NCERT Solutions for Class 9 Maths Chapter 12: Heron's Formula - Exercise 12.1

Exercise 12.1

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\[\mathbf{180}\text{ }\mathbf{cm}\] , what will be the area of the signal board?

Ans:

Side of traffic signal board$=a$ 

Perimeter of traffic signal board $=3\left( a \right)$ 

 By Heron’s formula, 

Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ 

Area of given triangle $=\sqrt{\frac{3}{2}\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)}$ 

$=\sqrt{\frac{3}{2}a\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)}$ 

$=\frac{\sqrt{3}}{2}{{a}^{2}}$ 

Area of given triangle$=\frac{\sqrt{3}}{2}{{a}^{2}}$ …..(1)

Perimeter of traffic signal board $=180cm$ 

Side of traffic signal board $\left( a \right)=\left( \frac{180}{3} \right)cm=60cm$

Using equation (1), area of traffic signal $=\frac{\sqrt{3}}{2}\left( 60\,c{{m}^{2}} \right)$ 

$=\left( \frac{3600}{4}\sqrt{3} \right)c{{m}^{2}}=900\sqrt{3}c{{m}^{2}}$ 

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are\[122\text{ }m,\text{ }22\text{ }m\] , and \[\mathbf{120}\text{ }\mathbf{m}\] (see the given figure). The advertisements yield an earning of \[\mathbf{Rs}.\text{ }\mathbf{5000}\text{ }\mathbf{per}\text{ }\mathbf{m2}\] per year. A company hired one of its walls for \[\mathbf{3}\] months. How much rent did it pay?

Flyover

Ans:

The sides of the triangle (i.e.,\[a,\text{ }b,\text{ }c\] ) are of \[122\text{ }m,\text{ }22\text{ }m,\] and \[120\text{ }m\] respectively.

Perimeter of triangle$=\left( 122+22+120 \right)m$ 

$2s=264m$ 

$s=132m$ 

By Heron’s formula,

Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{132\left( 132-122 \right)\left( 132-22 \right)\left( 132-120 \right)}{{m}^{2}}$ 

$=\sqrt{132\left( 10 \right)\left( 110 \right)\left( 12 \right)}$ 

Rent of $1\,{{m}^{2}}$ area per year$=Rs.5000$ 

Rent of $1\,{{m}^{2}}$ area per month$=Rs.\frac{5000}{12}$ 

Rent of \[1320\text{ }{{m}^{2}}\] area for \[3\]  months $=\frac{5000}{12}\left( 3 \right)\left( 1320 \right)=1650000$

Therefore, the company had to pay $Rs.1650000$  

3. The floor of a rectangular hall has a perimeter\[\mathbf{250}\text{ }\mathbf{m}\] . If the cost of panting the four walls at the rate of \[\mathbf{Rs}.\text{ }\mathbf{10}\text{ }\mathbf{per}\text{ }\mathbf{m2}\] is\[~\mathbf{Rs}.\mathbf{15000}\] , find the height of the hall.

Ans:

Let length, breadth, and height of the rectangular hall be\[l\text{ }m,\text{ }b\text{ }m\] , and \[h\text{ }m\] respectively. 

Area of four walls \[=2lh+2bh=2\left( l+b \right)h\] 

Perimeter of the floor of hall \[=2\left( l+b \right)=250m\] 

∴ Area of four walls \[=2\left( l+b \right)\text{ }h=250h\text{ }{{m}^{2}}\] 

Cost of painting per \[~{{m}^{2}}\] area \[=Rs.10\] 

Cost of painting \[250h\text{ }{{m}^{2}}\] area \[=Rs.\left( 250h\times 10 \right)=Rs.2500h\] 

However, it is given that the cost of paining the walls is\[Rs.\text{ }15000\] .

 ∴ \[15000=2500h\] 

\[h=6\text{ }\] 

Therefore, the height of the hall is\[6\text{ }m\] .

4. Find the area of a triangle two sides of which are \[\mathbf{18}\text{ }\mathbf{cm}\] and \[\mathbf{10}\text{ }\mathbf{cm}\] the perimeter is \[\mathbf{42}\text{ }\mathbf{cm}.\] 

Ans:

Let the third side of the triangle be\[x\] .

