Class 9 NCERT Solutions- Chapter 12 Heron's Formula - Exercise 12.2

Chapter 12 of the Class 9 NCERT Mathematics textbook introduces Heron's Formula a vital concept in the geometry used to calculate the area of a triangle when the lengths of all three sides are known. Exercise 12.2 focuses on applying Heron's Formula to solve problems involving the different triangles. This exercise helps in understanding the practical application of the formula and enhances problem-solving skills in geometry.

What is Heron's Formula?

Heron's Formula provides a method to calculate the area of the triangle when the lengths of its sides are known. The formula is named after the ancient Greek mathematician Hero of Alexandria. Given a triangle with sides of lengths a, b and c the area A can be calculated using the following steps:

Compute the semi-perimeter

Apply Heron's Formula to find the area

where s is the semi-perimeter of the triangle and a, b and c are the lengths of the sides. This formula is useful for the calculating the area of the triangle when the height is not known but the side lengths are provided.

Question 1. A park, in the shape of a quadrilateral ABCD, has ∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? 

Solution:

Given, a quadrilateral ABCD where ∠C = 90º.

Construction: Join diagonal BD.

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As we can see that, △DCB is right-angled at C

Hence, BC is the base and CD is height of △DCB, so

ar(△DCB) =  × Base × Height

=  × 12 × 5

= 30 m²..........................................(1)

As △DCB is right angle triangle we can calculate third side by Pythagoras theorem

BD² = CB² + CD²

BD² = 12² + 5²

BD = √(144+25)

BD = √169

BD =13 m

Now, Area of △DAB can be calculated by Heron's Formula, where 

AB = a = 9 m

AD = b = 8 m

BD = c = 13 m 

Semi Perimeter (s) = 

s = 

s = 15 m

ar(△DAB) = √s(s-a)(s-b)(s-c)

= √15(15-9)(15-8)(15-13)

= √15×(6)×(7)×(2)

= 6√35 m²............................................(2)

From (1) and (2), we can conclude that,

ar(ABCD) = ar(△DCB)+ar(△DAB)

= (30 + 6√35)

= (30 + 35.5)

≈ 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.

Solution:

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Here, we can notice that in △ABC

AC² = AB² + BC²

5² = 3² + 4²  

25 = 25

LHS = RHS

As this triangle is satisfying the Pythagoras theorem, Therefore, △ABC is a right angle triangle, 90° at B.

Hence, BC is the base and AB is height of △ABC. so

So, ar(△ABC) =  × Base × Height

=  × 4 × 3

= 6 cm².....................................(1)

Now, Area of △DAC can be calculated by Heron's Formula, where

AD = a = 5 cm

DC = b = 4 cm

AC = c = 5 cm

Semi Perimeter (s) = 

s = 

s = 7 cm

ar(△DAC) = √s(s-a)(s-b)(s-c)

= √7(7-5)(7-4)(7-5)

= √7×(2)×(3)×(2)

= 2√21 cm²............................................(2)

From (1) and (2), we can conclude that,

ar(ABCD) = ar(△ABC)+ar(△DAC)

= (6 + 2√21)

= (6 + 9.2(approx.))

≈ 15.2 cm² (approx.)

Question 3. Radha made a picture of an aeroplane with coloured paper as shown in (Fig). Find the total area of the paper used.

Solution:

Total area of the paper used = (Area I + Area II + Area III + Area IV + Area V)

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Now, Area of triangle can be calculated by Heron's Formula, where

a = 5 cm

b = 5 cm

c = 1 cm

Semi Perimeter (s) = 

s = 

s = 5.5 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √5.5(5.5-5)(5.5-5)(5.5-1)

= √5.5×(0.5)×(0.5)×(4.5)

= 0.5×0.5×3√11

= 0.75√11

 ≈ 2.5 cm² ............................................(1)

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Area of Rectangle = length×Breadth

= 1 × 6.5 = 6.5 cm².....................................(2)

= Area of parallelogram EFAO + △ AFD

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OD = 2cm

AD = OD-OA = 2-1 = 1 cm

Hence, △ AFD is equilateral.

PD =  AD =  cm

△ PFD is right angled at P, Pythagoras theorem is applicable 

1²=h² +()²

h = √(1-)

h = √ cm

Area III: 

= (Base × Height) + ( × Base × Height)

= (1 × ) + ( × 1 × )

= 

= = 1.29 cm².......................................(3)

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ar(△) =  × Base × Height

=  × 6 × 1.5

= 4.5 cm²

Area IV + Area V = 2×ar(△)

= 2×4.5

= 9 cm² .......................................(4)

Hence, Total area of the paper used = (Area I + Area II + Area III + Area IV + Area V)

= (1) + (2) + (3) + (4)

= 2.5 + 6.5 + 1.29 +9

= 19.29 cm²

Question 4. 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.

