Polygons Area of polygon can be calculated by dividing the polygon into triangles and using Heron’s formula for calculating area of each triangle.

Many a time the fields are in the shape of quadrilaterals. We need to divide the quadrilateral in triangular parts and then use the formula for area of the triangle.

In fig., diagonal BD ; divides the quadrilateral ABCD in two triangles

(i) ΔABD (ii) ΔBCD

∴ Area of quadrilateral ABCD

= Area of ΔABD + Area of ΔBCD

Solution.— In Fig., if we join BD, we observe that diagonal BD divides quadrilateral ABCD in two triangles (i) Right triangle BCD and (ii) ΔABD.

In rt. ΔBCD, rt. angle is at C.

Therefore Base; CD = 5 m

and Altitude, BC = 12 m

Area of rt. ΔBCD = 1/2 × CD × BC

= 1/2 × 5 × 12 m²

= 30 m²

To find the area of ΔABD; we require side BD, therefore to find it as follows :

In rt. ΔBCD, rt. angle at C,

Hence area occupied by the quadrilateral is 65.4 m².

Solution.— In quadrilateral ABCD; diagonal AC divides it in two triangles viz. ΔABC and ΔADC

Hence required area of quadrilateral of ABCD is 15.2 cm².

Solution.— Area of triangular part I using Heron’s formula

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

∴ Perimeter (2s) of triangle = (26 + 28 +30) cm

⇒ 2s = 84 cm

⇒ Semiperimeter s = 84/2

⇒ s = 42 cm

Using Heron’s Formula;

Area of triangle

= √42(42 – 26)(42 – 28)(42 – 30) cm²

= √42 × 16 × 14 × 12 cm²

= √2 × 3 × 7 × 2 × 2 × 2 × 2 × 2 × 7 × 2 × 2 × 3 cm²

= 2 × 2 × 2 × 2 × 3 × 7 cm²

= √2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

= 336 cm²

Given that area of parallelogram = Area of triangle

⇒ Base × Corresponding Height = 336 cm²

⇒ 28 cm × Height = 336 cm²

⇒ Height = 336/25 cm

⇒ Height = 12 cm

Hence corresponding height of parallelogram is 12 cm.

Solution.— Let rhombus shaped field be ABCD.

Therefore its each equal side; AB = BC = CD = DA = 30 m

and let its longer diagonal AC = 48 m. Diagonal AC divides the rhombus in two congruent triangles

(i) ΔABC and (ii) ΔACD. As we know that area of two congruent triangles are equal.

Therefore area (ΔABC) = area (ΔACD)

Now it is enough to find the area of ΔABC

whose sides are AB = BC = 30 m

and AC = 48 m

Perimeter (2s) of ΔABC = (30 + 30 + 48) m

⇒ 2s = 108 m

⇒ Semi-perimeter; s = 108/2 m

⇒ s = 54 m

Using Heron’s Formula; Area of ΔABC

= 54(54 – 30)(54 – 30)(54 – 48) m²

= √54 × 24 × 24 × 6 m²

= √3 × 3 × 6 × 24 × 24 × 6 m²

= √3 × 3 × 6 × 6 × 24 × 24 m²

= 3 × 6 × 24 m²

= 432 m²

Area of rhombus ABCD

= Area of ΔABC + Area of ΔACD

= ar. (ΔABC) + ar. (ΔABC)

                                  [∴ ar. (AABC) = ar. (AACD]

= 2 ar. (ΔABC)

= 2 × 432 m²

= 864 m²

Field available for 18 cows to graze the grass = 864 m²

Field available for 1 Cow to graze the grass = 864/18 m²

= 48 m²

Hence each cow will get 48 m² of the field.

Solution.— Sides of each of 10 triangular pieces two different colours are 20 cm, 50 cm and 50 cm.

∴ Perimeter (2s) of a triangular piece

= (20 + 50 + 50) cm

⇒ 2s = 120 cm

Semiperimeter ; s = 120/2 cm

⇒ s = 60 cm

By using Heron’s Formula,

Area of each triangular piece

= √60(60 – 20)(60 – 50)(60 – 50) cm²

= √60×40×10×10 cm²

= √6 × 10 × 4 × 10 × 10 × 10 cm²

= √2 × 2 × 10 × 10 × 10 × 10 × 6 cm²

= 2 × 10 × 10 √6 cm²

= 200 √6 cm²

According to the question there are 10 triangular pieces of cloth of two different colours.

Therefore there are 5 pieces of red colour and 5 pieces of green colour.

Cloth required for each piece = 200 √6 cm²

∴ Cloth required for 5 red pieces

= 5 × 200 √6 cm²

= 1000 √6 cm²

Similarly cloth required for 5 green pieces

= 5 × 200 √6 cm²

= 1000 √6 cm².

Solution.— Let ABCD be a square whose each sides is a cm and; diagonal