Perimeter of the given triangle \[=\text{ }42\text{ }cm\text{ }\]

\[18\text{ }cm+10\text{ }cm+x=42\] 

$x=14cm$ 

$s=\frac{perimeter}{2}=\frac{42}{2}=21cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{21\left( 21-18 \right)\left( 21-10 \right)\left( 21-14 \right)}c{{m}^{2}}$ 

$=\sqrt{21\left( 3 \right)\left( 11 \right)\left( 7 \right)}$

$=21\sqrt{11}c{{m}^{2}}$ 

5. Sides of a triangle are in the ratio of \[\mathbf{12}:\mathbf{17}:\mathbf{25}\] and its perimeter is\[\mathbf{540}\text{ }\mathbf{cm}\] . Find its area.

Ans:

Let the common ratio between the sides of the given triangle be\[x\] . Therefore, the side of the triangle will be \[12x,\text{ }17x,\] and\[25x\] .

Perimeter of this triangle \[=540\text{ }cm\] 

\[\Rightarrow 12x+17x+25x=540\text{ }cm\] 

\[\Rightarrow 54x=540\text{ }cm\] 

\[\Rightarrow x=10cm\]

Sides of the triangle will be\[120\text{ }cm,\text{ }170\text{ }cm,\text{ }250\text{ }cm\].

$s=\frac{perimeter\,of\,triangle}{2}=\frac{540\,cm}{2}=270cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)}c{{m}^{2}}$ 

$=\sqrt{270\left( 150 \right)\left( 100 \right)\left( 20 \right)}$

$=9000c{{m}^{2}}$ 

Therefore, the area of this triangle is $9000c{{m}^{2}}$.

6. An isosceles triangle has perimeter \[\mathbf{30}\text{ }\mathbf{cm}\] and each of the equal sides is\[\mathbf{12}\text{ }\mathbf{cm}\] . Find the area of the triangle.

Ans:

Let the third side of this triangle be\[x\] .

Perimeter of triangle \[=\text{ }30\text{ }cm\] 

\[\Rightarrow 12\text{ }cm+12\text{ }cm+x=30\text{ }cm\] 

\[\Rightarrow x=6\text{ }cm\] 

$s=\frac{perimeter\,of\,triangle}{2}=\frac{30\,cm}{2}=15cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{15\left( 15-12 \right)\left( 15-12 \right)\left( \left( 15-6 \right) \right)}c{{m}^{2}}$ 

$=\sqrt{15\left( 3 \right)\left( 3 \right)\left( 9 \right)}$

$=9\sqrt{15}c{{m}^{2}}$ 

Exercise (12.2)

1. A park, in the shape of a quadrilateral\[\mathbf{ABCD}\] , has \[\angle \mathbf{C90}{}^\circ \] °, \[\mathbf{AB}=\mathbf{9}\text{ }\mathbf{m},\text{ }\mathbf{BC}=\mathbf{12m},\mathbf{CD}=\mathbf{5m}\] and\[\mathbf{AD}\text{ }=\text{ }\mathbf{8}\text{ }\mathbf{m}\] . How much area does it occupy?

Ans:

Let us join\[BD\] 

In\[\vartriangle BCD\] , applying Pythagoras theorem, 

\[B{{D}^{2}}=B{{C}^{2}}+C{{D}^{2}}\] \[=\left( 12 \right)2+\left( 5 \right)2=144+25\] 

\[B{{D}^{2}}=169\] 

\[BD=13\text{ }m\] 

A Park in the Shape of a Quadrilateral

$Area\,of\,\vartriangle BCD=\frac{1}{2}\left( BC \right)\left( CD \right)=\frac{1}{2}\left( 12 \right)\left( 5 \right)=30{{m}^{2}}$ 

For $\vartriangle ABD,$

$s=\frac{perimeter\,of\,triangle}{2}=\frac{9+8+13}{2}=\frac{30\,}{2}=15cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{15\left( 15-9 \right)\left( 15-8 \right)\left( 15-13 \right)}{{m}^{2}}$ 

$\begin{align}

& =\sqrt{15\left( 6 \right)\left( 7 \right)\left( 2 \right)} \\ 

& =6\sqrt{35} \\ 

& =6\left( 5.496 \right) \\ 

& =35.496\,{{m}^{2}} \\ 

\end{align}$

Area of the park $=Area\,of\,\vartriangle ABD+\,Area\,of\,\vartriangle BCD$ 

$=35.496+30=65.496$ 

$=65.5{{m}^{2}}$ 

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.

