NCERT Solutions For Class 9 Maths Chapter 12

NCERT Solutions Class 9 Maths Heron's Formula

Ncert Solutions For Class 9 Maths Chapter 12 PDF Free Download

NCERT maths solutions for class 9 chapter 12 Heron’s formula provides a ready solution for all the questions available in the NCERT class 9 maths textbook. Students can practice by solving more questions from the NCERT solutions for class 9 maths chapter 12 either by downloading the PDF files or through online by visiting our website at BYJU’S. NCERT solutions for maths chapter 12  class 9 pdf includes set of solutions prepared by our team of subject experts under the guidance of CBSE board and the solutions are explained in a lucid manner along with the more examples, formulas, and other important questions. Heron was born in 10 AD possibly in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields. His geometrical works deal largely with problems on mensuration written in three books. These books deal with the area of squares, rectangles, triangles, trapezoids (trapezia), various other specialised quadrilaterals, the regular polygons, circles, surfaces of cylinders, cones, spheres etc. In this book, Heron has derived the famous formula for the area of a triangle in terms of its three sides. We have provided a comprehensive study material of NCERT Solutions For Class 9 Maths Heron’s Formula. The students can also download the pdfs from the link provided above.

NCERT Solutions For Class 9 Maths Chapter 12 Exercises

 

Why Heron’s Formula…?

When the triangle is a right angled triangle:

 

Area = \(\frac{1}{2}\times base\times altitude\)

Example:

1

Now, Area of \(\Delta ABC= \frac{1}{2}\times base\times altitude \)

\(Area\;of\;\Delta ABC= \frac{1}{2}\times 12\times 5 = 30\;cm^{2}\)

When the triangle is an equilateral triangle:

Consider an equilateral triangle ABC with each side = 10cm

2

If we drop AM perpendicular to BC then BM = 5cm,

Now, AB2 = AM2 + BM2    (Pythagoras Theorem)

AM2 = 102 – 52

Therefore, AM = \(5\sqrt{3}\) cm

Now, Area for triangle ABC = 2×(Area of triangle ABM)

And, \(Area\;\Delta ABM =\frac{1}{2}\times Base\times Altitude\)

Therefore, \(Area\;\Delta ABM =\frac{1}{2}\times 5\times 5\sqrt{3}\)

\(Area\;\Delta ABM =\frac{25\sqrt{3}}{2}\)cm2

Now, Area for triangle ABC = 2×(\(\frac{25\sqrt{3}}{2}\))cm2

Therefore, \(Area\; of \;\Delta ABC= 25\sqrt{3}\; cm^{2}\)

When the triangle is an isosceles triangle:         

Consider an isosceles triangle ABC with equal sides AB = AC = 13 cm and unequal side BC = 10cm

3

Drop AM perpendicular to BC. Therefore CM = 5cm

Now, AC2 = AM2 + CM2    (Pythagoras Theorem)

AM2 = 132 – 52

Therefore AM = 12cm

Now, Area for triangle ABC = 2×(Area of triangle ACM)

And, \(Area\;\Delta ACM =\frac{1}{2}\times Base\times Altitude\)

Therefore, \(Area\;\Delta ACM =\frac{1}{2}\times 5\times 12\)

\(Area\;\Delta ABM \)= 30 cm2

Now, Area for triangle ABC = 2× 30 cm2

Therefore, Area of triangle ABC = 60cm2

When the triangle is a scalene triangle:

Suppose that we know the lengths of sides of a scalene triangle and we have to find its area. Now, for applying the above formula we need to calculate its height. But in this case we have no idea about how to calculate the height of a scalene triangle and that is why it’s very difficult to find its area. Thus, Heron gave a formula to find the area of any triangle known as Heron’s Formula.