Solution:

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Now, Area of △AEB can be calculated by Heron's Formula, where

AE = a = 28 cm

EB = b = 30 cm

AB = c = 26 cm

Semi Perimeter (s) = 

s = (28+30+26)/2

s = 42 cm

ar(△AEB) = √s(s-a)(s-b)(s-c)

= √42(42-28)(42-30)(42-26)

= √42×(14)×(12)×(16)

= 336 cm²

As it is given, ar(△AEB) = ar(parallelogram ABCD)

336 = Base × Height

336 = 28 × h

h = 

h = 12 cm

Hence, the height of the parallelogram = 12 cm

Question 5. 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? 

Solution:

ABCD is a rhombus having diagonal AC = 48 cm

side AB=BC=CD=AD=30 cm

Diagonal of Rhombus divides the area into two equal parts.

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Now, ar(△ABC) can be calculated by Heron's Formula, where

AB = a = 30 m

BC = b = 30 m

AC = c = 48 m

Semi Perimeter (s) = 

s = 

s = 54 m

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √54(54-30)(54-30)(54-48)

= √54×(24)×(24)×(6)

= 432 m²

Hence, Area of rhombus = 2 × (ar(△))

= 2 × 432 = 864 m²

Area for 18 cows = 864 m²

Area for each cow = 864 / 18 = 48 m²

Question 6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution:

Let's consider for each triangle.

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Now, for each ar(△)can be calculated by Heron's Formula, where

a = 50 cm

b = 50 cm

c = 20 cm

Semi Perimeter (s) = 

s = 

s = 60 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √60(60-50)(60-50)(60-20)

= √60×(10)×(10)×(40)

= 200√6 cm²

Hence, the Total Area = 5×200√6

= 1000√6 cm²

Question 7. 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. How much paper of each shade has been used in it? 

Solution:

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As the area of kite is in the square, it area will be

Area of kite = ×(diagonal)²

= ×32×32

= 512 cm²

Diagonal divides area into equal areas.

512 = 2 × Area I

Area I =Area II = 256 cm²...................................(1)

Now, for each ar(△)can be calculated by Heron's Formula, where

a = 6 cm

b = 6 cm

c = 8 cm

Semi Perimeter (s) = 

s = 

s = 10 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √10(10-6)(10-6)(10-8)

= √10×(4)×(4)×(2)

= 8√5 cm²....................................(2)

Question 8. 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). Find the cost of polishing the tiles at the rate of 50p per cm².

Solution:

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Area for each triangle will be:

Now, for each ar(△)can be calculated by Heron's Formula, where

a = 9 cm

b = 28 cm

c = 35 cm

Semi Perimeter (s) = 

s = 

s = 36 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √36(36-9)(36-28)(36-35)

= √36×(27)×(8)×(1)

= 36√6 cm²

As there are 16 tiles, so total area = 16 × 36√6

= 1410.906 cm²

As 1 cm² = 50 p = ₹ 0.5

1410.906 cm² = ₹ 0.5 ×1410.906

= ₹ 705.45

Question 9. 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.

Solution:

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AB = 25 m

EB = AB-AE = 25-10 = 15 m

Now, for ar(△ECB) can be calculated by Heron's Formula, where

a = 13 m

b = 14 m

c = 15 m

Semi Perimeter (s) = 

s = 

s = 21 m

ar(△ECB) = √s(s-a)(s-b)(s-c)

= √21(21-3)(21-14)(21-15)

= √21×(18)×(7)×(6)

= 84 m² ...............................(1)

ar(△ECB) =  × Base × Height

84 m²=  × 15 × h

h =  m

h = 11.2 m

Total Area = Area of parallelogram AECD + ar(△ECB)

= (Base × Height) + 84m² 

= 10 × 11.2 + 84

Total Area = 196 m²

Conclusion

The Heron's Formula is a powerful tool for the calculating the area of the triangle using the only the lengths of its sides. Exercise 12.2 in Chapter 12 of the Class 9 NCERT textbook helps students practice this formula and apply it to solve the various geometric problems. Mastery of Heron's Formula enhances understanding of the triangle properties and improves overall problem-solving abilities in geometry.