Ans:

Quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm

In\[\vartriangle ABC\] , applying Pythagoras theorem, 

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] 

${{\left( 5 \right)}^{2}}={{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}$ 

Therefore, \[\vartriangle ABC\]is a right-angled triangle, right-angled at point\[B\] .

$Area\,of\,\vartriangle ABC=\frac{1}{2}\left( AB \right)\left( BC \right)=\frac{1}{2}\left( 3 \right)\left( 4 \right)=6{{m}^{2}}$ 

For $\vartriangle ADC,$

Perimeter \[=2s=AC+CD+DA=\left( 5+4+5 \right)=14\text{ }cm\] 

$s=\frac{perimeter\,of\,triangle}{2}=\frac{14}{2}=7cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{7\left( 7-5 \right)\left( 7-5 \right)\left( 7-4 \right)}{{m}^{2}}$ 

$\begin{align}

& =\sqrt{7\left( 2 \right)\left( 2 \right)\left( 3 \right)} \\ 

& =2\sqrt{21} \\ 

& =2\left( 4.583 \right) \\ 

& =9.166c{{m}^{2}} \\ 

\end{align}$

Area of the park $=Area\,of\,\vartriangle ABD+\,Area\,of\,\vartriangle ACD$ 

$=6+9.166=15.166$ 

$=15.2c{{m}^{2}}$ 

3. Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.

Aeroplane with Given Dimensions

Ans:

A Triangle with Base = 1 cm and One Side = 5cm

For triangle 1:

Perimeter $=2s=\left( 5+5+1 \right)=11cm$ 

$s=\frac{perimeter\,of\,triangle}{2}=\frac{11}{2}=5.5cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{5.5\left( 5.5-5 \right)\left( 5.5-5 \right)\left( 5.5-1 \right)}c{{m}^{2}}$ 

$\begin{align}

& =\sqrt{5.5\left( 0.5 \right)\left( 0.5 \right)\left( 4.5 \right)} \\ 

& =0.75\sqrt{11} \\ 

& =0.75\left( 3.317 \right) \\ 

& =2.488c{{m}^{2}} \\ 

\end{align}$

For quadrilateral II:

This quadrilateral is a rectangle.

\[~Area=l\times b=\left( 6.5\times 1 \right)c{{m}^{2}}=6.5\text{ }c{{m}^{2}}\] 

For quadrilateral III:

This quadrilateral is a trapezium. 

Perpendicular height of parallelogram $=\sqrt{{{1}^{2}}-{{\left( 0.5 \right)}^{2}}}cm$ 

$=\sqrt{0.75}=0.866cm$ 

Area = Area of parallelogram + Area of equilateral triangle 

$=\left( 0.866 \right)1+\frac{\sqrt{3}}{4}{{\left( 1 \right)}^{2}}$ 

$=0.866+0.433=1.299c{{m}^{2}}$ 

Trapezium with given dimensions

Area of triangle (IV) = Area of triangle in (V)

$=\left( \frac{1}{2}\times 1.5\times 6 \right)=4.5c{{m}^{2}}$ 

Total area of the paper used \[=2.488+6.5+1.299+4.5\times 2=19.287c{{m}^{2}}\] 

4. A triangle and a parallelogram have the same base and the same area. If the sides of triangle are \[\mathbf{26}\text{ }\mathbf{cm},\mathbf{28}\text{ }\mathbf{cm}\] and\[\mathbf{30}\text{ }\mathbf{cm}\] , and the parallelogram stands on the base\[\mathbf{28}\text{ }\mathbf{cm}\] , find the height of the parallelogram.

Ans:

Perimeter of triangle \[=\left( 26+28+30 \right)cm=84\text{ }cm\] 

\[2s=84\text{ }cm\] 

\[s=42\text{ }cm\] 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{42\left( 42-26 \right)\left( 42-28 \right)\left( 42-30 \right)}c{{m}^{2}}$ 

$\begin{align}

& =\sqrt{42\left( 16 \right)\left( 14 \right)\left( 12 \right)} \\ 

& =336c{{m}^{2}} \\ 

\end{align}$

Let the height of the parallelogram be\[h\] .

Area of parallelogram = Area of triangle 

\[h\times 28cm=336c{{m}^{2}}\]

\[h=12cm\] 

Therefore, the height of the parallelogram is \[12\text{ }cm\] .