Area of a Triangle \(ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

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

s = semi perimeter   (half of perimeter of triangle)

\(s=\frac{a+b+c}{2}\)

 

Finding square roots:

 (i) Find values of: \(\sqrt{72895}\)

4

 

(ii)Find values of \(\sqrt{745}\)

5

 

Exercise – 12.1

 

Q.1: The lengths of two sides of a triangle ABC are 10cm and 7cm and its perimeter is 28cm. Find the area of a triangle ABC.

Sol.

Given perimeter = 28cm

Therefore, s = \(\frac{perimeter}{2}\) = 14cm

And a = 10cm, b =7cm and c = 11cm (Since, perimeter = a + b + c)

 On applying Heron’s Formula;

\(Area\;of\;Triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of Triangle ABC

=\(\sqrt{14(14-10)(14-7)(14-11)}\)

(or)     =\(\sqrt{14\times 4\times 7\times 3}\)

= \(14\sqrt{6}\)

Therefore, area of triangle ABC = \(14\sqrt{6}\)cm2

 

Q.2: If perimeter of a triangular plot is 250m and the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9. Find its area.

Sol.

Since, the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9

Therefore, sides of triangular plot will be 4x, 7x and 9x

q2

Given perimeter = 250m

Therefore, 4x + 7x + 9x = 250

Hence x = 12.5, that gives:

a = 4x = 50m, b = 7x = 87.5m and c = 9x = 112.5m

And s = \(\frac{perimeter}{2}\)= 125m

Now on applying Heron’s Formula;

\(Area\;of\;plot\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of plot ABC:

=\(\sqrt{125(125-50)(125-87.5)(125-112.5)}\)

(or)     =\(\sqrt{125\times 75\times 37.5\times 12.5}\)

= \(937.5\sqrt{5}\)

=2096.3m2

Therefore, area of triangular plot ABC = 2096.3m2

 

Q.3: There is a triangular park in a locality having sides 110m, 90m and 40m. A gardener Fauzia has to plant grass inside. In how much area does she need to plant grass? Also Fauzia has to put a fence all around that triangular park. Find the cost of fencing that triangular park with barbed wire if the cost of barbed wire is Rs 25 per meter and also a space of 2m is to be left for a gate on any one side of that park.

Sol.

Given a = 110m, b = 90m and c = 40m

Therefore, perimeter = 110 + 90 + 40 = 240m

And s = \(\frac{perimeter}{2}\)= 120m

q3

Now applying Heron’s Formula;

\(Area\;of\;park\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of park ABC:

=\(\sqrt{120(120-110)(120-90)(120-40)}\)

(or)     =\(\sqrt{120\times 10\times 30\times 80}\)

= \(1200\sqrt{2}\)

= 1697.07m2 (Approx.)

Therefore, area of triangular plot ABC = 1697.07m2 (Approx.)

Since, perimeter = 240m

Therefore for fencing length of wire needed = 240 – 2 = 238m

So the total cost of fencing the triangular park = 238×25 = 5950Rs

 

Q.4: The side walls of a flyover are triangular in shape and are used for advertisements. Dimensions of side walls are: 120m, 140m and 50m. The cost of posting advertisements is 4500Rs per m2 per year. How much rent an automobile company paid if the company hired one of its walls for 4 months?

Sol.

Given a = 120m, b = 140m and c = 50m

Therefore, perimeter of each triangular wall = 120 + 140 + 50 = 310m

And s = \(\frac{perimeter}{2}\)= 155m

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of each triangular wall:

=\(\sqrt{155(155-120)(155-140)(155-50)}\)

(or)     =\(\sqrt{155\times 35\times 15\times 105}\)

= \(525\sqrt{31}\)

Therefore, area of each triangular wall = \(525\sqrt{31}\)m2

Since, rent of posting advertisements is 4500 per m2 per year (12 months)

Therefore rent for \(525\sqrt{31}\)m2 area = \(525\sqrt{31}\) × 4500

= \(2362500\sqrt{31}\)Rs per year(12 months).

Now, For 12 months’ rent =  \(2362500\sqrt{31}\) Therefore for 4months = \(\frac{4\times 2362500\sqrt{31}}{12}=787500\sqrt{31}\)

Hence, rent paid by company for hiring one of its walls for 4 months = \(787500\sqrt{31}\)Rs.