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

Ans:

A Rhombus ABCD with Each Side = 30cm

Let \[ABCD\] be a rhombus-shaped field. 

For\[\vartriangle BCD\] ,

Semi-perimeter, $s=\frac{48+30+30}{2}=54m$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{54\left( 54-48 \right)\left( 54-30 \right)\left( 54-30 \right)}{{m}^{2}}$ 

$\begin{align}

& =\sqrt{54\left( 6 \right)\left( 24 \right)\left( 24 \right)} \\ 

& =3\left( 6 \right)\left( 24 \right) \\ 

& =432{{m}^{2}} \\ 

\end{align}$

Area of the field $=2\left( Area\,of\,\vartriangle BCD \right)$ 

$=2\left( 432 \right){{m}^{2}}=864{{m}^{2}}$ 

Area for grazing for 1 cow $=\frac{864}{18}=48{{m}^{2}}$ 

Each cow will get \[48\text{ }{{m}^{2}}\] area of grass field.

6. An umbrella is made by stitching \[\mathbf{10}\]  triangular pieces of cloth of two different colours (see the given figure), each piece measuring\[\mathbf{20}\text{ }\mathbf{cm}\] , \[\mathbf{50}\text{ }\mathbf{cm}\] and \[\mathbf{50}\text{ }\mathbf{cm}\]. How much cloth of each colour is required for the umbrella?

Ans:

An Umbrella Made by Stitching Triangular Pieces

For each triangular piece,

Semi-perimeter, $s=\frac{20+50+50}{2}=60m$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{60\left( 60-50 \right)\left( 60-50 \right)\left( 60-20 \right)}c{{m}^{2}}$ 

$\begin{align}

& =\sqrt{60\left( 10 \right)\left( 10 \right)\left( 40 \right)} \\ 

& =200\sqrt{6}c{{m}^{2}} \\ 

\end{align}$

Since there are \[5\] triangular pieces made of two different coloured cloths, Area of each cloth required $=5\left( 200\sqrt{6} \right)=1000\sqrt{6}c{{m}^{2}}$ 

7. A kite in the shape of a square with a diagonal \[\mathbf{32}\text{ }\mathbf{cm}\] and an isosceles triangles of base \[\mathbf{8}\text{ }\mathbf{cm}\] and sides \[\mathbf{6}\text{ }\mathbf{cm}\] each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?

Ans:

A kite in the Shape of a Square

We know that

Area of square \[=\frac{1}{2}{{\left( diagonal \right)}^{2}}\] 

Area of the given kite $=\frac{1}{2}\left( 32 \right)=512c{{m}^{2}}$ = 

Area of 1st shade = Area of 2nd shade 

$=\frac{512}{2}=256c{{m}^{2}}$ 

Therefore, the area of paper required in each shape is\[256\text{ }c{{m}^{2}}\] .

For III triangle

Semi-perimeter, $s=\frac{6+6+8}{2}=10m$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{10\left( 10-6 \right)\left( 10-6 \right)\left( 10-8 \right)}c{{m}^{2}}$ 

$\begin{align}

& =\sqrt{10\left( 4 \right)\left( 4 \right)\left( 2 \right)} \\ 

& =8\sqrt{5} \\ 

& =8\left( 2.24 \right) \\ 

& =17.92c{{m}^{2}} \\ 

\end{align}$

Area of paper required for \[II{{I}^{rd}}\] shade \[=\text{ }17.92\text{ }c{{m}^{2}}\] 

8. A floral design on a floor is made up of \[\mathbf{16}\]  tiles which are triangular, the sides of the triangle being \[\mathbf{9}\text{ }\mathbf{cm},\text{ }\mathbf{28}\text{ }\mathbf{cm}\] and \[\mathbf{35}\text{ }\mathbf{cm}\] (see the given figure). Find the cost of polishing the tiles at the rate of\[\mathbf{50}p\,\mathbf{per}\text{ }\mathbf{c}{{\mathbf{m}}^{2}}\] .

Ans:

Floral Design on a Floor

It can be observed that,

Semi-perimeter of each triangular-shaped tile,

$s=\frac{35+28+9}{2}=36cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{36\left( 36-35 \right)\left( 36-28 \right)\left( 36-9 \right)}c{{m}^{2}}$ 

$\begin{align}

& =\sqrt{36\left( 1 \right)\left( 8 \right)\left( 27 \right)} \\ 

& =36\sqrt{6} \\ 

& =36\left( 2.45 \right) \\ 

& =88.2c{{m}^{2}} \\ 

\end{align}$ 

Cost of polishing per\[~c{{m}^{2}}\]  area \[=\text{ }50\text{ }p\] 

Cost of polishing \[1411.2c{{m}^{2}}\] area \[=Rs.\left( 1411.2\times 0.50 \right)=Rs.705.60\] Therefore, it will cost \[Rs.\text{ }705.60\] while polishing all the tiles.

9. A field is in the shape of a trapezium whose parallel sides are \[\mathbf{25}\text{ }\mathbf{m}\] and\[\mathbf{10}\text{ }\mathbf{m}\] . The non-parallel sides are \[\mathbf{14}\text{ }\mathbf{m}\] and\[\mathbf{13}\text{ }\mathbf{m}\] . Find the area of the field.

Ans:

A Field in the Shape of Trapezium

Draw a line \[BE\] parallel to \[AD\] and draw a perpendicular \[BF\] on\[CD\] .

It can be observed that \[ABED\] is a parallelogram.

\[BE=AD=13\text{ }m\] 

\[ED=AB=10\text{ }m\] 

\[EC=25-ED=15\text{ }m\]

Semi-perimeter,  $s=\frac{13+14+15}{2}=21cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{21\left( 21-13 \right)\left( 21-14 \right)\left( 21-15 \right)}{{m}^{2}}$ 

$\begin{align}

& =\sqrt{21\left( 7 \right)\left( 8 \right)\left( 6 \right)} \\ 

& =84c{{m}^{2}} \\ 

\end{align}$

Area of $\vartriangle BEC=\frac{1}{2}\left( CE \right)\left( BF \right)$ 

$BF=\frac{168}{15}=11.2m$ 

Area of \[ABED=BF\times DE=11.2\times 10=112{{m}^{2}}\] 

Area of the field \[=84+112=196\text{ }{{m}^{2}}\] 

An Overview of NCERT Solutions for Class 9 Maths Exercise 12.1

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula is a very important Math concept as it has many applications. Therefore, it is required to have a deep understanding of this fundamental concept of Mathematics. The NCERT Solutions of Exercise 12.1 of Class 9 Maths are provided in such an efficient way that by practising the sums provided in these NCERT Solutions, students will be able to easily learn this concept.

There are certain formulas to find the area of the triangles. Heron’s formula is one of them and it is very useful to calculate the area of a triangle. Exercise 12.1 of the NCERT Solutions for Class 9 Maths describes the step-by-step way of calculating the area of triangles using Heron’s Formula. This exercise consists of sums based on this important formula that helps the students to calculate the area of triangles if the length of all the three sides is given.

Class 9 Maths Exercise 12.1 Solution – All questions

What is Heron’s Formula: NCERT Solutions for Class 9 Maths Exercise 12.1

Heron’s formula is also referred to as Hero’s formula in geometry. The formula is named after the Hero of Alexandria. By implementing this formula, the area of a triangle can be calculated if the length of all three sides of the triangle is known. According to Heron’s formula, if a triangle has lengths a, b and c, then the area (A) will be:

Area = √[s(s - a)(s - b)(s - c)]

Where s = semi-perimeter of the triangle

s = (a + b + c)/2

Vedantu’s class 9 maths exercise 12.1 solution has many problems based on this formula - all solved and explained in a step-wise manner.

Class 9 Maths Ex 12.1: Question 1

Let us say there is an equilateral triangle signal board on the roadside with a perimeter of 180 cm. How can you find the area of this signal board with the help of Heron’s formula? Chapter 12 maths class 9 - 12.1 can help you in finding the answer.

Solution:

As you go through the solution of NCERT maths class 9 chapter 12 exercise 12.1, you will find a way to solve the question as there are similar questions also. The solution is as follows:

Let us imagine that each side of the triangular board is “a”.

Now, we know that the semi-perimeter of a triangle can be calculated by the following formula:

s = (a + b + c)/2

So, in this case, s = (a + a + a)/2 i.e. 3a/2

Now when we have the semi-perimeter, we can find the area by the implementation of the following formula:

Area = √[s(s - a)(s - b)(s - c)]

In place of s, we can put 3a/2. So, it will be:

√[s(s - a)(s - a)(s - a)]

= √[s(s - a)3]

= √[3a/2 (3a/2  - a)3]

As you continue calculating from here you will ultimately get the digit √(3a2)/4.