 

Q.5: If each side of a triangle is increased by six times then find the total percentage of increase in its area.

Sol.

Let a, b and c be the sides of the original triangle. Then,

Perimeter of original triangle = a + b + c and s1 = \(\frac{a + b + c}{2}\)

And sides of the new triangle will be 6a, 6b and 6c. Then,

Perimeter of new triangle = 6a + 6b + 6c and s2 = \(\frac{6a + 6b + 6c}{2}\) = 3(a + b + c) = 6s1   (Since, 2s1 = a + b + c)

Let A1 and A2 be the areas of the original and new triangle respectively. Then,

        Area of triangle A1=\(\sqrt{s_{1}(s_{1} – a)(s_{1} – b)(s_{1} – c)}\)

 

And  Area of triangle A2=\(\sqrt{s_{2}(s_{2} – a)(s_{2} – b)(s_{2} – c)}\)

Since, s2 = 6s1

Therefore, Area of triangle A2=\(\sqrt{6s_{1}(6s_{1} – a)(6s_{1} – b)(6s_{1} – c)}\)

Area of triangle A2=\(\sqrt{12s_{1}(s_{1} – a)(s_{1} – b)(s_{1} – c)}\)

Therefore, A2 = 12A1

Therefore, increase in the area of triangle = A2 – A1

= 12A1 – A1 = 11A1

Therefore, percentage increase in area = \(\frac{11A_{1}}{A_{1}}\times 100\) = 1100%

Hence, area of the new triangle is increased by 1100%

 

Q.6: If perimeter of a triangular plot is 490 and the lengths of the sides of a triangular plot are in the ratio of 22 : 16 : 32. Find its area.

Sol.

Since, the lengths of the sides of a triangular plot are in the ratio of 22 : 16 : 32

Therefore, sides of triangular plot will be 22x, 16x and 32x

q6

Given perimeter = 490m

Therefore, 22x + 16x + 32x = 490

Hence x =7 , that gives:

a =22x = 154m, b = 16x = 112m and c = 32x = 224m

Therefore perimeter = a + b + c = 490m

And s = \(\frac{perimeter}{2}\)= 245m

Now on applying Heron’s Formula;

\(Area\;of\;plot\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of plot ABC:

=\(\sqrt{245(245-154)(245-112)(245-224)}\)

(or)     =\(\sqrt{245\times 91\times 133\times 21}\)

 = 7891.13m2 (Approx.)

Therefore, area of triangular plot ABC = 7891.13 m2 (Approx.)

 

Q.7: The length of equal sides of an isosceles triangle ABC is 8cm and its perimeter is 28cm. Find the area of a triangle ABC.

Sol.

Given perimeter = 28cm

Therefore, s = \(\frac{perimeter}{2}\) = 14cm

And a = 8cm, b =8cm and c = 12m (Since, perimeter = a + b + c)

On applying Heron’s Formula;

\(Area\;of\;Triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of Triangle ABC

=\(\sqrt{14(14-8)(14-8)(14-12)}\)

(or)     =\(\sqrt{14\times 6\times 6\times 2}\)

= \(12\sqrt{7}\)

=31.75cm2 (Approx.)

Therefore, area of triangle ABC = 31.75cm2(Approx.)

 

Exercise – 12.2

 

Q.1 Ankesh grows rice in a triangular field with sides 260m, 180m and 350m. Adjacent to this field there is another field with sides 260m, 280m and 380m where he wants to grow carrot and ladyfinger. He decided to divide this field in to two parts by joining the mid-point of the longest side to the opposite vertex. Now in one part he will be growing carrot and in other part he will be growing lady finger. Find how much area (in hectare) is used for growing rice, carrot and ladyfinger.