The question already said that the perimeter is 180 cm. The perimeter of a triangle means the sum of its 3 sides. So, according to the question,

a + a + a = 180 cm

or, 3a = 180 cm

So, a = 60 cm (as 180 is divided by 3).

Now, you can put the value of “a” in  √(3a2)/4.

This will lead you to the result that the area of that triangular signboard is 900√3 cm3.

Heron’s Formula Exercise 12.1: Question 2

Suppose there is a triangle with sides in the ratio of 2:7:5 and a perimeter of 140 cm. Now, you are to calculate the area of the triangle referring to class 9th maths chapter 12 exercise 12.1.

Solution:

According to the ratio, we can suppose that the sides of the triangle are 2x, 7x and 5x. 

So, according to the question, 

2x + 7x + 5x = 140

Or, 14x = 140

So, x = 10

Thus, the three sides of the triangle are:

2x = (2 × 10) = 20 cm, 

7x = (7 × 10) = 70 cm and 

5x = (5 × 10) = 50 cm

NCERT solutions class 9 maths Herons formula exercise 12.1 has similar problem practicing which will make your ideas clearer.

Now, semi-perimeter of the triangle is,

s = (a + b + c)/2

Or, s = (20 + 70 + 50)/2 cm

Or, s = 140/2= 70 cm

Now, the area can be calculated according to the guidance found in class 9th maths chapter 12 exercise 12.1. Solve the math yourself and check whether you proceed along the right path.

Class 9 Maths Ch 12 Ex 12.1: Question 3

In one of the questions from the exercise 12.1 class 9 maths, you will be introduced to areas shaped in the form of a quadrilateral with one right angle and sides of varying measurements. You will again be asked to calculate the area of that shape.

Solution:

In solving this, the maths class 9 chapter 12 exercise 12.1 solutions can be of great help to you. You have to draw an imaginary line so that you get a right-angled triangle for the ease of calculation.

Class 9 maths Heron’s formula exercise 12.1 teaches you about Pythagoras theorem and how you can implement it in the calculation. Applying that, you will get the measurement of the sides that you did not know.

From here, NCERT class 9 maths chapter 12 exercise 12.1 will give you the sides. And when you have them, there is no limitation in calculating the perimeter, semi-perimeter, and area.

In this particular question of class 9 maths NCERT solutions chapter 12 exercise 12.1, you will finally get the result that the area of the quadrilateral is approximately 65.5 m2.

Ex 12.1 Class 9 Maths NCERT Solutions: Question 4

Referring to NCERT solutions for class 9 maths ex 12.1 you will come across questions asking for the area of a triangle. Suppose two sides of a triangle are 20 m and 14 cm and it has a perimeter of 50 cm. So, what will be the area?

Solution:

Let the third side of the triangle be “x” cm.

According to the question,

20 + 14 + x = 50 cm

So, x = 50 – (20 +14) m = 16 cm

So, the semi perimeter s = 50/2 cm = 25 cm.

Now, you have to apply the formula Area = √[s(s - a)(s - b)(s - c)] as you have learnt in exercise 12.1 maths class 9 and find out the area of the triangle.

These are just a few examples. You will find many such solved problems in class 9 maths chapter 12 exercise 12.1 solution.

CBSE Class 9 Maths Chapter 12 Other Exercise

Chapterwise NCERT Solutions for Class 9 Maths 

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 12 of CBSE Class 9 Maths Solutions –

• Chapter 12: Important Questions

• Chapter 12: Important Formulas

• Chapter 12: Revision Notes

• Chapter 12: NCERT Exemplar Solutions

• Chapter 12: RD Sharma Solutions

Why Choose Vedantu?

Vedantu has a pool of experienced teachers who provide easy and smooth learning from home. Sample papers, as well as their solutions, are provided on the official website. Vedantu helps you to choose teachers of your own choice. Sitting at home and preparing for exams have become easier now with Vedantu. If you choose online tuitions, you can get access to a whole range of latest online resources just like the one discussed here - NCERT solutions for class 9 maths chapter 12 exercise 12.1. Online tuitions like Vedantu are comfortable and can be availed of under parent’s supervision.