Sol.

 e12.2 q1

Consider triangle ABC,

Here a = 350m, b = 260m and c = 180m

Now, s = \(\frac{350 + 260 + 180}{2}\)

Therefore, s = 395

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle ABC :

=\(\sqrt{395(395-350)(395-260)(395-180)}\)

(or)     =\(\sqrt{395\times 45\times 135\times 215}\)

= \(\sqrt{515919375}\)

Therefore, area of triangle ABC = 22713.86m2 (approx.)

Since 1 hectare = 10000 m2

Hence area used for growing rice = 2.27 hectares (approx.)

Consider triangle ACD,

Here a = 260m, b = 280m and c = 380m

Now, s = \(\frac{260 + 280 + 380}{2}\)

Therefore, s = 460

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ACD=\sqrt{s(s – a)(s – b)(s – c)}\)

Therefore, area of triangle ACD:

=\(\sqrt{460(460-260)(460-280)(460-380)}\)

(or)     =\(\sqrt{460\times 200\times 180\times 80}\)

= \(\sqrt{1324800000}\)

Therefore, area of triangle ABC = 36397.80m2 (approx.)

Since 1 hectare = 10000 m2

Hence area used for growing Carrot and ladyfinger = 3.63 hectares (approx.)

Since, M is the mid-point of AD and CM divides the triangle CAD in two halves equal in area. Therefore, area for growing carrot = area for growing ladyfinger = 1.815 hectares.

 

Q.2 A plot is in the shape of trapezium whose non parallel sides are 12m and 14m and parallel sides are 10m and 20m. Find the area of park.

Sol.

e12.2 q2

Here area of park = area of trapezium ABCD

Area of Trapezium = \(\frac{1}{2}\)× (AD + BC) × BM

BM =???

Draw BM perpendicular to AD and BN parallel to CD

Since BC is parallel to DN and BN is parallel to CD therefore, BCDN forms a parallelogram.

Therefore, BN = 12cm (opposite side of parallelogram)

AN = 10cm    (20 – DN = 10cm) and

AB = 14cm (given)

Perimeter = (a + b + c) = 36m

Therefore s = \(\frac{perimeter}{2}\)= 18m

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ABN=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle ABN:

=\(\sqrt{18(18-12)(18-14)(18-10)}\)

(or)     =\(\sqrt{18\times 6\times 4\times 8}\)

= \(24\sqrt{6}\)m2

Therefore, area of triangle ABN = \(24\sqrt{6}\)m2

In triangle ABN, area of triangle ABC = \(\frac{1}{2}\)×base × altitude

Therefore   \(24\sqrt{6}\) = \(\frac{1}{2}\)×10×BM

Therefore, 10 BM = 2×\(24\sqrt{6}\)

Therefore BM = \(\frac{24\sqrt{6}}{5}\)m

Now, area of Trapezium ABCD = \(\frac{1}{2}\)× (AD + BC) × BM

= \(\frac{1}{2}\)× (20 + 10) × \(\frac{24\sqrt{6}}{5}\)

= \(72\sqrt{6}\)m2

=176.36m2 (Approx.)

Therefore, area of park = area of trapezium ABCD = 176.36m2 (Approx.)

 

Q.3 Eight Triangular pieces of two different colours are stitched together to make an umbrella. Each piece measuring 10cm, 40cm and 40cm. Find how much cloth of each colour is used to make the umbrella.

Sol.

e12.2 q3

Consider triangle ABC:

Here a = 40cm, b = 40cm and c = 10cm

Therefore, perimeter = 40 + 40 +10 = 90cm

Hence s = \(\frac{perimeter}{2}\)= 45cm

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle ABC:

=\(\sqrt{45(45-40)(45-40)(45-10)}\)

(or)     =\(\sqrt{45\times 5\times 5\times 35}\)

= \(75\sqrt{7}\)cm2

Therefore, area of triangle ABC = \(75\sqrt{7}\)cm2

Since there are 4 triangular pieces of two different colours each

Therefore, area of each clothes required:

=\(75\sqrt{7}\)×4

=\(300\sqrt{7}\)cm2

=793.73 cm2 (Approx.)

Q.4 There is a rhombus shaped field that has to be divided in to 5 sons. If each side of that field is 40m and its longer diagonal is 48m, how much area will each son be getting?

Sol.

 e12.2 q4

Let ADCB be a rhombus shaped field.

In triangle ABC

Since, AB = BC = 40m and AC = 48m

Therefore perimeter = (40 + 40 + 48) = 128m

Hence s = \(\frac{perimeter}{2}\)= 64m

Now applying Heron’s Formula;

\(Area\;of\;triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle ABC:

=\(\sqrt{64(64-40)(64-40)(64-48)}\)

(or)     =\(\sqrt{64\times 24\times 24\times 16}\)

= 768m2

Therefore, area of field = 2× area of triangle ABC

Therefore area of field = 1536m2

Since there are 4 sons, therefore each son will get 384m2 of field.

 

Q.5 Three Students of Saint Thomas School and three students of Saint Marry School participated in a cleanliness campaign. There were two groups of each school, 1st group was of Saint Thomas School and walked through the lanes PS, SR and RP and 2nd group was of Saint Marry School and walked through the lanes PQ, QR and RP. Then they cleaned the area enclosed within their lanes (As shown in figure). Find the total area cleaned in this campaign by the students of both school. Also find which school cleaned more area and by how much??

e12.2 q5 

Sol.

Area cleaned by Saint Marry School = Area of triangle PQR

Since PQR is a right angled triangle

Therefore area of triangle PQR:

= \(\frac{1}{2}\times Base\times Altitude\)

=\(\frac{1}{2}\times 24\times 7\)

Therefore area of triangle PQR = 84m2

Hence area cleaned by Saint Marry School (A1) = 84m2

Area cleaned by Saint Thomas School = Area of triangle PRS

Since   PR2 = RQ2 + PQ2 (Pythagoras Theorem)

PR2 = 72 + 242

Therefore, PR = 25            m

Now, In triangle PRS;

a = 20m, b = 9m and c = 25m

Therefore, Perimeter = a + b + c = 54m

Hence s = \(\frac{perimeter}{2}\)= 27m

Now applying Heron’s Formula;

\(Area\;of\;triangle\;PRS=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle PRS:

=\(\sqrt{27(27 – 20)(27 – 9)(27 – 25)}\)

(or)     =\(\sqrt{27\times 7 \times 18 \times 2}\)

=\(18\sqrt{21}\)

= 82.486m2 (Approx.)

Therefore, Area cleaned by Saint Thomas School (A2) = 82.486m2 (Approx.)

So, total area cleaned = A1 + A2 = 84 + 82.48 = 166.48m2

Since, A1 > A2 and A1 – A2 = 1.52m2 (Approx.)

Therefore, area cleaned by students of Saint Marry School is greater than area cleaned by students of Saint Thomas School by 1.52m2 (Approx.)

 

 Q.6 Find the total area of given figure.

Sol.

 e12.2 q6

Area of figure-1:

Area DCMK represents a rectangle, with DC = 1cm and DK = 10cm

Therefore area of figure DCMK = length × breadth

= 1 × 10

Therefore area of figure-1 = 10cm2

 

Area of figure-4:

Triangle GKH represents a right angled triangle

From figure 4; base = 2cm and height = 8cm

Therefore area of triangle GKH

= \(\frac{1}{2}\times base\times altitude\)

=\(\frac{1}{2}\times 2\times 8\)

Therefore area of figure-4 = 8cm2

Since figure-4 and figure-5 are similar

Therefore area of figure-5 = 8cm2

Area of figure-3:

In triangle KLM; a = 5cm, b = 5cm and c = 1cm

Therefore, Perimeter = a + b + c = 11cm

Hence s = \(\frac{perimeter}{2}\)= 5.5cm

Now applying Heron’s Formula;

\(Area\;of\;triangle\;KLM=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle KLM:

=\(\sqrt{5.5(5.5 – 5)(5.5 – 5)(5.5 – 1)}\)

(or)     =\(\sqrt{5.5\times 0.5\times 0.5\times 4.5}\)

= \(\frac{3\sqrt{11}}{4}\)cm2

Therefore, area of figure- 3 = 2.487cm2

Area of figure-2:

Figure ABCD represents a trapezium

e12.2 q6.2

From figure: In triangle AMD

AD = 8cm (given)

AM + NB = AB – MN = 1cm

Therefore, AM = 0.5cm

Now,

AD2 = AM2 + MD2

MD2= 12 – 0.52

MD = 0.866cm

Now, area of trapezium = \(\frac{1}{2}\times (base_{2}+base_{1})\times height\)

=\(\frac{1}{2}\times (2+1)\times 0.866\)

Hence, area of trapezium = 1.299cm2

Therefore, area of figure-2 = 1.299cm2

Hence, total area of given figure = figure-1 + figure-2 + figure-3 + figure-4 + figure-5

= 10 +1.299 + 2.487 + 8 + 8

Therefore, total area of given figure = 29.786cm2

 

Q.7: Find the Area of a quadrilateral PQRS in which PQ = 10cm, QR = 7cm, RS =9cm, PS = 8cm and SQ = 14cm

Sol.

e12.2 q7

Consider triangle PQS;

Here a = 10cm, b = 8cm and c = 14cm

Therefore Perimeter = 10 + 8 + 14 = 32cm

Hence s = \(\frac{perimeter}{2}\)= 16cm

Now applying Heron’s Formula;

\(Area\;of\;triangle\;PQS=\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle PQS:

=\(\sqrt{16(16 – 10)(16 – 8)(16 – 14)}\)

(or)     =\(\sqrt{16\times 6\times 8\times 2}\)

= \(16\sqrt{6}\)cm2

                  = 39.19cm2 (Approx.)

Therefore, area of triangle PQS = 39.19cm2 (Approx.)

Consider triangle QRS;

Here a = 7cm, b = 9cm and c = 14cm

Therefore Perimeter = 7 + 9 + 14 = 30cm

Hence s = \(\frac{perimeter}{2}\)= 15cm

Now applying Heron’s Formula;

Area of triangle QRS=\(\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle QRS:

=\(\sqrt{15(15 – 7)(15-9)(15-14)}\)

(or)     =\(\sqrt{15\times 8\times 6\times 1}\)

= \(12\sqrt{5}\)cm2

                  = 26.83 cm2 (Approx.)

Therefore, area of triangle QRS = 26.83cm2 (Approx.)

Now, area of quadrilateral = area of triangle PQS + area of triangle QRS

= 39.19+26.83 = 66.02cm2 (Approx.)

Therefore area of quadrilateral PQRS = 66.02cm2 (Approx.)

 

Q.8: A parallelogram and a triangle with sides 12cm, 15cm and 9cm have the same area and the same base. The Base of the triangle and the parallelogram is taken as 12cm. Find the height of the parallelogram.

Sol.  

e12.2 q8

From triangle ABC;

Here a = 12cm, b = 9cm and c = 15cm

Therefore Perimeter = 12 + 9 + 15 = 36cm

Hence s = \(\frac{perimeter}{2}\)= 18cm

Now applying Heron’s Formula;

Area of triangle ABC=\(\sqrt{s(s-a)(s-b)(s-c)}\)

Therefore, area of triangle ABC:

=\(\sqrt{18(18 – 12)(18-9)(18-15)}\)

(or)     =\(\sqrt{18\times 6\times 9\times 3}\)

= 54cm2

Therefore, area of triangle ABC = 54cm2

Area of parallelogram = area of triangle (Given)

Therefore 54 = base × height

Height = 4.5cm

Therefore height of parallelogram = 4.5cm

 

In conclusion, NCERT Solutions For Class 9 Maths Chapter 12  talks about Heron’s formula. This is an important concept as it is required for all fundamental operation in modern geometry.